Regex using grep in bash gives unwanted output at the end

Question

I am iterating through a text file line by line in bash and printing out the word segments that match the relevant regex pattern. This process is done by accepting the text file as an argument. My code works fine but at the end, in the last line, an unwanted output gets printed. I am new to bash and I have no idea why this is happening. Shown below is my code:

#!/bin/bash
pat='[a-z][a-zA-Z0-9_]*\.png'
while IFS='' read -r line || [[ -n $line ]];do
   echo $line | grep -o -P $pat
done < "$1"

shown below is the output(the unwanted line is the highlighted one):

I want to get rid of this last line of unwanted output.

You may directly use `grep`: `grep -oE $pat "$1"`. Or directly `grep -oE $pat file` — Wiktor Stribiżew, Mar 13 '19 at 19:06
when you say "directly" does that mean that the while loop is not needed? — Suleka_28, Mar 13 '19 at 19:26
If you are extracting every single match from a file a loop is redundant. Note you are using `P` option but the regex is POSIX BRE compliant, you may omit `P`. Use `grep -o '[a-z][a-zA-Z0-9_]*\.png' file` or - to save matches in a file - `grep -o '[a-z][a-zA-Z0-9_]*\.png' file > newfile` — Wiktor Stribiżew, Mar 13 '19 at 19:28
@WiktorStribiżew I just need to list the files. So basically this would do? echo | grep -o $pat $1 — Suleka_28, Mar 13 '19 at 19:40
Is there a particular reason for putting $1 inside inverted commas? — Suleka_28, Mar 13 '19 at 19:45

score 0 · Answer 1 · answered Mar 13 '19 at 19:43

The problem is that you're running the script with sh instead of bash. [[ is a bash extension, it's not available in sh. So change

sh test2.sh summary.md

to

bash test2.sh summary.md

Or, since the script starts with #!/bin/bash, just use:

./test.sh summary.md

You should also quote all the variables in the script. See I just assigned a variable, but echo $variable shows something else

Regex using grep in bash gives unwanted output at the end

1 Answers1