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

I would like to extract only the CN information, till the first , for ex:

> HP_NetworkSupport
> Review users

in the above case to another file.

What would be command for doing the same.

Solution 1

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

Solution 2

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

Solution 3

Using sed:

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

Solution 4

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
>

View more solutions

Comments

[Birei]

Does this work for you? I think that sed can't use non-greedy quantifiers and parentheses must be escaped to do grouping.

[mvp]

Unfortunately, it will fail for cases like CN="Smith, John",OU="My Organization"

[fedorqui]

Yes, but the input is not double quotes surrounded.

[mvp]

Well, this is part of software engineering job - predict what customers would use. And they most likely will

[fedorqui]

I know, but the input and the explanation does not show that. What if the parameters are not in the CN, OU, DC order? It would fail on your answer.

[mvp]

Granted, for many regex questions correct answer is NOT regex, but proper parser

[fedorqui]

What do you mean? I don't understand it.

[mvp]

I mean that this is one nice example that to answer this question 100% correctly one must use LDAP parser. Technically, LDAP CN or OU could include text CN=blah (probably quoted) inside of it. How's that? This is similar to premise that you cannot use regex to parse XML.

[fedorqui]

Yes, I agree :) In fact I wasn't aware that this was a piece of LDAP config file. Let's see if our attempts are enough for the requirements, cheers!

[Chris Seymour]

Good solution is GNU Grep is available +1. To fixed @mvp criticism you could do include positive lookbehind grep -Po '(?<=CN=).*(?=,OU=)' file

[fedorqui]

Thanks @sudo_O! I will not update my answer, as the accepted one is not using it, so the OP may have enough with the simple comma checking. Anyway, thanks for teaching again! :)

[Admin]

Thank you for providing regular-expressions.info/lookaround.html link.

[Big McLargeHuge]

I'm curious, if you're searching for HTML or XML tags, would you need to escape the characters like this? grep -Po '(?<=\<string name\=")[^"]*'

[fedorqui]

@DavidKennedy give it a try ;-) I tried and my GNU grep says it is not necessary: grep -Po '(?<=<hola>).*(?=</hola>)' <<< "<hola>adeu</hola>" returns "adeu" without a problem. Of course, you can escape every single character, but this one is not necessary.