substitution of string in brackets

Shell Programming and Scripting
1

Hi friends!
I have a tab delimited file with two columns :
GB_45_DRB SP:0139466(mrmi sisignm)|SP:3674(fllflg_itoioh)|SP:68954779(RMTKLGF to emmdm-roomto)
GB_45_DRD SP:475928(mgmdksi rikgkg)|SP:587959(roykgl tiic-tm)|SP:0139466(mrmi sisignm)|SP:3674(fllflg_itoioh)|SP:68954779(RMTKLGF to emmdm-roomto)
GB_45_DRCD
GB_45_P5294 SP:3735599(fkkfdlk)

I want to remove all occurences of stuff between brackets(inclusive of brackets) in second column and replace all "|" with ";"

So that it looks like SP:0139466;SP:3674;SP:68954779 in second col. Problem is that second col might be empty, as in third row here.
Ur help would be appraciated..
Thanks

2

Try this one:

awk -F"[()]" '{
     gsub(/\|/, ";", $0);
     n=split($0, arr, /[()]/);}
     {for (i=1; i<=n; i++)
        if(i%2 == 1)
                printf ("%s", $i)
      }
      {print "";}
        ' yourfilename

The output I got is:

GB_45_DRB       SP:0139466;SP:3674;SP:68954779
GB_45_DRD       SP:475928;SP:587959;SP:0139466;SP:3674;SP:68954779
GB_45_DRCD
GB_45_P5294      SP:3735599
3

The first example below does the 'eliminate (stuff)' and the second also does the substitution of delimiters.
I thought there would have been a way to use [[:print:]] or [[:graph:]] to select the letters and sumbols inside the (), but I could not get that to work properly.

> echo "GB_45_DRB SP:0139466(mrmi sisignm)|SP:3674(fllflg_itoioh)|SP:68954779(RMTKLGF to emmdm-roomto)" | sed "s/([A-Za-z _\-]*)//g"
GB_45_DRB SP:0139466|SP:3674|SP:68954779

> echo "GB_45_DRB SP:0139466(mrmi sisignm)|SP:3674(fllflg_itoioh)|SP:68954779(RMTKLGF to emmdm-roomto)" | sed "s/([A-Za-z _\-]*)//g" | tr "|" ";"
GB_45_DRB SP:0139466;SP:3674;SP:68954779

>
4

Here is another simple one:

sed -e 's/([^)]*)//g' 's/\|/;/g' filename
5

Thanks guys..That works fine..

In suggestion from the previous reply, I need to substitute in second step ";" for "|", but that's okay...