awk error with "[" and "]" delimiters

Question

I have string looks likes this

string="xxxxx.yyyyy[2].zzzzz"

I want to extract the number between the [ ]. I used the following awk command

echo $string | awk -F'[]' '{print $2}'

But this awk command returns error:

awk: bad regex '[]': Unmatched [ or [^

How to fix that?

score 6 · Accepted Answer · answered Nov 06 '14 at 11:06

6

This should work:

echo "xxxxx.yyyyy[2].zzzzz" | awk -F '[][]' '{print $2}'
2

Order of ] before [ inside character class is important here.

This shall also work by double escaping [ and ] using alternation regex:

echo "xxxxx.yyyyy[2].zzzzz" | awk -F '\\[|\\]' '{print $2}'
2

OR:

echo "xxxxx.yyyyy[2].zzzzz" | awk -F '[\\[\\]]' '{print $2}'
2

answered Nov 06 '14 at 11:06

anubhava

1

+1 I'm surprised that the `][` are interpreted as part of one character class, rather than the whole thing being interpreted as two empty classes. Another option would be to use `']|\\['`. – Tom Fenech Nov 06 '14 at 11:14
@TomFenech I have seen it someplace in the `awk` manual that best way to get the brackets, is the `-F'[][]'` – Jotne Nov 06 '14 at 11:28
Or with double quote: `awk -F "\\\[|\\\]" '{print $2}'` or `awk -F "[][]" '{print $2}'` – Jotne Nov 06 '14 at 11:30
1

@TomFenech its specifically defined by awk language that if you want `]` in a bracket expression (it's not a character class) it must be the first character or be escaped `\]`. Other characters that have meaning for bracket expressions (e.g. `-` as used in a range like `0-9`) have similar requirements on position/escaping within the expression. – Ed Morton Nov 06 '14 at 13:57

score 0 · Answer 2 · answered Nov 06 '14 at 11:10

0

With sed:

echo "xxxxx.yyyyy[2].zzzzz" | sed -r 's/.*\[(.*)\].*/\1/g'

With awk:

 echo "xxxxx.yyyyy[2].zzzzz" | awk 'BEGIN {FS="[";RS="]"} {print $2}'

answered Nov 06 '14 at 11:10

Arjun Mathew Dan

score 0 · Answer 3 · answered Nov 06 '14 at 11:14

0

With grep

echo "xxxxx.yyyyy[2].zzzzz" | grep -oE '[0-9]'

2

answered Nov 06 '14 at 11:14

Arnab Nandy

score 0 · Answer 4 · answered Nov 06 '14 at 11:26

Here is a more generic approach. It will get any number out of a text file:

echo "xxxxx.yyyyy[2].zzzzz" | awk -F "[^0-9]*" '{for (i=2;i<=NF;i++) printf "%s%s",$i,(i==NF?RS:" ")}'
2

Other example

cat file
"C 56","Home athletics center","LAC"
"D001","50M Run","50M"
"D003","10.505M Run","100M","42K"

awk -F "[^0-9.]*" '{for (i=2;i<=NF;i++) printf "%s%s",$i,(i==NF?RS:" ")}' file
56
001 50 50
003 10.505 100 42

4 Answers4