Testing regex from stdin using grep|sed|awk

Question

Sometimes, I want to test is my regex correct.

How can I do reverse regex match from stdin?

F.e. I can match a string with provided regex, like:

grep "\(foo\)-bar"
foo
bar
foo-bar
foo-bar #Match found

What I would like to do, is the opposite, something like this:

$ grep "This is one string"
\(This\) #Will send "This" to stdout
This?.*  #Will send full match

Is this somehow possible without much scripting?

Answer 1

You can use - as the "file" to search, which will use standard input as the "haystack" to search for matching "needles" in:

$ grep -oE '[aeiou]+' -
This is a test  < input
i               > output
i               > output
a               > output
e               > output
whaaaat?        < input
aaaa            > output

Use Ctrl-D to send EOF and end the stream.

I don't believe, though, that you can do the same to use standard input for the -f switch which reads a list of patterns from a file. However, if you have a lot of patterns to text on one corpus, you can:

grep -f needle-patterns haystack.txt

where needle-patterns is a plaintext file with one regular expression per line.

Answer 2

Define the following function in your shell (you can just type it in, or put it in your ~/.bashrc):

testregex() {
  [ "$#" -eq 1 ] || return 1
  while IFS= read -r line; do
    printf '%s\n' "$1" | grep -Eoe "$line"
  done
}

Then you can test a regex as follows:

$ testregex 'This is a line'
This            <--input
This            <--output
This?.*         <--input
This is a line  <--output
slkdjflksdj     <--input with no output (no match)
s.*             <--input
s is a line     <--output
$               <--I pressed Ctrl-D to end the test