Shell Programming and Scripting

I want to finde a no in a fine

My line look like this

Code:

 “SWD_BUILD ?= 207 # can be 0-999- Build nr”

and I only want 1 output. the no 207
My command rigtht now is

Code:

 grep -A0 “SWD_BUILD” Makefile | grep -m1 -Eoe “[0-9][0-9][0-9]|[0-9][0-9]|[0-9]”

But I get
207
0
999

How to get only the 207 out ??

Moderator's Comments:

Please use code tags for your code and data, next time

It is quite simple: you use the wrong tool you get the wrong result.

grep is a tool for extracting LINES out a file, not for extracting certain numbers or other parts of a line. So the first grep has a single line as a result (the "-A0" is superfluous), the second grep has the same line as a result and the "-o" (which i suppose is a GNU-extension and therefore better avoided anyway) matches 1,2 or 3 consecutive numerical digits. Look at your sample line, there are 3 places which match this: the three numbers which the output consists of.

Btw.: the extended regexp "[0-9][0-9][0-9]|[0-9][0-9]|[0-9]" is much more simpler written as a standard regexp: "[0-9][0-9]*[0-9]*".

So, after this lengthy explanation of why that didn't work: what you need is sed:

Code:

sed -n '/^SWD_BUILD \?=/
                 {
                     s/^[^=]*= //
                     s/^\([0-9][0-9]*\).*/\1/p
                 }' /path/to/input.file

This looks for lines starting with "SWD_BUILD ?=", all the other lines are silently ignored (the "-n" switch). On these lines the following is done: first, cut off everything from the beginning to the first "="-character and a following blank. The number you are interested in is now at the beginning of the line. In the next step everything not numerical after this initial number is cut off too so that the number remains. This is finally printed.

I hope this helps.

bakunin

If we fix the quotes in your sample code and add another element to your pipeline to just take the 1st line of output, you can get what you want with:

Code:

 grep -A0 "SWD_BUILD" Makefile | grep -m1 -Eoe "[0-9][0-9][0-9]|[0-9][0-9]|[0-9]"|head -n 1

But, a three element pipeline for this seems like overkill, and if there is more than one line containing the string SWD_BUILD, it will only show you the results from the 1st line containing that string.

A more efficient alternative, using awk, is:

Code:

awk '/SWD_BUILD/{match($0, /[[:digit:]]{1,3}/);print substr($0, RSTART, RLENGTH)}' Makefile

which will print up to three characters from the 1st string of digits found on every line containing the string SWD_BUILD.

However, if there are lines in Makefile containing SWD_BUILD that do not contain any digits, the above awk command will print an empty line for each of those input lines. If this is a problem, the following version will ignore those lines:

Code:

awk '/SWD_BUILD/{if(match($0, /[[:digit:]]{1,3}/)) print substr($0, RSTART, RLENGTH)}' Makefile

If you want to try either of these on a Solaris/SunOS system, change awk to /usr/xpg4/bin/awk or nawk.

Thanks Don Cragun, iw works fine

Note1: nawk on Solaris knows neither [[:digit:]] nor {1,3}, use /usr/xpg4/bin/awk instead

Note2: when using GNU awk version < 4 , use awk --re-interval or --posix to make {1,3} work

Note3: mawk also does not know {1,3}

Yes it dos watt I want
But now I face watt I thought was a simple thing. I know this is not the right forum, but maybe you can tel me where to ask then.

Watt I need is to set a Jenkins Environment Variable to this value.
I have running the script from Jenkins and start a Cygwin shell.
I can see that in the cygwin shell the env is set but when I return to the Jenkins shell it is gone.

I have also try "Gawk for Windows" but dit not succeed.

Who to ask ??