creating multiple user named and numbered files without loop from bash - linux

Totally new in bash programming and got a problem like this:
N empty files should be created. The file sum N and also the file name should be given from users via command line.
no loop and getopts should be used.
I've tried something like this which i found from google but it doesn't works. I got confused between linux and windows bash scripts. Hope someone can help me with the problem. Thank you!
#!/bin/bash
echo $N
:start
if [ $N > 0 ]
then
touch xyzfile_$N
$N = $N - 1
pause
goto start
else
then
echo "no data will be created"
fi
The command line and the result should be (for example if N=7) look like this:
command line:
./createfiles -n filename 7
expected result:
filename_1,filename_2,filename_3...filename_7

You can do this using a recursive shell function:
fn() {
# add some sanity checks to check parameters
touch "$1_${2}"
(($2 > 1)) && fn "$1" $(($2 - 1))
}
This part is a recursive call:
(($2 > 1)) && fn "$1" $(($2 - 1))
Where it basically calls itself by decrementing 1 from 2nd argument as long as $2 is greater than 1.
Then just call it as:
fn filename 7
It will create 7 files as:
filename_1 filename_2 filename_3 filename_4 filename_5 filename_6 filename_7

I ended up the question with a for-loop like this:
createfile() {
name="$1"
num="$2"
for((i=1;i<=num;i++))
do
touch "$1_${i}"
done
}
createfile $1 ${2}
It works great actually. And now I tried to solve the problem with getopts and it seems a function call doesn't work in the case option

Related

Changing global var inside function doesnt mutate global variable [duplicate]

I'm working with this:
GNU bash, version 4.1.2(1)-release (x86_64-redhat-linux-gnu)
I have a script like below:
#!/bin/bash
e=2
function test1() {
e=4
echo "hello"
}
test1
echo "$e"
Which returns:
hello
4
But if I assign the result of the function to a variable, the global variable e is not modified:
#!/bin/bash
e=2
function test1() {
e=4
echo "hello"
}
ret=$(test1)
echo "$ret"
echo "$e"
Returns:
hello
2
I've heard of the use of eval in this case, so I did this in test1:
eval 'e=4'
But the same result.
Could you explain me why it is not modified? How could I save the echo of the test1 function in ret and modify the global variable too?
When you use a command substitution (i.e., the $(...) construct), you are creating a subshell. Subshells inherit variables from their parent shells, but this only works one way: A subshell cannot modify the environment of its parent shell.
Your variable e is set within a subshell, but not the parent shell. There are two ways to pass values from a subshell to its parent. First, you can output something to stdout, then capture it with a command substitution:
myfunc() {
echo "Hello"
}
var="$(myfunc)"
echo "$var"
The above outputs:
Hello
For a numerical value in the range of 0 through 255, you can use return to pass the number as the exit status:
mysecondfunc() {
echo "Hello"
return 4
}
var="$(mysecondfunc)"
num_var=$?
echo "$var - num is $num_var"
This outputs:
Hello - num is 4
This needs bash 4.1 if you use {fd} or local -n.
The rest should work in bash 3.x I hope. I am not completely sure due to printf %q - this might be a bash 4 feature.
Summary
Your example can be modified as follows to archive the desired effect:
# Add following 4 lines:
_passback() { while [ 1 -lt $# ]; do printf '%q=%q;' "$1" "${!1}"; shift; done; return $1; }
passback() { _passback "$#" "$?"; }
_capture() { { out="$("${#:2}" 3<&-; "$2_" >&3)"; ret=$?; printf "%q=%q;" "$1" "$out"; } 3>&1; echo "(exit $ret)"; }
capture() { eval "$(_capture "$#")"; }
e=2
# Add following line, called "Annotation"
function test1_() { passback e; }
function test1() {
e=4
echo "hello"
}
# Change following line to:
capture ret test1
echo "$ret"
echo "$e"
prints as desired:
hello
4
Note that this solution:
Works for e=1000, too.
Preserves $? if you need $?
The only bad sideffects are:
It needs a modern bash.
It forks quite more often.
It needs the annotation (named after your function, with an added _)
It sacrifices file descriptor 3.
You can change it to another FD if you need that.
In _capture just replace all occurances of 3 with another (higher) number.
The following (which is quite long, sorry for that) hopefully explains, how to adpot this recipe to other scripts, too.
The problem
d() { let x++; date +%Y%m%d-%H%M%S; }
x=0
d1=$(d)
d2=$(d)
d3=$(d)
d4=$(d)
echo $x $d1 $d2 $d3 $d4
outputs
0 20171129-123521 20171129-123521 20171129-123521 20171129-123521
while the wanted output is
4 20171129-123521 20171129-123521 20171129-123521 20171129-123521
The cause of the problem
Shell variables (or generally speaking, the environment) is passed from parental processes to child processes, but not vice versa.
If you do output capturing, this usually is run in a subshell, so passing back variables is difficult.
Some even tell you, that it is impossible to fix. This is wrong, but it is a long known difficult to solve problem.
There are several ways on how to solve it best, this depends on your needs.
Here is a step by step guide on how to do it.
Passing back variables into the parental shell
There is a way to pass back variables to a parental shell. However this is a dangerous path, because this uses eval. If done improperly, you risk many evil things. But if done properly, this is perfectly safe, provided that there is no bug in bash.
_passback() { while [ 0 -lt $# ]; do printf '%q=%q;' "$1" "${!1}"; shift; done; }
d() { let x++; d=$(date +%Y%m%d-%H%M%S); _passback x d; }
x=0
eval `d`
d1=$d
eval `d`
d2=$d
eval `d`
d3=$d
eval `d`
d4=$d
echo $x $d1 $d2 $d3 $d4
prints
4 20171129-124945 20171129-124945 20171129-124945 20171129-124945
Note that this works for dangerous things, too:
danger() { danger="$*"; passback danger; }
eval `danger '; /bin/echo *'`
echo "$danger"
prints
; /bin/echo *
This is due to printf '%q', which quotes everything such, that you can re-use it in a shell context safely.
But this is a pain in the a..
This does not only look ugly, it also is much to type, so it is error prone. Just one single mistake and you are doomed, right?
Well, we are at shell level, so you can improve it. Just think about an interface you want to see, and then you can implement it.
Augment, how the shell processes things
Let's go a step back and think about some API which allows us to easily express, what we want to do.
Well, what do we want do do with the d() function?
We want to capture the output into a variable.
OK, then let's implement an API for exactly this:
# This needs a modern bash 4.3 (see "help declare" if "-n" is present,
# we get rid of it below anyway).
: capture VARIABLE command args..
capture()
{
local -n output="$1"
shift
output="$("$#")"
}
Now, instead of writing
d1=$(d)
we can write
capture d1 d
Well, this looks like we haven't changed much, as, again, the variables are not passed back from d into the parent shell, and we need to type a bit more.
However now we can throw the full power of the shell at it, as it is nicely wrapped in a function.
Think about an easy to reuse interface
A second thing is, that we want to be DRY (Don't Repeat Yourself).
So we definitively do not want to type something like
x=0
capture1 x d1 d
capture1 x d2 d
capture1 x d3 d
capture1 x d4 d
echo $x $d1 $d2 $d3 $d4
The x here is not only redundant, it's error prone to always repeate in the correct context. What if you use it 1000 times in a script and then add a variable? You definitively do not want to alter all the 1000 locations where a call to d is involved.
So leave the x away, so we can write:
_passback() { while [ 0 -lt $# ]; do printf '%q=%q;' "$1" "${!1}"; shift; done; }
d() { let x++; output=$(date +%Y%m%d-%H%M%S); _passback output x; }
xcapture() { local -n output="$1"; eval "$("${#:2}")"; }
x=0
xcapture d1 d
xcapture d2 d
xcapture d3 d
xcapture d4 d
echo $x $d1 $d2 $d3 $d4
outputs
4 20171129-132414 20171129-132414 20171129-132414 20171129-132414
This already looks very good. (But there still is the local -n which does not work in oder common bash 3.x)
Avoid changing d()
The last solution has some big flaws:
d() needs to be altered
It needs to use some internal details of xcapture to pass the output.
Note that this shadows (burns) one variable named output,
so we can never pass this one back.
It needs to cooperate with _passback
Can we get rid of this, too?
Of course, we can! We are in a shell, so there is everything we need to get this done.
If you look a bit closer to the call to eval you can see, that we have 100% control at this location. "Inside" the eval we are in a subshell,
so we can do everything we want without fear of doing something bad to the parental shell.
Yeah, nice, so let's add another wrapper, now directly inside the eval:
_passback() { while [ 0 -lt $# ]; do printf '%q=%q;' "$1" "${!1}"; shift; done; }
# !DO NOT USE!
_xcapture() { "${#:2}" > >(printf "%q=%q;" "$1" "$(cat)"); _passback x; } # !DO NOT USE!
# !DO NOT USE!
xcapture() { eval "$(_xcapture "$#")"; }
d() { let x++; date +%Y%m%d-%H%M%S; }
x=0
xcapture d1 d
xcapture d2 d
xcapture d3 d
xcapture d4 d
echo $x $d1 $d2 $d3 $d4
prints
4 20171129-132414 20171129-132414 20171129-132414 20171129-132414
However, this, again, has some major drawback:
The !DO NOT USE! markers are there,
because there is a very bad race condition in this,
which you cannot see easily:
The >(printf ..) is a background job. So it might still
execute while the _passback x is running.
You can see this yourself if you add a sleep 1; before printf or _passback.
_xcapture a d; echo then outputs x or a first, respectively.
The _passback x should not be part of _xcapture,
because this makes it difficult to reuse that recipe.
Also we have some unneded fork here (the $(cat)),
but as this solution is !DO NOT USE! I took the shortest route.
However, this shows, that we can do it, without modification to d() (and without local -n)!
Please note that we not neccessarily need _xcapture at all,
as we could have written everyting right in the eval.
However doing this usually isn't very readable.
And if you come back to your script in a few years,
you probably want to be able to read it again without much trouble.
Fix the race
Now let's fix the race condition.
The trick could be to wait until printf has closed it's STDOUT, and then output x.
There are many ways to archive this:
You cannot use shell pipes, because pipes run in different processes.
One can use temporary files,
or something like a lock file or a fifo. This allows to wait for the lock or fifo,
or different channels, to output the information, and then assemble the output in some correct sequence.
Following the last path could look like (note that it does the printf last because this works better here):
_passback() { while [ 0 -lt $# ]; do printf '%q=%q;' "$1" "${!1}"; shift; done; }
_xcapture() { { printf "%q=%q;" "$1" "$("${#:2}" 3<&-; _passback x >&3)"; } 3>&1; }
xcapture() { eval "$(_xcapture "$#")"; }
d() { let x++; date +%Y%m%d-%H%M%S; }
x=0
xcapture d1 d
xcapture d2 d
xcapture d3 d
xcapture d4 d
echo $x $d1 $d2 $d3 $d4
outputs
4 20171129-144845 20171129-144845 20171129-144845 20171129-144845
Why is this correct?
_passback x directly talks to STDOUT.
However, as STDOUT needs to be captured in the inner command,
we first "save" it into FD3 (you can use others, of course) with '3>&1'
and then reuse it with >&3.
The $("${#:2}" 3<&-; _passback x >&3) finishes after the _passback,
when the subshell closes STDOUT.
So the printf cannot happen before the _passback,
regardless how long _passback takes.
Note that the printf command is not executed before the complete
commandline is assembled, so we cannot see artefacts from printf,
independently how printf is implemented.
Hence first _passback executes, then the printf.
This resolves the race, sacrificing one fixed file descriptor 3.
You can, of course, choose another file descriptor in the case,
that FD3 is not free in your shellscript.
Please also note the 3<&- which protects FD3 to be passed to the function.
Make it more generic
_capture contains parts, which belong to d(), which is bad,
from a reusability perspective. How to solve this?
Well, do it the desparate way by introducing one more thing,
an additional function, which must return the right things,
which is named after the original function with _ attached.
This function is called after the real function, and can augment things.
This way, this can be read as some annotation, so it is very readable:
_passback() { while [ 0 -lt $# ]; do printf '%q=%q;' "$1" "${!1}"; shift; done; }
_capture() { { printf "%q=%q;" "$1" "$("${#:2}" 3<&-; "$2_" >&3)"; } 3>&1; }
capture() { eval "$(_capture "$#")"; }
d_() { _passback x; }
d() { let x++; date +%Y%m%d-%H%M%S; }
x=0
capture d1 d
capture d2 d
capture d3 d
capture d4 d
echo $x $d1 $d2 $d3 $d4
still prints
4 20171129-151954 20171129-151954 20171129-151954 20171129-151954
Allow access to the return-code
There is only on bit missing:
v=$(fn) sets $? to what fn returned. So you probably want this, too.
It needs some bigger tweaking, though:
# This is all the interface you need.
# Remember, that this burns FD=3!
_passback() { while [ 1 -lt $# ]; do printf '%q=%q;' "$1" "${!1}"; shift; done; return $1; }
passback() { _passback "$#" "$?"; }
_capture() { { out="$("${#:2}" 3<&-; "$2_" >&3)"; ret=$?; printf "%q=%q;" "$1" "$out"; } 3>&1; echo "(exit $ret)"; }
capture() { eval "$(_capture "$#")"; }
# Here is your function, annotated with which sideffects it has.
fails_() { passback x y; }
fails() { x=$1; y=69; echo FAIL; return 23; }
# And now the code which uses it all
x=0
y=0
capture wtf fails 42
echo $? $x $y $wtf
prints
23 42 69 FAIL
There is still a lot room for improvement
_passback() can be elmininated with passback() { set -- "$#" "$?"; while [ 1 -lt $# ]; do printf '%q=%q;' "$1" "${!1}"; shift; done; return $1; }
_capture() can be eliminated with capture() { eval "$({ out="$("${#:2}" 3<&-; "$2_" >&3)"; ret=$?; printf "%q=%q;" "$1" "$out"; } 3>&1; echo "(exit $ret)")"; }
The solution pollutes a file descriptor (here 3) by using it internally.
You need to keep that in mind if you happen to pass FDs.
Note thatbash 4.1 and above has {fd} to use some unused FD.
(Perhaps I will add a solution here when I come around.)
Note that this is why I use to put it in separate functions like _capture, because stuffing this all into one line is possible, but makes it increasingly harder to read and understand
Perhaps you want to capture STDERR of the called function, too.
Or you want to even pass in and out more than one filedescriptor
from and to variables.
I have no solution yet, however here is a way to catch more than one FD, so we can probably pass back the variables this way, too.
Also do not forget:
This must call a shell function, not an external command.
There is no easy way to pass environment variables out of external commands.
(With LD_PRELOAD= it should be possible, though!)
But this then is something completely different.
Last words
This is not the only possible solution. It is one example to a solution.
As always you have many ways to express things in the shell.
So feel free to improve and find something better.
The solution presented here is quite far from being perfect:
It was nearly not tested at all, so please forgive typos.
There is a lot of room for improvement, see above.
It uses many features from modern bash, so probably is hard to port to other shells.
And there might be some quirks I haven't thought about.
However I think it is quite easy to use:
Add just 4 lines of "library".
Add just 1 line of "annotation" for your shell function.
Sacrifices just one file descriptor temporarily.
And each step should be easy to understand even years later.
Maybe you can use a file, write to file inside function, read from file after it. I have changed e to an array. In this example blanks are used as separator when reading back the array.
#!/bin/bash
declare -a e
e[0]="first"
e[1]="secondddd"
function test1 () {
e[2]="third"
e[1]="second"
echo "${e[#]}" > /tmp/tempout
echo hi
}
ret=$(test1)
echo "$ret"
read -r -a e < /tmp/tempout
echo "${e[#]}"
echo "${e[0]}"
echo "${e[1]}"
echo "${e[2]}"
Output:
hi
first second third
first
second
third
What you are doing, you are executing test1
$(test1)
in a sub-shell( child shell ) and Child shells cannot modify anything in parent.
You can find it in bash manual
Please Check: Things results in a subshell here
I had a similar problem when I wanted to remove temporary files I had created automatically. The solution I came up with was not to use command substitution, but rather to pass the name of the variable, that should take the final result, into the function. E.g.
#!/usr/bin/env bash
# array that keeps track of tmp-files
remove_later=()
# function that manages tmp-files
new_tmp_file() {
file=$(mktemp)
remove_later+=( "$file" )
# assign value (safe form of `eval "$1=$file"`)
printf -v "$1" -- "$file"
}
# function to remove all tmp-files
remove_tmp_files() { rm -- "${remove_later[#]}"; }
# define trap to remove all tmp-files upon EXIT
trap remove_tmp_files EXIT
# generate tmp-files
new_tmp_file tmpfile1
new_tmp_file tmpfile2
So, adapting this to the OP, it would be:
#!/usr/bin/env bash
e=2
function test1() {
e=4
printf -v "$1" -- "hello"
}
test1 ret
echo "$ret"
echo "$e"
Works and has no restrictions on the "return value".
Assuming that local -n is available, the following script lets the function test1 modify a global variable:
#!/bin/bash
e=2
function test1() {
local -n var=$1
var=4
echo "hello"
}
test1 e
echo "$e"
Which gives the following output:
hello
4
I'm not sure if this works on your terminal, but I found out that if you don't provide any outputs whatsoever it gets naturally treated as a void function, and can make global variable changes.
Here's the code I used:
let ran1=$(( (1<<63)-1)/3 ))
let ran2=$(( (1<<63)-1)/5 ))
let c=0
function randomize {
c=$(( ran1+ran2 ))
ran2=$ran1
ran1=$c
c=$(( c > 0 ))
}
It's a simple randomizer for games that effectively modifies the needed variables.
It's because command substitution is performed in a subshell, so while the subshell inherits the variables, changes to them are lost when the subshell ends.
Reference:
Command substitution, commands grouped with parentheses, and asynchronous commands are invoked in a subshell environment that is a duplicate of the shell environment
A solution to this problem, without having to introduce complex functions and heavily modify the original one, is to store the value in a temporary file and read / write it when needed.
This approach helped me greatly when I had to mock a bash function called multiple times in a bats test case.
For example, you could have:
# Usage read_value path_to_tmp_file
function read_value {
cat "${1}"
}
# Usage: set_value path_to_tmp_file the_value
function set_value {
echo "${2}" > "${1}"
}
#----
# Original code:
function test1() {
e=4
set_value "${tmp_file}" "${e}"
echo "hello"
}
# Create the temp file
# Note that tmp_file is available in test1 as well
tmp_file=$(mktemp)
# Your logic
e=2
# Store the value
set_value "${tmp_file}" "${e}"
# Run test1
test1
# Read the value modified by test1
e=$(read_value "${tmp_file}")
echo "$e"
The drawback is that you might need multiple temp files for different variables. And also you might need to issue a sync command to persist the contents on the disk between one write and read operations.
You can always use an alias:
alias next='printf "blah_%02d" $count;count=$((count+1))'

Can I only call a function from the terminal rather the whole bash script?

I have a bash script that looks as such:
#!/bin/bash
function one {
echo "I am function one!!"
}
function two {
echo "I am function two!!"
}
one
two
If I simply do bash test.sh both functions are being executed.
What I'd like to do is to call the script from the terminal while also specifying one of the two functions, and executing only it.
Maybe something like: bash test.sh$one() and it should only print out
I am function one!!
Is this possible and if so, how will I go about achieving it?
Thanks!
=========================
EDIT: As per #Waqas suggestion I ended up implementing the below which did the trick for me:
function main {
if [ -z "$1" ]
then
some commands
# else run the given function only
else
$1
fi
}
main "$#"
Thanks!!!
There are many ways to write the code in order to fulfill your requirement. The way I will write the code for this, is the following:
#!/bin/bash
function main {
# If the argument is empty then run both functions else only run provided function as argument $1.
[ -z "$1" ] && { one; two; } || $1
}
function one {
echo "I am function one!!"
}
function two {
echo "I am function two!!"
}
main "$#"
If you only execute the script without passing argument then both functions will run and with passing argument only single function will work.
Example1 (Both functions will run): bash script_name
Example2 (Only function one will run): bash script_name one
Example3 (Only function two will run): bash script_name two
You better separate the files: Move the function definitions in a separate file, say ~/lib/testlib.src. Your test.sh then becomes
#!/bin/bash
. ~/lib/testlib.src
one
two
If you need the definitions in your interactive shell, either do there a . ~/lib/testlib.src manually, or if you want to have them always available, put this statement into your ~/.bashrc.
You could do this by writing a case statement after defining the functions but before any other lines of code.
#!/bin/bash
function one {
echo "I am function one!!"
}
function two {
echo "I am function two!!"
}
case $1 in
one)
one
;;
two)
two
;;
*)
one
two
;;
esac
Which could then be used as:
$ ./test.sh one
# I am function one!!
$ ./test.sh two
# I am function two!!
In the above example I put the body of your script under the *) option, but if it better suits your needs, you could instead have the one) and two) options "exit" after calling their single function:
case $1 in
one)
one
exit 0
;;
two)
two
exit 0
;;
esac
one
two
This is all assuming you are not passing any other arguments to the script and that $1 would be used for the desired function. The case statement would become more complex otherwise.

Bash script : $x=$x+2 is not getting recognised

When I am executing the below script, I am getting the following error :-
The script executes infintely and below line is printed everytime.
"line 9: 1=1+2: command not found". Why?
#!/bin/bash
echo "Script 1 - Linux Scripting Book"
x=1
while [ $x -le 45 ]
do
echo x : $x
$x=$x+2
done
echo "End Of Script 1"
exit 0
Also if I change the $x=$x+2 to x+$x+2 then also I am getting the below error.
line 6: [: 1+2: integer expression expected
Same script when executed like this runs fine.
#!/bin/bash
echo "Script 1 - Linux Scripting Book"
x=1
while [ $x -le 45 ]
do
echo x : $x
let x=x+2
done
echo "End Of Script 1"
exit 0
You get line 9: 1=1+2: command not found because 1=1+2 is what $x=$x+2 is expanded into.
Use expr or let or ((...)) for integer calculations and bc for floating point:
let x=x+2
((x=x+2)) #same as above
((x+=2)) #same
((x++)) #if adding just one
((++x)) #if adding just one
x=$((x+2))
x=`expr $x + 2` #space before and after +
x=$(echo $x+2|bc) #using bc
x=$(echo $x+2.1|bc) #bc also works with floating points (numbers with decimals)
Since this part of the question isn't cleared yet, and not fine to post in a comment, I add this partial answer:
x=1; for i in 1 2 3 ; do x=$x+2; echo $x; done
1+2
1+2+2
1+2+2+2
As a side note: Don't use exit 0 at the end of your script without a good reason. When the script is done, it exits by itself without your help. The exit status will be the exit status of the last command performed, in your case a simple echo, which will almost always succeed. In the rare cases it fails, you will probably without intention hide that failure.
If you source the script, the exit will throw you out of your running shell.
But you can rewrite your while loop like this:
x=0
while (($((x)) < 9))
do
echo x : $x
x=$x+2
done
echo $((x))
x : 0
x : 0+2
x : 0+2+2
x : 0+2+2+2
x : 0+2+2+2+2
10
Because that's not the Bourne shell syntax for setting a variable; it looks more like Perl or PHP. The $ is used for parameter expansion and is not part of the variable name. Variable assignment simply uses =, and let evaluates arithmetic expressions (much like $((expression))). Another syntax that should work is x=$((x+2)). Note that these arithmetic evaluations are a bash feature; standard unix shells might require use of external tools such as expr.

Conditional statement not working as expected

I am using Konsole on kubuntu 14.04.
I want to take arguments to this shell-script, and pass it to a command. The code is basically an infinite loop, and I want one of the arguments to the inner command to be increased once every 3 iterations of the loop. Ignoring the actual details, here's a gist of my code:
#!/bin/bash
ct=0
begin=$1
while :
do
echo "give: $begin as argument to the command"
#actual command
ct=$((ct+1))
if [ $ct%3==0 ]; then
begin=$(($begin+1))
fi
done
I am expecting the begin variable to be increased every 3 iterations, but it is increasing in the every iteration of the loop. What am I doing wrong?
You want to test with
if [ $(expr $cr % 3) = 0 ]; then ...
because this
[ $ct%3==0 ]
tests whether the string $ct%3==0, after parameter substitution, is not empty. A good way for understanding this is reading the manual for test and look at the semantics when it is given 1, 2, 3 or more arguments. In your original script, it only sees one argument, in mine it sees three. White space is very important in the shell. :-)
In BASH you can completely utilize ((...)) and refactor your script like this:
#!/bin/bash
ct=0
begin="$1"
while :
do
echo "give: $begin as argument to the command"
#actual command
(( ct++ % 3 == 0)) && (( begin++ ))
done

Is there a "goto" statement in bash?

Is there a "goto" statement in bash ? I know It is considered bad practice, but I need specifically "goto".
If you are using it to skip part of a large script for debugging (see Karl Nicoll's comment), then if false could be a good option (not sure if "false" is always available, for me it is in /bin/false):
# ... Code I want to run here ...
if false; then
# ... Code I want to skip here ...
fi
# ... I want to resume here ...
The difficulty comes in when it's time to rip out your debugging code. The "if false" construct is pretty straightforward and memorable, but how do you find the matching fi? If your editor allows you to block indent, you could indent the skipped block (then you'll want to put it back when you're done). Or a comment on the fi line, but it would have to be something you'll remember, which I suspect will be very programmer-dependent.
No, there is not; see §3.2.4 "Compound Commands" in the Bash Reference Manual for information about the control structures that do exist. In particular, note the mention of break and continue, which aren't as flexible as goto, but are more flexible in Bash than in some languages, and may help you achieve what you want. (Whatever it is that you want . . .)
It indeed may be useful for some debug or demonstration needs.
I found that Bob Copeland solution http://bobcopeland.com/blog/2012/10/goto-in-bash/ elegant:
#!/bin/bash
# include this boilerplate
function jumpto
{
label=$1
cmd=$(sed -n "/$label:/{:a;n;p;ba};" $0 | grep -v ':$')
eval "$cmd"
exit
}
start=${1:-"start"}
jumpto $start
start:
# your script goes here...
x=100
jumpto foo
mid:
x=101
echo "This is not printed!"
foo:
x=${x:-10}
echo x is $x
results in:
$ ./test.sh
x is 100
$ ./test.sh foo
x is 10
$ ./test.sh mid
This is not printed!
x is 101
You can use case in bash to simulate a goto:
#!/bin/bash
case bar in
foo)
echo foo
;&
bar)
echo bar
;&
*)
echo star
;;
esac
produces:
bar
star
If you're testing/debugging a bash script, and simply want to skip forwards past one or more sections of code, here is a very simple way to do it that is also very easy to find and remove later (unlike most of the methods described above).
#!/bin/bash
echo "Run this"
cat >/dev/null <<GOTO_1
echo "Don't run this"
GOTO_1
echo "Also run this"
cat >/dev/null <<GOTO_2
echo "Don't run this either"
GOTO_2
echo "Yet more code I want to run"
To put your script back to normal, just delete any lines with GOTO.
We can also prettify this solution, by adding a goto command as an alias:
#!/bin/bash
shopt -s expand_aliases
alias goto="cat >/dev/null <<"
goto GOTO_1
echo "Don't run this"
GOTO_1
echo "Run this"
goto GOTO_2
echo "Don't run this either"
GOTO_2
echo "All done"
Aliases don't usually work in bash scripts, so we need the shopt command to fix that.
If you want to be able to enable/disable your goto's, we need a little bit more:
#!/bin/bash
shopt -s expand_aliases
if [ -n "$DEBUG" ] ; then
alias goto="cat >/dev/null <<"
else
alias goto=":"
fi
goto '#GOTO_1'
echo "Don't run this"
#GOTO1
echo "Run this"
goto '#GOTO_2'
echo "Don't run this either"
#GOTO_2
echo "All done"
Then you can do export DEBUG=TRUE before running the script.
The labels are comments, so won't cause syntax errors if disable our goto's (by setting goto to the ':' no-op), but this means we need to quote them in our goto statements.
Whenever using any kind of goto solution, you need to be careful that the code you're jumping past doesn't set any variables that you rely on later - you may need to move those definitions to the top of your script, or just above one of your goto statements.
Although others have already clarified that there is no direct goto equivalent in bash (and provided the closest alternatives such as functions, loops, and break), I would like to illustrate how using a loop plus break can simulate a specific type of goto statement.
The situation where I find this the most useful is when I need to return to the beginning of a section of code if certain conditions are not met. In the example below, the while loop will run forever until ping stops dropping packets to a test IP.
#!/bin/bash
TestIP="8.8.8.8"
# Loop forever (until break is issued)
while true; do
# Do a simple test for Internet connectivity
PacketLoss=$(ping "$TestIP" -c 2 | grep -Eo "[0-9]+% packet loss" | grep -Eo "^[0-9]")
# Exit the loop if ping is no longer dropping packets
if [ "$PacketLoss" == 0 ]; then
echo "Connection restored"
break
else
echo "No connectivity"
fi
done
This solution had the following issues:
Indiscriminately removes all code lines ending in a :
Treats label: anywhere on a line as a label
Here's a fixed (shell-check clean and POSIX compatible) version:
#!/bin/sh
# GOTO for bash, based upon https://stackoverflow.com/a/31269848/5353461
goto() {
label=$1
cmd=$(sed -En "/^[[:space:]]*#[[:space:]]*$label:[[:space:]]*#/{:a;n;p;ba};" "$0")
eval "$cmd"
exit
}
start=${1:-start}
goto "$start" # GOTO start: by default
#start:# Comments can occur after labels
echo start
goto end
# skip: # Whitespace is allowed
echo this is usually skipped
# end: #
echo end
There is one more ability to achieve a desired results: command trap. It can be used to clean-up purposes for example.
There is no goto in bash.
Here is some dirty workaround using trap which jumps only backwards:)
#!/bin/bash -e
trap '
echo I am
sleep 1
echo here now.
' EXIT
echo foo
goto trap 2> /dev/null
echo bar
Output:
$ ./test.sh
foo
I am
here now.
This shouldn't be used in that way, but only for educational purposes. Here is why this works:
trap is using exception handling to achieve the change in code flow.
In this case the trap is catching anything that causes the script to EXIT. The command goto doesn't exist, and hence throws an error, which would ordinarily exit the script. This error is being caught with trap, and the 2>/dev/null hides the error message that would ordinarily be displayed.
This implementation of goto is obviously not reliable, since any non-existent command (or any other error, for that manner), would execute the same trap command. In particular, you cannot choose which label to go-to.
Basically in real scenario you don't need any goto statements, they're redundant as random calls to different places only make your code difficult to understand.
If your code is invoked many times, then consider to use loop and changing its workflow to use continue and break.
If your code repeats it-self, consider writing the function and calling it as many times as you want.
If your code needs to jump into specific section based on the variable value, then consider using case statement.
If you can separate your long code into smaller pieces, consider moving it into separate files and call them from the parent script.
I found out a way to do this using functions.
Say, for example, you have 3 choices: A, B, and C. A and Bexecute a command, but C gives you more info and takes you to the original prompt again. This can be done using functions.
Note that since the line containg function demoFunction is just setting up the function, you need to call demoFunction after that script so the function will actually run.
You can easily adapt this by writing multiple other functions and calling them if you need to "GOTO" another place in your shell script.
function demoFunction {
read -n1 -p "Pick a letter to run a command [A, B, or C for more info] " runCommand
case $runCommand in
a|A) printf "\n\tpwd being executed...\n" && pwd;;
b|B) printf "\n\tls being executed...\n" && ls;;
c|C) printf "\n\toption A runs pwd, option B runs ls\n" && demoFunction;;
esac
}
demoFunction
This is a small correction of the Judy Schmidt script put up by Hubbbitus.
Putting non-escaped labels in the script was problematic on the machine and caused it to crash. This was easy enough to resolve by adding # to escape the labels. Thanks to Alexej Magura and access_granted for their suggestions.
#!/bin/bash
# include this boilerplate
function goto {
label=$1
cmd=$(sed -n "/$#label#:/{:a;n;p;ba};" $0 | grep -v ':$')
eval "$cmd"
exit
}
start=${1:-"start"}
goto $start
#start#
echo "start"
goto bing
#boom#
echo boom
goto eof
#bang#
echo bang
goto boom
#bing#
echo bing
goto bang
#eof#
echo "the end mother-hugger..."
A simple searchable goto for the use of commenting out code blocks when debugging.
GOTO=false
if ${GOTO}; then
echo "GOTO failed"
...
fi # End of GOTO
echo "GOTO done"
Result is-> GOTO done
My idea for creating something like "goto" is to use select with case and assign a variable, which I then check in an if statement. Not perfect, but may help in some cases
Example:
#!/usr/bin/env bash
select goto in Ubuntu Debian Quit ; do
case $goto in
Ubuntu) { CHOICE="Ubuntu" ; break ; } ;;
Debian) { CHOICE="Debian" ; break ; } ;;
Quit) { echo "Bye" ; exit ; } ;;
*) { echo "Invalid selection, please try again..." ; } ;;
esac
done
if [ "$CHOICE" == "Ubuntu" ]; then
echo "I'm in Ubuntu"
fi
if [ "$CHOICE" == "Debian" ]; then
echo "I'm in Debian"
fi
Why don't anyone just use functions directly ?
BTW functions are easier to deal with than making a new thing
My style :
#!/bin/bash
# Your functions
function1 ()
{
commands
}
function2 ()
{
commands
}
:
:
functionn ()
{
commands
}
# Execute 1 to n in order
for i in {1..n}
do
function$i
done
# with conditions
for i in {1..n}
do
[ condition$i ] && function$i
done
# Random order
function1
functionn
function5
:
:
function3
Example for above style :
#!/bin/bash
# Your functions
function1 ()
{
echo "Task 1"
}
function2 ()
{
echo "Task 2"
}
function3 ()
{
echo "Task 3"
}
function1
function3
function2
Output :
Task 1
Task 3
Task 2
Drawbacks :
Script in an organized way.
Less problems and not prone to errors.
You can make function inside a existing function.
Move back and forth without any problems.

Resources