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.

Answer 1

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[@]}

Answer 2

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

Answer 3

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

string="TestT100String"

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

Removes all non-digits.

Answer 4

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.

Answer 5

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]}"

Answer 6

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
}