shell - Parsing the required field from an URL in awk -

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i have requirement parse required field url. purpose need traverse url letter letter since there no space separation between words . not able . have tried below

awk '{fs = "";for (i = 2; <= nf; i++) {print $i}}'  file

the file contains

https://stackoverflow.com/questions/ask/submit?s=sddasdsadsomefield=8171wantedfield=121212somefield=1201212...

so idea traverse url letter letter , reach required field .. not able that. suggestions please

this work, assuming pattern = , *.

echo "http://stackoverflow.com/questions/ask/submit?s=sddasdsadsomefield=8171*wantedfield=121212*somefield=1201212..." |awk -f "[*=]" -vfieldname="wantedfield"  '{for (i=1;i<=nf;i++){if(match($i,fieldname)){printf "%s\t", $(i+1)}};printf "\n"}'

output:

121212

changing somefield target field:

echo "http://stackoverflow.com/questions/ask/submit?s=sddasdsadsomefield=8171*wantedfield=121212*somefield=1201212..." | awk -f "[*=]" -vfieldname="somefield"  '{for (i=1;i<=nf;i++){if(match($i,fieldname)){printf "%s\t", $(i+1)}};printf "\n"}'

results in:

8171    1201212...

Comments

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