CODEWITHSUNDEEP
awk
Sayalic
Sayalic

Reputation: 7650

Awk:How to get a part of a field?

I got this data:

aaa:123 bbb:**234** ccc:345

eee:435 ddd:**231** xxx:897

...

How to get the bolded part by using Awk?

Upvotes: 3

Views: 7543

Answers (5)

Lucian Enache
Lucian Enache

Reputation: 2520

You can use something like this

BEGIN { FS = ":" }
{  
   split($(NF-1),a," ");
   $2=a[2]; 
   print $2;   
}

Supposed that your column is is always the 2th and the number of columns is 3.

Upvotes: 1

Jotne
Jotne

Reputation: 41446

I would have used this:

awk '{split($2,a,":");print a[2]}' file

Why? Its not easy to see if its space or tab that separates the filed. Using default setting of FS in awk is any blank, so it will work even if its tab or spaces. Then we from second field take out part we need, using split by :

Upvotes: 4

Michael J. Barber
Michael J. Barber

Reputation: 25032

A general approach to obtaining portions of fields is to use split:

awk -F':' '{ split($3, subfield, " "); print subfield[1] }'

Others have mentioned using a regex to treat both spaces and colons as field separators. This is fine, so long as the number of spaces is predictable. But it will fail otherwise, e.g., for input like this:

aaa:123 bbb:234 ccc:345
eee:435 ddd:231 xxx:897
fff:214 ads 23423 fds:183 eee:234

Pick whichever approach better matches your data.

Upvotes: 16

devnull
devnull

Reputation: 123458

It's rather straightforward. Did you try

awk -F'[: ]' '{print $4}' inputfile

Upvotes: 3

Birei
Birei

Reputation: 36262

Use a regular expression of spaces plus colon and count fields:

awk -F'[[:blank:]:]+' '{ print $4 }' infile

It yields:

234
231

Upvotes: 6

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