Just out of curiosity:
A="echo hi"
B="echo hello"
C="$A && $B"
echo $C // prints "echo hi && echo hello"
$C
echo "$A && $B" // prints "echo hi && echo hello"
$A && $B
I though this would work in a same manner. But it shows different results as this:
echo hi && echo hello
hi && echo hello
echo hi && echo hello
hi
hello
Why?
Very Short Answer
Read BashFAQ #50.
Slightly Short Answer
Variable expansions do not go through all parsing steps (indeed, if they did, it would be impossible to write scripts securely processing untrusted data in bash). Unquoted expansions go through string-splitting and glob expansion only -- these stages split strings into pieces, and evaluate those pieces as globs, but don't recognize any of those pieces as syntax. Quoted expansions are substituted literally, meaning they stay strings, period.
Thus:
echo $C # runs: "echo" "echo " "hi" "&&" "echo" "hello"
$C # runs: "echo " "hi" "&&" "echo" "hello"
echo "$A && $B" # runs: "echo" "echo hi && echo hello"
$A && $B # runs: "echo" "hi" && "echo" "hello"
Notice how in the last example, the && wasn't quoted? That represents that it was parsed as syntax, not data. As syntax, && acts as a command separator (whereas as data, it just gets passed to echo).
Related
I am searching for a command, that separates all given parameters with a specific delimiter, and outputs them quoted.
Example (delimiter is set to be a colon :):
somecommand "this is" "a" test
should output
"this is":"a":"test"
I'm aware that the shell interprets the "" quotes before passing the parameters to the command. So what the command should actually do is to print out every given parameter in quotes and separate all these with a colon.
I'm also not seeking for a bash-only solution, but for the most elegant solution.
It is very easy to just loop over an array of these elements and do that, but the problem is that I have to use this inside a gnu makefile which only allows single line shell commands and uses sh instead of bash.
So the simpler the better.
How about
somecommand () {
printf '"%s"\n' "$#" | paste -s -d :
}
Use printf to add the quotes and print every entry on a separate line, then use paste with the -s ("serial") option and a colon as the delimiter.
Can be called like this:
$ somecommand "this is" "a" test
"this is":"a":"test"
apply_delimiter () {
(( $# )) || return
local res
printf -v res '"%s":' "$#"
printf '%s\n' "${res%:}"
}
Usage example:
$ apply_delimiter hello world "how are you"
"hello":"world":"how are you"
As indicated in a number of the comments, a simple "loop-over" approach, looping over each of the strings passed as arguments is a fairly straight-forward way to approach it:
delimit_colon() {
local first=1
for i in "$#"; do
if [ "$first" -eq 1 ]; then
printf "%s" "$i"
first=0
else
printf ":%s" "$i"
fi
done
printf "\n"
}
Which when combined with a short test script could be:
#!/bin/bash
delimit_colon() {
local first=1
for i in "$#"; do
if [ "$first" -eq 1 ]; then
printf "%s" "$i"
first=0
else
printf ":%s" "$i"
fi
done
printf "\n"
}
[ -z "$1" ] && { ## validate input
printf "error: insufficient input\n"
exit 1
}
delimit_colon "$#"
exit 0
Test Input/Output
$ bash delimitargs.sh "this is" "a" test
this is:a:test
Here a solution using the z-shell:
#!/usr/bin/zsh
# this is "somecommand"
echo '"'${(j_":"_)#}'"'
If you have them in an array already, you can use this command
MYARRAY=("this is" "a" "test")
joined_string=$(IFS=:; echo "$(MYARRAY[*])")
echo $joined_string
Setting the IFS (internal field separator) will be the character separator. Using echo on the array will display the array using the newly set IFS. Putting those commands in $() will put the output of the echo into joined_string.
function ctrace {
echo "+ $#"
"$#"
}
ctrace echo "hi"
How would I get this function to output (with quotes):
echo "hi"
In this version the quotes are lost echo hi... Here is another example:
a=b
ctrace echo $a
This should output echo $a instead of echo b
The problem is not the function, but the caller.
In the first case the quotes are stripped out before the function gets the parameters. In the second, $a substitution is done before it gets to the function.
Try:
ctrace 'echo "hi"'
ctrace 'echo $a'
You need to enclose the string within single quotes
ctrace 'echo "hi"'
ctrace 'echo $a'
This function works:
source foo.bash && foo -n "a b c.txt"
The problem is, no matter what I've tried, I couldn't get the last line echo "$CMD" (or echo $CMD) to generate exactly this output:
cat -n "a b c.txt"
How to achieve that?
# foo.bash
function foo() {
local argv=("$#");
local OUT=`cat "${argv[#]}"`
local CMD=`echo cat "${argv[#]}"`
echo "--------------------------"
echo "$OUT"
echo "--------------------------"
echo "$CMD"
}
The output is instead:
cat -n a b c.txt
With this command: foo -n \"a b c.txt\" it does work for the display of the command, but it gives errors for the execution via the backtick.
The file "a b c.txt" is a valid, small, text file.
You need to escape quotes inside of the assignment:
local CMD="cat \"${argv[#]}\""
Also, echo is not needed to concatenate strings.
There you go, with the help of number of tokens in bash variable I've come up with the right solution.
I've almost forgot WHY we actually need quoting for one argument, it's because it has multiple words!
function foo() {
local argv=( "$#" );
local OUT=`cat "${argv[#]}"`
echo "--------------------------"
echo "$OUT"
echo "--------------------------"
local CMD="cat"
for word in "${argv[#]}"; do
words="${word//[^\ ]} "
if [[ ${#words} > 1 ]]; then
local CMD="$CMD \"${word}\""
else
local CMD="$CMD $word"
fi
done
echo "$CMD"
}
Hope it helps someone.
I know i can test a string whether it is empty with -z and test a string whether it is non empty with -n. So I write a script in ubuntu 10.10:
#!/bin/bash
A=
test -z $A && echo "A is empty"
test -n $A && echo "A is non empty"
test $A && echo "A is non empty"
str=""
test -z $str && echo "str is empty"
test -n $str && echo "str is non empty"
test $str && echo "str is non empty"
To my surprise, it output :
A is empty
A is non empty
str is empty
str is non empty
which I thing it should be
A is empty
str is empty
Could any Linux expert explain why ?
Thank you.
This is a consequence of the way Bash command lines are parsed. Variable substitution happens before constructing the (rudimentary) syntax tree, so the -n operator doesn't get an empty string as an argument, it gets no argument at all! In general, you must enclose any variable reference into "" unless you can be positively sure it isn't empty, precisely to avoid this and similar problems
The 'problem' comes from this:
$ test -n && echo "Oh, this is echoed."
Oh, this is echoed.
i.e. test -n without an argument returns 0/ok.
Change that to:
$ test -n "$A" && echo "A is non empty"
and you'll get the result you expect.
This one will work:
#!/bin/bash
A=
test -z "$A" && echo "A is empty"
test -n "$A" && echo "A is non empty"
test $A && echo "A is non empty"
str=""
test -z "$str" && echo "str is empty"
test -n "$str" && echo "str is non empty"
test $str && echo "str is non empty"
Just $A or $str as empty string will not be a parameter for test and then test status is always true with one parameter (string with nonzero length). The last line works right because test without parameters returns always false.
I have bash script where i have echo before every command showing what is happening.
But i need to disbale echo when setting as cron job and then enable again if do some testing.
i find it very hard to go to each line and then add/remove comment
is there anything which i can include at top something like
enable echo or disable echo
so that i don't have to waste time
The absolute easiest would be to insert the following line after the hashbang line:
echo() { :; }
When you want to re-enable, either delete the line or comment it out:
#echo() { :; }
If you're not using echo but printf, same strategy, i.e.:
printf() { :; }
If you absolutely need to actually echo/printf something, prepend the builtin statement, e.g.:
builtin echo "This 'echo' will not be suppressed."
This means that you can do a conditional output, e.g.:
echo () {
[[ "$SOME_KIND_OF_FLAG" ]] && builtin echo $#
}
Set the SOME_KIND_OF_FLAG variable to something non-null, and the overridden echo function will behave like normal echo.
EDIT: another alternative would be to use echo for instrumenting (debugging), and printf for the outputs (e.g., for piping purposes). That way, no need for any FLAG. Just disable/enable the echo() { :; } line according to whether you want to instrument or not, respectively.
Enable/Disable via CLI Parameter
Put these lines right after the hashbang line:
if [[ debug == "$1" ]]; then
INSTRUMENTING=yes # any non-null will do
shift
fi
echo () {
[[ "$INSTRUMENTING" ]] && builtin echo $#
}
Now, invoking the script like this: script.sh debug will turn on instrumenting. And because there's the shift command, you can still feed parameters. E.g.:
Without instrumenting: script.sh param1 param2
With instrumenting: script.sh debug param1 param2
The above can be simplified to:
if [[ debug != "$1" ]]; then
echo () { :; }
shift
fi
if you need the instrumenting flag (e.g. to record the output of a command to a temp file only if debugging), use an else-block:
if [[ debug != "$1" ]]; then
echo () { :; }
shift
else
INSTRUMENTING=yes
fi
REMEMBER: in non-debug mode, all echo commands are disabled; you have to either use builtin echo or printf. I recommend the latter.
Several things:
Don't use echo at all
Instead use set -xv to set debug mode which will echo each and every command. You can set PS4 to the desired prompt: for example PS4='$LINENO: ' will print out the line number on each line. In BASH, I believe it's the same. Then, you don't have to clean up your script. To shut off, use set +xv.
Example:
foo=7
bar=7
PS4='$LINENO: '
set -xv #Begin debugging
if [ $foo = $bar ]
then
echo "foo certainly does equal bar"
fi
set +xv #Debugging is off
if [ $bar = $foo ]
then
echo "And bar also equals foo"
fi
Results:
$ myprog.sh
if [ $foo = $bar ]
then
echo "foo certainly does equal bar"
fi
5: [ 7 = 7 ]
7: echo 'foo certainly does equal bar'
foo certainly does equal bar
set +xv #Debugging is off
And bar also equals foo
Use a function
Define a function instead of using echo:
Example:
function myecho {
if [ ! -z "$DEBUG" ]
then
echo "$*"
fi
}
DEBUG="TRUE"
my echo "Will print out this line"
unset DEBUG
myecho "But won't print out this line"
Use the nop command
The colon (:) is the nop command in BASH. It doesn't do anything. Use an environment variable and define it as either echo or :. When set to a colon, nothing happens. When set to echo, the line prints.
Example:
echo=":"
$echo "This line won't print"
echo="echo"
$echo "But this line will."
Building on Matthew's answer, how about something like this:
myEcho = "/bin/true"
if [ ! "$CRON" ]: then
myEcho = "/bin/echo"
fi
and then use $myEcho instead of echo in your script?
You can do one better. If you setup your crontab as detailed in another answer, you can then check if you are running in cron and only print if you are not. This way you don't need to modify your script at all between different runs.
You should then be able to use something like this (probably doesn't quite work, I'm not proficient in bash):
if [ ! "$CRON" ]; then
echo "Blah blah"
fi
Try set -v at the top to echo each command. To stop echoing change it to set +v.
Not sure if I miss the below solution to use a variable (e.g. debug) at the start of the bash script.
Once you set the debug=true, any conditional-if will enable or disable multiple “echo statements” in bash script.
typeset debug=false # set to true if need to debug
...
if [ $debug == "true" ]; then
echo
echo "Filter"
read
fi
...
if [ $debug == "true" ]; then
echo
echo "to run awk"
fi
Couldn't post a code block in a comment, so I'll post this as an answer.
If you're a perfectionist (like I am) and don't want the last set +x line to be printed... and instead print Success or FAIL, this works:
(
set -e # Stop at first error
set -x # Print commands
set -v # Print shell input lines as they are read
git pull
// ...other commands...
) && echo Success || echo FAIL
It will create a sub process, though, which may be an overkill solution.
If you're running it in cron, why not just dump the output? Change your crontab entry so that it has > /dev/null at the end of the command, and all output will be ignored.