CODEWITHSUNDEEP
linuxshellawkash
MOHAMED
MOHAMED

Reputation: 43558

awk error with "[" and "]" delimiters

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?

Upvotes: 2

Views: 998

Answers (4)

Jotne
Jotne

Reputation: 41460

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

Upvotes: 0

Arnab Nandy
Arnab Nandy

Reputation: 6712

With grep

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

2

Upvotes: 0

Arjun Mathew Dan
Arjun Mathew Dan

Reputation: 5298

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}'

Upvotes: 0

anubhava
anubhava

Reputation: 785856

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

Upvotes: 6

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