UNIX for Dummies Questions & Answers

I have a script that does the following. It searches a listing of directories with specific extensions and then formats a wc on those files. The code looks like this

Code:

find <directory>  -name '*.js' -o -name '*.html' | awk '{print \"wc -l \"$1}' > file \n"

The result is a file with the "wc -l" command preceeding the file directory listing ---> wc -l <file directory listing>

Is there a way to do the same thing but use

Code:

 | awk '{print \"egrep -cv ^$|^[/]\$1'}

Using egrep at the CLI helps to eliminate the blanks and commented code. I just can't seem to get it to work in the script. Thanks in advance

Do you mean something like this?

Code:

... | awk '{print "egrep -cv \^$|\^[/]\\" $1}'

Regards

Something like that but I am still getting a syntax error.

Give some lines of the input and the desired output of the awk command.

The script I was given is an AWK script and It has alot of output string commands. This AWK script creates a .sh script with all the listing and various functions. Once, the AWK script is complete, the .sh script runs. The first piece of code, finds all the files of a particular type and then that output is piped to an AWK statement that prints the letters 'wc -l' proceeding each listed directory file (the listings are in a .sh file).

There is alot to post but I just want to get the correct expression to accurately capture lines of code by omitting blank lines and comments. Egrep seems to do it best but I don't know how to add that to the script.

You can try this with find:

Code:

find <directory>  -name '*.js' -o -name '*.html' -exec egrep <options> {} \;

This didn't work either. I will send you a PM.