Shell Programming and Scripting

Hi all,

I'm trying to get 6 variable characters before a set of 4 fixed charaters in a text file and send it in a variable.
Ex:
Output log file 123456.txt is full creating new log.

What I want from this file is to get the 123456.txt into my variable. How do I grep this from the file when I don't know what the number will be. I know that it's gonna be a .txt file... so how can I grep the 6 chars before?

Thanks

Try:

Code:

var=$(sed -n 's/.*Output log file \(.*\)\.txt.*/\1/p' file)

Code:

var=$(awk -F'[ \t.]*'  '/Output log .* full/{print $4}' file)

thanks for the answer Scrutinizer. I know it can be done either by cutting the text or eleminate the text around it but it's not the way I want to do it, I know the text around the file name can vary. Know another way I could do it?

Please give us an example of input and desired output. Scrutinizer gave you exactly what you asked for in the first post.

Try this

Code:

var=$(sed -n 's/.* \([^ ][^ ]*\)\.txt.*/\1/p' file)

We have a winner... ctsgnb thanks

Code:

awk '{gsub(/.txt/,"",$4);print $4}' text.file