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
}

Solution 1

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))

Solution 2

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]}"

Solution 3

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

string="TestT100String"

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

Removes all non-digits.

Solution 4

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

Solution 5

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

View more solutions

Comments

[OMG Ponies]

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.

[Mic]

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.

[Mint]

Added in the input method in the original question. (it’s input is in the forum of a string)

[ghostdog74]

i don't think he wants to match. He says he needs the 100

[ghostdog74]

so using grep's -o should do the job

[Mint]

Yep, I need the 100, working solution found though, thanks to your guys (especially Dennis)

[James Andino]

it takes a real monkey to figure this out

[Jake Biesinger]

But see stackoverflow.com/questions/218156/bash-regex-with-quotes for issues with bash quoting in different BASH versions.

[Yuval]

I agree, only it should be said: SED is more than sufficient, while grep IS sufficient :)

[Mint]

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.

[i30817]

Bash is so terrible that there are people looking to do things like this all the time.