CODEWITHSUNDEEP
bashtextawk
user8034918
user8034918

Reputation: 439

How to extract column in text file in bash

I have a text file, the format is below, and I would like to only capture the numbers after the .txt. I did awk '{print $2}' filename and it gave me the wrong result.

For some of the lines, it gave me : instead of the number. For example, in the second line I will get : instead of 914.

Is there any other way that I can extract the numbers after .txt? I am not referring to numbers in the rgbpart.

image/Subject01.txt:1310 : image/Subject01/Scene4/Color/rgb7
image/Subject01.txt: 914 : image/Subject01/Scene4/Color/rgb3
...

Upvotes: 1

Views: 2871

Answers (4)

RavinderSingh13
RavinderSingh13

Reputation: 133428

Could you please try following.

awk '
match($0,/\.txt:[^:]*/){
  val=substr($0,RSTART,RLENGTH)
  sub(/[^0-9]+/,"",val)
  print val
}
' Input_file

2nd solution: Using field separator.

awk 'BEGIN{FS="[.:]"} $2=="txt"{print $3+0}'  Input_file

Upvotes: 1

Phil
Phil

Reputation: 26

You can also use the cut command

cut -d ':' -f 2 filename

This will set the [d]elimiter to [:] and then will take the [2]nd [f]ield

Upvotes: 1

Benjamin W.
Benjamin W.

Reputation: 52112

You could use GNU grep for this:

$ grep -Po '\.txt: *\K[[:digit:]]+' infile
1310
914

-P enables Perl-compatible regular expressions (required for the \K), and -o retains only the match.

The regular expression looks for the string .txt:, followed by any number of blanks (including zero); this part of the match is discarded (\K), and then we match as many digits as we can find.

Upvotes: 0

oguz ismail
oguz ismail

Reputation: 50750

You forgot specifying a custom field separator. Like

awk -F ' *: *' '{print $2}' file

Upvotes: 4

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