How can I get to stdout all commands that run in bash script?

That is output must contain commands output AND commands themselves.

I found

#!/bin/bash -x

but this is not exactly the same because instead of

 mysql -v dbname < dump.sql | grep "CREATE TABLE"

it shows

+ mysql -v dbname
+ grep 'CREATE TABLE'
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2 Answers

up vote 4 down vote accepted

Use bash -v.

This is the script:

#!/bin/bash -v

echo "Hello, World" | sed 's|Hello|Goodbye|'

echo "Done."

This is the output:

#!/bin/bash -v

echo "Hello, World" | sed 's|Hello|Goodbye|'
Goodbye, World

echo "Done."
Done.

Unfortunately, there is no special marker like PS4 for printing expanded commands. You could combine both though to quickly identify commands:

#!/bin/bash -vx

echo "Hello, World" | sed 's|Hello|Goodbye|'
+ echo 'Hello, World'
+ sed 's|Hello|Goodbye|'
Goodbye, World

echo "Done."
+ echo Done.
Done.
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set -x is other way of doing it.

$ cat a.sh
#!/bin/bash

set -x
echo "Hello, World" | sed 's|Hello|Goodbye|'
echo "Done."

Output will be:

sh a.sh
+ echo 'Hello, World'
+ sed 's|Hello|Goodbye|'
Goodbye, World
+ echo Done.
Done.
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Which is kind of what the OP dismissed as useless right in his question... – Daniel Beck Feb 7 at 18:12
@Ashok Could you please explain? I can't see the difference: here p.s.: GNU bash, version 4.1.5(1)-release (i486-pc-linux-gnu) Ubuntu 10.04.3 – Putnik Feb 8 at 13:49
@Putnik It's the same basic thing, but you can set -x anywhere in the script, and even deactivate it again. So if you only want it on the echo "Done" line, put set -x just before that one. – Daniel Beck Feb 8 at 15:09
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