findstr confused by colon?

Question

If I run from cmd.exe:

findstr "\"A\" : \"B\"" c:\temp\sample.json

echo %errorlevel%

with the contents of sample.json being

{
    "Abad" : "B"
}

errorlevel shows as 0.

If I replace ':' with '.', I again get 0. But, if I instead use two '.'s:

findstr "\"A\" ..\"B\"" c:\temp\sample.json

findstr correctly returns an errorlevel of 1. What is findstr trying to do?

Answer 1

RTFM.

With a space between strings, FINDSTR looks for string1 OR string2 OR string3...

Your FINDSTR is thus looking for "A" OR : OR "B"

To look for a string containing spaces, use (eg)

 findstr /c:"\"A\" : \"B\""