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So I want to do some logging and therefor, want to put a date in front of a bash script's output. The problem is that is has multiple lines of output. I am able to only put the date before the whole output. But then I have a line without a date in the logs. Of course I can assume the date from the line above is the same, but I was hoping there is a solution. Thanks in advance!

This is my script that calls another script:

echo $(date "+%F %T") : starting script
echo $(date "+%F %T") : $(./
echo $(date "+%F %T") :script ended

This is the output:

2012-07-26 15:34:12 : starting script
2012-07-26 15:35:14 : First line of output
second line of output
2012-07-26 15:35:17 : script ended

And thats what I would like to have:

2012-07-26 15:34:12 : starting script
2012-07-26 15:35:14 : First line of output
2012-07-26 15:35:15 : second line of output
2012-07-26 15:35:17 : script ended
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I don't suppose you could just just change the second script? – jmetz Jul 26 '12 at 14:32
Nah, Unfortunately not. That's why I came here. – tzippy Jul 26 '12 at 14:33
up vote 7 down vote accepted

According to a similar question on Stack Overflow, there are 2 great options.

awk (Answer)

<command> | awk '{ print strftime("%Y-%m-%d %H:%M:%S"), $0; }'

annotate (Answer)

annotate is a small bash script, that can either be directly obtained through the link provided here, or, on Debian based systems, through the package devscripts (in the form of annotate-output).

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I think that you can use awk for this:

./ | awk '{ print strftime()" : "$0; }'

(see for formatting the date returned by strftime())

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awk: calling undefined function strftime -- is this specific to gawk? – Daniel Beck Jul 26 '12 at 14:50
Now that you mention it, yes it seems to be :/ – epingle Jul 26 '12 at 15:29

You can use awk

./ | awk '{ print d,$1}' "d=$(date "+%F %T")"

awk takes an input stream and modifies it's output. This script is saying "print d followed by the original output", then then populates d with the date.

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Doesn't this evaluate the date expression only once as well, so that all such lines print the same time? – Daniel Beck Jul 26 '12 at 14:42
@DanielBeck Yes it does - I had assumed it was picking up the date each time but just finished my test quickly enough – Paul Jul 26 '12 at 14:45

This will print the current date and time before each line of output of ./

set -f
./ | while read -r LINE; do echo $(date "+%F %T") : "$LINE"; done
set +f

How it works

  • set -f turns off bash expansion (or echo * wouldn't print an actual asterisk).

  • while read -r LINE; do ... done saves one line of output in the variable $LINE and executes ..., until all lines are processed.

  • echo $(date "+%F %T") : "$LINE" prints the line with the current time and date.

  • set +f turns bash expansion back on, so it won't interfere with the rest of your bash script.

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It gets evaluated only once, so every sed inserts the same date, from just before starting the second script. Likely not what the user wants... – Daniel Beck Jul 26 '12 at 14:37
I don't think this attempt will work either. AFAIK, the whole subshell gets evaluated completely first, then the loop is traversed. So now the times are from after the execution. – Daniel Beck Jul 26 '12 at 14:49
Your example doesn't work, since you only delay output, not execution of ls. Just try your script with the following #!/bin/bash (newline) echo 1 ; sleep 1 ; echo 2 ; sleep 2 ; echo 3 ; sleep 3 ; echo 4 – Daniel Beck Jul 26 '12 at 14:56
Use read with a while loop instead of a for loop. – chepner Jul 26 '12 at 15:01
@chepner: Good idea. – Dennis Jul 26 '12 at 15:28

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