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

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.

Related

creating multiple user named and numbered files without loop from bash

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

Syntax error near unexpected token `done' whis done written

I have a problem with the done.
It says I have some typo error but I can't figure what's wrong at all.
Here is the code:
#./bin/bash
until [$err == 0];
do
java -Xms512m -Xmx512m -cp lib/*:lib/uMad/*:mysql-connector-java-5.1.15-bin.jar:l2jfrozen-core.jar com.l2jfrozen.gameserver.GameServer
err=$?
sleep 5
done
Your shebang line is wrong. #./bin/bash will not execute bash.
It should read #!/bin/bash. You are probably using a shell other than bash to invoke this script.
Also, beware that the [$err == 0] line expands the value of $err, which is probably an empty string, unless it has been exported. If it's empty, this will result in an error, because Bash will be interpreting [ == 0].
The safest approach is this:
unset err
until [ "$err" == "0" ];
do
# etc...
done
From my experience when working with brackets and if loops, you need proper spacing and double, not single brackets. There needs to be space on each side of the double brackets with the exception of the semi-colon. Here is an example block:
#!/bin/bash
err=5
until [[ $err == 0 ]]; do
((err-=1));
echo -e "$err\n";
sleep 3
done
I do not see why the same would not apply to a do until loop.
You're probably aware but your heading has a period in it instead of a shebang.
#./bin/bash

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 ยง3.2.4 "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.

ksh function return value in parentheses

In the following very simple ksh script example, I need to ask if func1 results equal to 4 ,
This is what I did in the example but this script does not print the "function result = 4" as I expected it to.
What do I need to change in the [[......]] in order to print the "function result = 4"
Remark - func1 must be in the [[.....]]
#!/bin/ksh
func1()
{
return 4
}
[[ ` func1 ` = ` echo $? ` ]] && print "function result = 4"
You need
#!/bin/ksh
func1()
{
print -- 4
}
[[ $(func1) = 4 ]] && print "function result = 4"
OR
#!/bin/ksh
func1()
{
return 4
}
func1 ; [[ $? == 4 ]] && print "function result = 4"
There are several issues in the code that you present, so let me try to explain (You're making it more complicated than it need be).
No. 1 is your use of back-ticks for command substitution, these have been deprecated in the ksh language since ~ 1995! Use $( ... cmd ) for modern cmd-substitution. We often see backticks listed as a nod to portability, but only scripts written for systems where the Bourne shell is the only shell available require the use of backticks. (well, I don't know about dash or ash, so maybe those too).
No 2. is that $? gets set after ever function or command or pipeline is executed and is the return code of that last command. It is a value between 0-255. When you have code like cmd ; rc=$? ; echo $? ; you're now echoing the status of the assignment of rc=$? (which will almost always be 0), AND that is why you will see experienced scriptors save the value of $? before doing anything else with it.
Recall that command-substitution uses what ever is the output of the $( ... cmd ...) or backtics enclosed command while return sets the value of $? (until the very next command execution resets that value).
I hope this helps.
Function does return 4. The operator `` (backticks) ignores the result value, and returns the function's stdout instead (in your case an empty string, since func1 did not print anything to stdout).
And
`echo $?`
is just over-complicated way of saying
$?

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