How to search a pattern only first line using awk

Hi ,

I know this is a noob question but I am not able to find a solution .

I have many scripts which are either shell or perl scripts .
I want to count the number of perl scripts . I am searching the first line from all the files

This is my idea

find . -type f -perm -700 | xargs awk 'NR == 1 && /perl/ {print $0}'

This is not giving any output .

How can I search a pattern only in the first line of several files using awk.


Thanks.

Try a while loop or -exec with find

Find . Blah ¦ while read I;do awk etcetc "$I"; done

Sorry crappy formatting I'm on my phone

Hi
You can use a simple perl program:

#!/usr/bin/perl
@aa=`find . -t-name '*'.sh -o '*'.pl `;
$dd=scalar(@aa);$dd will give no of perl files or script files
foreach $zx(@aa)
{
$line1=`cat $zx|head -n 1`; # $line1 gives the first line
print $line1 if $line1 =~ /pattern/; # substitute the pattern you want to match
}

Hope this solves the problem.
For further solutions you can visit my blogspot:
http://linux-forum-karthik.blogspot.com/

Thanks for the reply.

But I want to avoid any loop , using exec is the same as using xargs as I did in my code .

Any other idea.

@Karthink : I am sure this can be done in that way , I am looking for one line , just wondering if awk has a quick solution or not . Certainly there are other ways also.

Since xargs will pass several file names to your AWK program, I suspect that your NR==1 test will only succeed for the first record of the first file in any batch. Perhaps a FNR==1 might be a better choice. And, if I were trying this, I'd want to print the FILENAME rather than the first record of the file. Perhaps you always keep the file name in there?

Originally Posted by debianD using exec is the same as using xargs as I did in my code .

This is not true. xargs executes the command once and puts the arguments all together, whereas -exec executes the command multiple times, once per each file. In this case NR == 1 will be okay.

find . -type f -perm -700 -exec awk 'NR == 1 && /perl/ {print FILENAME}' {} \;

If you use xargs, you need FNR == 1, as PTrenholme already explained:

find . -type f -perm -700 | xargs awk 'FNR == 1 && /perl/ {print FILENAME}'

May I also suggest adding 'nextfile' after your print statement, otherwise you will scan the full length of every file:

find . -type f -perm -700 | xargs awk 'FNR == 1 && /perl/ {print FILENAME}{nextfile}'

@colucix : Thanks , I will keep that in mind.

@grail : Does not the search is only restricted in the first line because of FNR == 1 , do I really need to do nextfile ?








Thanks.
D

FNR == 1 says that only to do what is between the curly braces for this expression. That means for all lines that do not match the expression it will do nothing, but this does not stop them from being read.