How to determine if string contains substring with awk

Question

I have the following bash script segment that uses grep to find out whether input line contains certain pattern or not.

echo $input_line | grep -q "$pattern"
if [ $? -eq 0 ];
then
    # Input line contains pattern
    # Execute some bash commands
else
    # Input line DOES NOT contain pattern
    # Execute other bash commands
fi

I need to rewrite it with awk without grep. I tried several variants, but none of them works, for example, both of the following always print MATCH, no matter contains input line pattern or not.

echo $input_line | awk -v b="$pattern" '/pattern/ { print "MATCH" }'

echo $input_line | awk -v b="$pattern" '/$0 ~ pattern/ { print "MATCH" }'

Update:

Removing / as recommended resolved the described issue. Now I have the problem with returning result of awk command for usage in if statement. I wrote the following "ugly" snippet:

pattern="abc"  
input_line="xyz123 678q we uqa abc asd"  
RES=$(echo $input_line | awk -v pat="$pattern" '$0 ~ pat { print "1" }')  
if [ $RES -eq 1 ];  
then  
    echo Input line $input_line contains pattern $pattern  
else  
    echo Input line $input_line DOES NOT contains pattern $pattern  
fi  

input_line="tult uil7665 5444tu l098 7tu"  
RES=$(echo $input_line | awk -v pat="$pattern" '$0 ~ pat { print "1" }')  
if [ $RES -eq 1 ];  
then  
    echo Input line $input_line contains pattern $pattern  
else  
    echo Input line $input_line DOES NOT contains pattern $pattern  
fi

So I've got the following output, because in second case when input line doesn't contain pattern the variable RES is not set:

Input line xyz123 678q we uqa abc asd contains pattern abc  
./test.sh: line 21: [: -eq: unary operator expected  
Input line tult uil7665 5444tu l098 7tu DOES NOT contains pattern abc

Any ideas how to improve this "ugly" script snippet with if statement?
Thanks

Just take the `/` of your last attempt, `$0 ~ pattern {print...` — Eric Renouf, Jul 09 '17 at 11:52
When you're editting the question, please also include sample input and output to help show what "didn't work" means — Eric Renouf, Jul 09 '17 at 15:51

Raman Sailopal · Answer 1 · 2017-07-09T23:02:32.147

1

To use awk with a referenced variable for the pattern match you would need to do:

if [[ "$(input_line | awk '/'$pattern'/ { print "MATCH" }')" == "MATCH" ]]
then
         pattern matches
else
         No pattern matches
fi

I'm assuming here that the regular expression is a dynamic variable referenced by the variable pattern.

Place particular attention to the position of the single quotes in the awk statement.

edited Jul 09 '17 at 23:02

answered Jul 09 '17 at 12:17

Raman Sailopal

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Never do this, it's wrong in multiple fundamental shell and awk ways. – Ed Morton Jul 09 '17 at 16:41
If you care to explain? – Raman Sailopal Jul 09 '17 at 21:07
3

A couple of the things you are doing that you should never do are not quoting a shell variable and letting the value of a shell variable expand to become part pf the text of an awk script. See any shell intro text and http://cfajohnson.com/shell/cus-faq-2.html#Q24, or google. You're now additionally trying to execute a variable and so will instead pick up any command of the same name in your PATH. You're also using awk to to a regexp match that bash could do itself, and generating text from awk to test as a comparison when you should just use an exit status. The OPs code was closer to correct. – Ed Morton Jul 09 '17 at 23:48
nice and clean solution – mati kepa Jul 27 '21 at 07:09

How to determine if string contains substring with awk

1 Answers1