Shell Programming and Scripting

I have some input file, which contains some lines which are comma separated. Eg.

a,b,id=999],d
d,f,g,id=345],x
x,y,x,s,id=677],y

I run a loop to read the lines one by one. What i want is to extract the value on the right of id=. I cannot do it by Awk, since the column number is not fixed. So what i need is to first search and match "id=" pattern, and then print the number following it, until ']' is encountered.
Any way to do this??

Try this

Code:

sed 's/.*=\(.*\)].*/\1/' file

Stupendous!
Works like a charm, Thanks!

If you dont mind, can you just explain a little bit of that sed command you used, agn?

I have a beginner level of understanding. What i read in a book about sed like if there is an expression,
sed 's/\([a-z]*\).*/\1/'
then \1 remembers the second part of the pattern matched.

But in your answer you have used 3 *s. so that means three patterns? Also could you explain why we use '\' before parenthesis? And what is the significance of the expression inside parenthesis?

sed 's/.*=\(.*\)].*/\1/'

s/find/replace/ - substitute 'find' with 'replace'
s/.*=\(.*\)].*/\1/ - match till '='
s/.*=\(.*\)].*/\1/ - match till ']' . We escape the parenthesis as () is used for grouping in regex.
s/.*=\(.*\)].*/\1/ - match the rest of the string
s/.*=\(.*\)].*/\1/ - replace the pattern (the 'find' part) with what's inside the parenthesis (in the 'find' part). \1 contains the string that matched the regex inside the first parenthesis. \2 contains the match found by second parenthesis, \3 the third and so on.

Hope that helps.

if its not fixed, you can go through the fields one by one

Code:

awk -F"," '
{
 for(i=1;i<=NF;i++) {
    if ( $i ~ /id/) {
      gsub(/[^0-9]+/,"",$i)
      print $i
    }
 }
}
' file

#

Another one with awk: this also doesn't look for coulumn to be of fixed size.