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regex - display all occurrences of match

 
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# 8  
Old 02-13-2009
Quote:
Originally Posted by danmauer
Code:
 
#!/usr/bin/nawk -f
 
BEGIN {
pat="AI_SQL\/[a-zA-Z_]+\.sql"
}
 
{
    while (match($0, pat)) {
       printf substr($0, RSTART, RLENGTH) OFS;
       $0=substr($0, RSTART+RLENGTH)
    }
}

you don't need a 'pat' in a BEGIN block - you're going to define on a command line when you call a script - unless you want to 'hard-wire' it in a script.....
# 9  
Old 02-13-2009
Thanks.Smilie
# 10  
Old 02-16-2009
I am trying to accomplish something similar. Will this work with awk too. We do not have nawk on our server.

When I run it from $ prompt, it seems to work with the 'foo' example. But when I write it to a file and execute as

test.awk -v pat='foo[0-9][0-9]*' myfile.txt , I get the error

ksh: test.awk: not found

test.awk
Code:
#! /usr/bin/awk -f
{
while (match($0, pat)) {
printf("%s\n", substr($0, RSTART, RLENGTH) OFS);
       $0=substr($0, RSTART+RLENGTH)
       }
}
Quote:
Originally Posted by vgersh99
you don't need a 'pat' in a BEGIN block - you're going to define on a command line when you call a script - unless you want to 'hard-wire' it in a script.....
# 11  
Old 02-16-2009
make sure you have execute premissions on the file.
# 12  
Old 02-16-2009
I do.. Actually the file has 777 permissions. Is by any chance awk vs nawk making the difference ?

Quote:
Originally Posted by danmauer
make sure you have execute premissions on the file.
# 13  
Old 02-16-2009
how do you run the script?
Do you have '.' in your PATH?
Specify an absolute/relative path to the script.
# 14  
Old 02-16-2009
sorry, but i'm not sure... still learning myself.
 
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