5

I have a file which contains pipe delimited data. I want to print the lines which have only 5 columns. I tried the below. It does not work. It simply prints all the lines. Could you please let me know how to do it correctly. Thank you!

Contents of the file:

10413300|Maintenance Stock Clerk A - Michelle Hanley | Michelle Hanley (1680654)|Filled||10402112
10413301|Maintenance Stock Clerk - dasda|space||10402122
MOM Parkcentrum F;Alkotás u. 53.||34535|34345|asd
Business Center Naberezhnaya Tower" Block C"|||3453|ffg|345

Here I'm trying to print lines which have 5 columns only.

$ gawk -F "|" '{ if (NF=5) print $0 }' data3
10413300 Maintenance Stock Clerk A - Michelle Hanley   Michelle Hanley (1680654) Filled
10413301 Maintenance Stock Clerk - dasda space  10402122
MOM Parkcentrum F;Alkotás u. 53.  34535 34345 asd
Business Center Naberezhnaya Tower Block C   3453 ffg

The output shows that 2 out 4 lines have more than 5 columns.

$ gawk -F "|" '{print NF}' data3
6
5
5
6
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1 Answer 1

Reset to default
15

NF=5 is an assignment, testing values is done with ==. Like this:

awk -F \| 'NF==5' data3
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  • 2
    Confusing assignment and comparison operators is one of the most-common problems in "programming" (in the broadest sense of the term). It's very easy, especially in a *nix command-line environment, where Bourne shell and descendants interpret Variable=value as assignment but Variable = value as comparison
    – Monty Harder
    Jan 31, 2017 at 16:39

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