Return a regex match in a Bash script, instead of replacing it

Question

I just want to match some text in a Bash script. I've tried using sed but I can't seem to make it just output the match instead of replacing it with something.

echo -E "TestT100String" | sed 's/[0-9]+/dontReplace/g'

Which will output TestTdontReplaceString.

Which isn't what I want, I want it to output 100.

Ideally, it would put all the matches in an array.

edit:

Text input is coming in as a string:

newName()
{
 #Get input from function
 newNameTXT="$1"

 if [[ $newNameTXT ]]; then
 #Use code that im working on now, using the $newNameTXT string.

 fi
}

score 57 · Answer 1 · answered Dec 14 '09 at 01:59

57

You could do this purely in bash using the double square bracket [[ ]] test operator, which stores results in an array called BASH_REMATCH:

[[ "TestT100String" =~ ([0-9]+) ]] && echo "${BASH_REMATCH[1]}"

answered Dec 14 '09 at 01:59

John Kugelman

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10

it takes a real monkey to figure this out – Tegra Detra Feb 22 '13 at 10:31
1

But see http://stackoverflow.com/questions/218156/bash-regex-with-quotes for issues with bash quoting in different BASH versions. – Jake Biesinger Aug 27 '13 at 18:47

Dennis Williamson · Accepted Answer · 2009-12-14T02:24:37.460

41

echo "TestT100String" | sed 's/[^0-9]*\([0-9]\+\).*/\1/'

echo "TestT100String" | grep -o  '[0-9]\+'

The method you use to put the results in an array depends somewhat on how the actual data is being retrieved. There's not enough information in your question to be able to guide you well. However, here is one method:

index=0
while read -r line
do
    array[index++]=$(echo "$line" | grep -o  '[0-9]\+')
done < filename

Here's another way:

array=($(grep -o '[0-9]\+' filename))

edited Dec 14 '09 at 02:24

answered Dec 14 '09 at 02:00

Dennis Williamson

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Added in the input method in the original question. (it’s input is in the forum of a string) – Mint Dec 14 '09 at 02:29

score 31 · Answer 3 · answered Dec 16 '09 at 17:33

31

Pure Bash. Use parameter substitution (no external processes and pipes):

string="TestT100String"

echo ${string//[^[:digit:]]/}

Removes all non-digits.

answered Dec 16 '09 at 17:33

Fritz G. Mehner

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score 9 · Answer 4 · answered Apr 12 '19 at 10:58

I Know this is an old topic but I came her along same searches and found another great possibility apply a regex on a String/Variable using grep:

# Simple
$(echo "TestT100String" | grep -Po "[0-9]{3}")
# More complex using lookaround
$(echo "TestT100String" | grep -Po "(?i)TestT\K[0-9]{3}(?=String)")

With using lookaround capabilities search expressions can be extended for better matching. Where (?i) indicates the Pattern before the searched Pattern (lookahead), \K indicates the actual search pattern and (?=) contains the pattern after the search (lookbehind).

https://www.regular-expressions.info/lookaround.html

The given example matches the same as the PCRE regex TestT([0-9]{3})String

Wow, can't believe people still come here, after 10 years. I would never have guessed it would be viewed this much when I was asking it. I thought I was the only one at the time, who didn't know how to do it. — Mint, May 02 '19 at 07:31
Bash is so terrible that there are people looking to do things like this all the time. — i30817, Dec 13 '21 at 13:00

score 4 · Answer 5 · answered Dec 14 '09 at 01:56

4

Use grep. Sed is an editor. If you only want to match a regexp, grep is more than sufficient.

answered Dec 14 '09 at 01:56

Mic

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SED doesn't write to the source file - only way for that to happen is if you write the output with the same filename in order to overwrite the file. – OMG Ponies Dec 14 '09 at 01:57
1

True, but sed is still an editor. Its name is even short for stream editor. If he only needs to match he can use grep. – Mic Dec 14 '09 at 01:58
1

i don't think he wants to match. He says he needs the 100 – ghostdog74 Dec 14 '09 at 02:45
1

so using grep's -o should do the job – ghostdog74 Dec 14 '09 at 02:46
Yep, I need the 100, working solution found though, thanks to your guys (especially Dennis) – Mint Dec 14 '09 at 05:13
1

I agree, only it should be said: SED is more than sufficient, while grep IS sufficient :) – Yuval Sep 29 '16 at 14:33

score 1 · Answer 6 · answered Dec 14 '09 at 02:38

1

using awk

linux$ echo -E "TestT100String" | awk '{gsub(/[^0-9]/,"")}1'
100

answered Dec 14 '09 at 02:38

ghostdog74

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score 1 · Answer 7 · answered Dec 14 '09 at 17:30

1

I don't know why nobody ever uses expr: it's portable and easy.

newName()
{
 #Get input from function
 newNameTXT="$1"

 if num=`expr "$newNameTXT" : '[^0-9]*\([0-9]\+\)'`; then
  echo "contains $num"
 fi
}

answered Dec 14 '09 at 17:30

ephemient

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score 0 · Answer 8 · answered Dec 14 '09 at 02:06

Well , the Sed with the s/"pattern1"/"pattern2"/g just replaces globally all the pattern1s to pattern 2.

Besides that, sed while by default print the entire line by default . I suggest piping the instruction to a cut command and trying to extract the numbers u want :

If u are lookin only to use sed then use TRE:

sed -n 's/.*\(0-9\)\(0-9\)\(0-9\).*/\1,\2,\3/g'.

I dint try and execute the above command so just make sure the syntax is right. Hope this helped.

ghostdog74 · Answer 9 · 2009-12-14T02:58:21.153

using just the bash shell

declare -a array
i=0
while read -r line
do
        case "$line" in
            *TestT*String* )
            while true
            do
                line=${line#*TestT}
                array[$i]=${line%%String*}
                line=${line#*String*}
                i=$((i+1))
                case "$line" in
                    *TestT*String* ) continue;;
                    *) break;;
                esac
            done
            esac
done <"file"
echo ${array[@]}

Return a regex match in a Bash script, instead of replacing it

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