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).

See more tips here