CODEWITHSUNDEEP
bashshellunixshcut
sailesh
sailesh

Reputation: 93

how to extract a field containing delimiter from file using cut command

In Unix, suppose a file contains 5 fields & data such as:

"112233"|"Roshan"|"25"|" FAX 022 3987789 \| TEL 77766288892 \| abc "|"Male"

need to extract 4th field. using below

column_value=`echo $line | cut -f4 -d'|'

This only gives us - " FAX 022 3987789 \

but need " FAX 022 3987789 \| TEL 77766288892 \| abc " as 4th column value.

Effective delimiter should be -

"|"

Upvotes: 0

Views: 1907

Answers (2)

Inian
Inian

Reputation: 85600

cut is not the right tool for the job when it involves a multi-character de-limiter needed for parsing input string/file.

You can use GNU Awk with FPAT which defines how each field in a record should look like. You can write FPAT as a regular expression constant in which case something like below should work.

FPAT = "(\"[^\"]+\")"

Using this in the Awk command,

line='"112233"|"Roshan"|"25"|" FAX 022 3987789 \| TEL 77766288892 \| abc "|"Male"'
awk '
BEGIN {
    FPAT = "(\"[^\"]+\")"
}{print $4}' <<<"$line"

produces an output as

" FAX 022 3987789 \| TEL 77766288892 \| abc "

Regular Expression - Test results

Upvotes: 1

Ian Kenney
Ian Kenney

Reputation: 6426

you can add the two extra fields as follows 

echo $line | cut -f 4,5,6 -d\|

alternatively you could use sed to replace the "|" delimiter with a different char (for example a tab)

echo $line | sed s/\"\|\"/\t/g | cut -f 4

Upvotes: 0

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