How to make bash scripts print out every command before it executes?

Question

For example, I have a simple bash file

#!/bin/bash
cd ~/hello
ls

How can I make it display every command before executing it? Just the opposite effect of "@echo off" in windows batch scripting.

Answer 1

bash -x script

or

set -x

in the script

Answer 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.

Answer 3

This should also work:

#!/bin/bash -x
cd ~/hello
ls

Answer 4

This should work:

set -o verbose #echo on
...
set +o verbose #echo off

Answer 5

goes like

language -x script

language = python, perl, bash -x = operator script = filename

hope it helps.

Answer 6

There are a few ways.

#!/usr/bin/env bash -x

as the shebang line.

Including set -x in the script itself will enable the functionality while set +x will disable it. Both of these methods will also work with the more portable sh shell.

If I remember correctly perl also has the -x option.

Answer 7

#! /bin/bash -x does what you want.