how to check if I find a pattern with sed in loop

mierdatuti · 09-10-2020, 12:28 PM

Hi,
I have these code, to find values of a array in a file. The script only search in some lines number that the are in other file

Code:

  


while read -r Line; do
        
        echo "Line $Line"


      for i in "${patrones[@]}"
          do
        


            if [[ -n $(sed "${Line}!d;/'"${i}"'/I!d" $rutaFich/titles) ]]; then
                    echo "find it pattern $i en la linea $Line"
                   exit 0
            else
                   echo "No encontrado el patron $i en linea $Line"
             fi

       done

done < $rutaFich/differences

But not works. If find a pattern always goes to the else clause.
Any help please?
Thanks

pan64 · 09-10-2020, 12:32 PM

since we have no any idea about the content of $rutaFich/differences or "${patrones[@]}" or $rutaFich/titles hard to say anything.
would be also nice to know what should be the condition to that if.

mierdatuti · 09-10-2020, 03:16 PM

Quote:

Originally Posted by pan64

since we have no any idea about the content of $rutaFich/differences or "${patrones[@]}" or $rutaFich/titles hard to say anything.
would be also nice to know what should be the condition to that if.

Sorry. I'm trying to explain better.

$rutaFich/differences have line number where I search with sed (I get these line numbers of other code). For example

cat $rutaFich/differences
4

"${patrones[@]}" is array with all paterns that I must find with sed

Titles is text file where I must find (in these example at line 4).

So, in this example I must find in line number 4, all patterns of my array and if I find some pattern I say : "find it pattern $i en la linea $Line"

Thanks!

berndbausch · 09-10-2020, 03:57 PM

This command

Code:

sed "${Line}!d;/'"${i}"'/I!d"

first outputs line $Line, then outputs lines that contain $i enclosed in single quotes. E.g. if $i has the value mypattern, the line is printed if it contains 'mypattern', with quote characters. Is that the intention?

A potential problem is the fact that ! is a special character. You should escape it with a backslash \!.

A simpler(?) solution:

Code:

sed -n ${Line}p $rutaFich/titles | grep -i "'$i'"

or, if the single quotes are not desired

Code:

sed -n ${Line}p $rutaFich/titles | grep -i "$i"

and to test for presence

Code:

if sed -n ${Line}p $rutaFich/titles | grep -i -q "'$i'"
then echo pattern \'$i\' is in line $Line
else echo not found
fi

pan64 · 09-11-2020, 02:45 AM

yes, you can try to print the result of both:

Code:

sed "${Line}!d" $rutaFich/titles
sed "/'"${i}"'/I!d" $rutaFich/titles

to see what are they doing exactly. I would say the two sed commands will delete everything, most probably because of the pattern or the second sed is not exactly what you want (see post #4).
From the other hand running the first sed in a loop is just waste of time and cpu, but probably it is not that important in your case.

For readability probably I would use awk

Code:

while read -r Line; do
        
        echo "Line $Line"


      for i in "${patrones[@]}"
          do
            found=$(awk -v Line=$Line -v pattern="$i" ' NR != Line { next } $0~pattern ' $rutaFich/titles)
            if [[ -n $found ]]; then
                    echo "find it pattern $i en la linea $Line"
                   exit 0
            else
                   echo "No encontrado el patron $i en linea $Line"
             fi

       done

done < $rutaFich/differences

Not tested, just an idea. It will still read the file $rutaFich/titles in loop, so still better:

Code:

while read -r Line; do
        
        echo "Line $Line"
        data="$(sed -n ${Line}p $rutaFich/titles)"

      for i in "${patrones[@]}"
          do
            if [[ $data =~ $i ]]; then
                    echo "find it pattern $i en la linea $Line"
                   exit 0
            else
                   echo "No encontrado el patron $i en linea $Line"
             fi

       done

done < $rutaFich/differences

or something similar