Cutting a length of specific string with grep

Question

Let's say we have a string "test123" in a text file.
How do we cut out "test12" only or let's say there is other garbage behind "test123" such as test123x19853 and we want to cut out "test123x"?

I tried with grep -a "test123.\{1,4\}" testasd.txt and so on, but just can't get it right.
I also looked for example, but never found what I'm looking for.

Answer 1

What you need is -o which print out matched things only:

$ echo "test123x19853"|grep -o "test.\{1,4\}"
test123x

$ echo "test123x19853"|grep -oP "test.{1,4}"
test123x

-o, --only-matching show only the part of a line matching PATTERN

Answer 2

expr:

kent$  x="test123x19853" 
kent$  echo $(expr "$x" : '\(test.\{1,4\}\)')
test123x

Answer 3

If you are ok with awkthen try following(not this will look for continuous occurrences of alphabets and then continuous occurrences of digits, didn't limit it to 4 or 5).

echo "test123x19853" |  awk 'match($0,/[a-zA-Z]+[0-9]+/){print substr($0,RSTART,RLENGTH)}'

In case you want to look for only 1 to 4 digits after 1st continuous occurrence of alphabets then try following(my awk is old version so using --re-interval you could remove it in case you have latest version of ittoo).

echo "test123x19853" | awk --re-interval 'match($0,/[a-zA-Z]+[0-9]{1,4}/){print substr($0,RSTART,RLENGTH)}'