To grep for a word before matching pattern

Question

I have the below output. I want to extract the number before the text. Like for example, I grep for TrainIdentifyBusinessError and I want 1612 to be displayed. I grep for TrainIdentifyTechnicalError and I want 3 to be displayed.

1612 TrainIdentifyBusinessError 252 TrainIdentifySuccess 3 TrainIdentifyTechnicalError

Answer 1

You could use sed instead... for example

$ sed -nr 's/.*( |^)([0-9]+) TrainIdentifyBusinessError.*/\2/p' file
1612

or

$ sed -nr 's/.*( |^)([0-9]+) TrainIdentifySuccess.*/\2/p'
252

or

$ sed -nr 's/.*( |^)([0-9]+) TrainIdentifyTechnicalError.*/\2/p'
3

• -n don't print anything until we ask for it
• -r use ERE
• .* any number of any chars on the line
• ( |^) space or start of line
• ([0-9]+) one or more digits and (save this)
• \2 back reference to the second (saved pattern)
• p print the edited line

Afterthought... if you need to do this regularly you could make a shell function (add to your shell's ~/.*rc file, for example, ~/.bashrc if you use bash), for example:

getnum() { sed -nr 's/.*( |^)([0-9]+) TrainIdentify'"$1"'.*/\2/p' "$2" ; }

Usage example (specify the field and the filename on the command line - if the file is always the same file, you could put the full path to it inside the function instead of "$2"):

$ getnum BusinessError file
1612
$ getnum TechnicalError file
3
$ getnum Success file
252

Answer 2

This can be solved using the Perl extension for grep (the -P flag). To get 3 from TrainIdentifyTechnicalError:

$ echo "1612 TrainIdentifyBusinessError 252 TrainIdentifySuccess 23 TrainIdentifyTechnicalError" | grep -Po "[[:digit:]]+ *(?=TrainIdentifyTechnicalError)"
23 

To get 1612 from TrainIdentifyBusinessError

$ echo "1612 TrainIdentifyBusinessError 252 TrainIdentifySuccess 23 TrainIdentifyTechnicalError" | grep -Po "[[:digit:]]+ *(?=TrainIdentifyBusinessError)"
1612