CODEWITHSUNDEEP
bashgrep
Uxio
Uxio

Reputation: 33

Select match from second column where filter is in first column

I need to find this number: The '3' on the second column where there is a '3' in the first column.

(This is an example. I also could need to find the '25' on the second column where there is a '36' on the first column).

The numbers on the first column are unique. There is no other row starting with a '3' on the first column.

This data is in a text file, and I'd like to use bash (awk, sed, grep, etc.)

I need to find a number on the second column, knowing the (unique) number of the first column.

In this case, I need to grep the 3 below the 0, on the second column (and, in this case, third row):

108 330
132 0
3   3
26  350
36  25
43  20
93  10
101 3
102 3
103 1

This is not good enough, because the grep should apply only to the first column elements, so the output would by a single number:

cat foo.txt | grep -w '3' | awk '{print $2}'

Although it is a text file, let's imagine it is a MySQL table. The needed query would be:

SELECT column2 WHERE column1='3'

In this case (text file), I know the input value of the first column (eg, 132, or 93, or 3), and I need to find the value of the same row, in the second column (eg, 0, 10, 3, respectively).

Regards,

Upvotes: 1

Views: 1505

Answers (2)

tripleee
tripleee

Reputation: 189749

Assuming you mean "find rows where the first column contains a specific string exactly, and the second contains that string somewhere within";

awk -v val="3" '$1 == val && $2 ~ $1 { print $2 }' foo.txt

Notice also how this avoids a useless use of cat.

Upvotes: 3

Thor
Thor

Reputation: 47189

You can find repeated patterns by grouping the first match and looking for a repeat. This works for your example:

grep -wE '([^ ]+) +\1' infile

Output:

3   3

Upvotes: 0

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