How to make bash scripts print out every command before it executes?
Solution 1:
bash -x script
or
set -x
in the script.
You can unset the option again with set +x
. If you just want to do it for a few commands you can use a subshell: `(set -x; command1; command; ...;)
Solution 2:
These also work:
set -v
or
#!/bin/bash -v
But -v doesn't print the PS4 string before each script line and it doesn't trace the steps of a "for" statement (for example) individually. It does echo comments while -x doesn't.
Here's an example of the output using -v:
#!/bin/bash -v
# this is a comment
for i in {1..4}
do
echo -n $i
done
1234echo
echo hello
hello
Here's the result of the same script with -x:
+ for i in '{1..4}'
+ echo -n 1
1+ for i in '{1..4}'
+ echo -n 2
2+ for i in '{1..4}'
+ echo -n 3
3+ for i in '{1..4}'
+ echo -n 4
4+ echo
+ echo hello
hello
Note that I included "echo -n" to add emphasis to the differences between -v and -x. Also, -v is the same as "-o verbose", but the latter seems not to work as part of a shebang.
Solution 3:
This should also work:
#!/bin/bash -x
cd ~/hello
ls
Solution 4:
This should work:
set -o verbose #echo on
...
set +o verbose #echo off
Solution 5:
set -o xtrace
and set +o xtrace
are your friends (it is more verbose, than -o verbose
, and the output goes to STDERR, as opposed to verbose, which seems to log to STDOUT).
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