extracting date field from the lines

Question

I have a file like this

Sampleoutput1[Fri Jun 10 06:11:59 2016]Local/ESSBASE0///139929633446208/Error(10
Sampleoutput1Local/ESSBASE0///139929633446208/[Fri Jun 11 06:11:59 2016]Error(10

I need to extract date field and add this date as column with delimited ($).Date can be present anywhere in the file.

[Fri Jun 10 06:11:59 2016]$Sampleoutput1[Fri Jun 10 06:11:59 2016]Local/ESSBASE0///139929633446208/Error(10
[Fri Jun 11 06:11:59 2016]$Sampleoutput1Local/ESSBASE0///139929633446208/[Fri Jun 11 06:11:59 2016]Error(10

Answer 1

Assuming the date is always enclosed in [ ... ] and that's the first [ on the line we can use a number of methods. Two common ones I use:

Using two cut commands (the lazy option, but it's very easy to understand what it does):

cut -d'[' -f2 $srcfile | cut -d ']' -f1

The standard sed option, using a simple regex:

sed 's/.*\[\([^]]*\)].*/\1/' $srcfile

If your output is more complicated with multiple [ characters in it then you'll need to use the sed option with a more complicated string:

sed 's/.*\[\(... ... .. ..:..:.. ....\)].*/\1/' $srcfile

EDIT: The revised question wants this added to the front of the string, leaving the rest of the line untouched. So we use the sed variant with a minor change:

sed 's/\(.*\[\(... ... .. ..:..:.. ....\)].*\)/[\2]$\1/' $srcfile

Result:

[Fri Jun 10 06:11:59 2016]$Sampleoutput1[Fri Jun 10 06:11:59 2016]Local/ESSBASE0///139929633446208/Error(10
[Fri Jun 11 06:11:59 2016]$Sampleoutput1Local/ESSBASE0///139929633446208/[Fri Jun 11 06:11:59 2016]Error(10