Matching a quotation in the beginning of a field

[AwkWeasel]

I need to print every record whose first field begins with a double quotation.

I began by trying to set my Record Separator as any word with a quotation in it. (since quotes appear later in the record also)

I can't seem to match the quotation in the $1.

The text looks like this:

Wards Room Source Site

"SMITH, JOHN L 01/30/02 373369 040618 EM



Here is what I have:

Begin { RS = "\" *[[:upper]]+ " }

$1 ~ /^\"/ { print $0 > "mfinal.txt" }


Thanks for looking

 

[Annihilannic]

Since $1 is on the beginning of the line, you don't really need to examine the individual fields anyway, therefore no need to play with RS. As an aside, note that BEGIN should be in capitals.

Try:

awk '/^"/' inputfile > mfinal.txt

(Note, if you're working on Windows, you may have fun with quotes, so let us know...)

Annihilannic.

 

[AwkWeasel]

I am working with windows, so that may be why '/^"/' isn't matching anything.

I'll have to try a workaround.

Thank you for the input, though.

 

[Annihilannic]

A simple workaround is to create a separate AWK script file, say myscript.awk, containing simply:

/^"/

Then the following should work:

awk -f myscript.awk inputfile > mfinal.txt

Annihilannic.