Extracting a part of String using grep/sed

Question

I have a file in linux with similar entries as below

dn: CN=HP_NetworkSupport,OU=groups,DC=HDFCSLDM,DC=COM
dn: CN=Review users,OU=groups,DC=HDFCSLDM,DC=COM
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Answer 1

This is one way with lookahead:

grep -Po '(?<=CN=)[^,]*' file > new_file

It gets all text from CN= (not included) until it finds a comma ,. The idea of [^,]* is to fetch any character that is not a comma.

Test

$ grep -Po '(?<=CN=)[^,]*' file
HP_NetworkSupport
Review users

Answer 2

Using sed:

$ sed -r 's/.*CN=([^,]*),.*/\1/' inputfile
HP_NetworkSupport
Review users

Answer 3

perl -lne 'print $1 if(/CN=([^\,]*),/)' your_file

Tested Below:

> cat temp
dn: CN=HP_NetworkSupport,OU=groups,DC=HDFCSLDM,DC=COM
dn: CN=Review users,OU=groups,DC=HDFCSLDM,DC=COM
> perl -lne 'print $1 if(/CN=([^\,]*),/)' temp
HP_NetworkSupport
Review users
>

Answer 4

Using awk

awk -F"=|," '{print $2}' file
HP_NetworkSupport
Review users

or

awk -F[=,] '{print $2}' file
HP_NetworkSupport
Review users

Set the delimiter to , or =, then print second field.

---

To handel field with comma within, you should use a parser for LDAP, but this should work.

echo file
dn: CN=HP_NetworkSupport,OU=groups,DC=HDFCSLDM,DC=COM
dn: CN="Review, users",OU=groups,DC=HDFCSLDM,DC=COM

awk -F"CN=|,OU" '{print $2}' file
HP_NetworkSupport
Review, users

Answer 5

Pipe it through this command:

sed -E "s/.*CN=(.+?),OU=.*/\\1/g"