bash - variable storing multiple lines of file

Question

This is my code:

grep $to_check $forbidden >${dir}variants_of_interest;

cat ${dir}variants_of_interest | (while read line; do 
     #process ${line} and echo result
done;
)

Thank to grep I get lines of data that I then process separately in loop. I would like to use variable instead of using file variants_of_interest.

Reason for this is that I am afraid that writing to file thousands of time (and consequently reading from it) rapidly slows down computation, so I am hoping that avoiding writing to file could help. What do you think?

I have to do thousands of grep commands and variants_of_interest contains up to 10 lines only.

Thanks for your suggestions.

lynxlynxlynx · Accepted Answer · 2012-11-19 13:51:42Z

3

You can just make grep pass its output directly to the loop:

grep "$to_check" "$forbidden" | while read line; do 
  #process "${line}" and echo result
done

I removed the explicit subshell in your example, since it is already in a separate one due to the piping. Also don't forget to quote the $line variable to prevent whitespace expansion on use.

answered Nov 19, 2012 at 13:51

lynxlynxlynx

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2 Comments

Perlnika Over a year ago

Thank you very much, this is working solution. Because you and Lynch were just few seconds apart, I chose the answer randomly.

lynxlynxlynx Over a year ago

Ideally you should pick the best answer as the final one — these two are not identical.

Lynch · Accepted Answer · 2012-11-19 13:50:31Z

2

You dont have to write a file. Simply iterate over the result of grep:

grep $to_check $forbidden | (while read line; do 
     #process ${line} and echo result
done;
)

answered Nov 19, 2012 at 13:50

Lynch

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1 Comment

Perlnika Over a year ago

Thank you, this worked:) I ticked the answer randomly as you and lynxlynxlynx were few seconds apart.

potong · Accepted Answer · 2012-11-19 14:28:04Z

0

This might work for you:

OIFS="$IFS"; IFS=$'\n'; lines=($(grep $to_check $forbidden)); IFS="$OIFS"
for line in "${lines[@]}"; do echo $(process ${line}); done

The first line places the results of the grep into the variable array lines.

The second line processes the array lines placing each line into the variable line

answered Nov 19, 2012 at 14:28

potong

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