I have a text file and want to write a batch file to verify that it only contains one line with a specific content without any newline at the end of said line.
Comparing the content of the line is not difficult, as I can just read it into a variable and then compare it to what I expect.
My problem is with
asserting that there are no superfluous newlines around. As long as there are only Windows line endings (CRLF), I can do this quite easily by counting the number of lines ending with a newline, similarly to here:
Set num_of_endlines=0
For /f "tokens=1 delims=:" %%i in ('FINDSTR /r /n "$" %filepath%') DO (
Set num_of_endlines=%%i
)
If that num_of_endlines is 0 then all is good.
My problem is now, that this method does not work for Unix line endings. I also tried counting the line starts ("^") instead of the line endings ("$"). That improved the result as now leading newlines are captured, regardless whether they are CRLF or LF. But this does not capture a newline at the end of the first line (Not even a CRLF). Is there any way to count the number of CRLF and LF in a text file using a batch file?
lit :
Windows FINDSTR is not fully regex capable. But, PowerShell can do it. Note the use of -Raw to get the contents of the file. That is required to see the newline bytes.\n\nSET \"THEFILE=t.bash\"\n\nFOR /F %%a IN ('powershell -NoLogo -NoProfile -Command ^\n \"(Get-Content -Path '%THEFILE%' -Raw | Select-String -Pattern '\\x0a|\\x0d' | Measure-Object).Count\"') DO (\n SET \"LINECOUNT=%%a\"\n)\n\nECHO LINECOUNT is %LINECOUNT%\n",
2019-04-17T12:20:17