grep regex在两个已知字符串之间拉出一个字符串

Question

I have a string of text in a file that I am parsing out, I almost got it but not sure what I am missing

basic expression I am using is

cat cred.txt | grep -m 1 -o '&CD=[^&]*'

I am getting a results of

&CD=u8AA-RaF-97gc_SdZ0J74gc_SdZ0J196gc_SdZ0J211

I do not want the &CD= part in the resulting string, how would I do that.

The string I am parsing from is:

webpage.asp?UserName=username&CD=u8AA-RaF-97gc_SdZ0J74gc_SdZ0J196gc_SdZ0J211&Country=USA

Answer 1

If your grep knows Perl regex:

grep -m 1 -oP '(?<=&CD=)[^&]*' cred.txt

If not:

sed '1s/.*&CD=\([^&]*\).*/\1/' cred.txt

Answer 2

Many ways to skin this cat.

Extend your pipe:

grep -o 'CD=[^&]*' cred.txt | cut -d= -f2

Or do a replacement in sed:

sed -r 's/.*[&?]CD=([^&]*).*/\1/' cred.txt

Or get really fancy and parse the actual QUERY_STRING in awk:

awk -F'?' '{ split($2, a, "&"); for(i in a){split(a[i], kv, "="); out[kv[1]]=kv[2];} print out["CD"];}'