Reputation: 5031
in my script i need to loop through lines in a file, once i find some specific line i need to save it to variable so later on i can use it outside the loop, i tried the following but it wont' work:
count=0
res=""
python my.py -p 12345 |
while IFS= read -r line
do
count=$((count+1))
if [ "$count" -eq 5 ]; then
res=`echo "$line" | xargs`
fi
done
echo "$res"
it output nothing, i also tried this,
res=""
... in the loop...
res=$res`echo "$line" | xargs`
still nothing. please help. thanks.
Update: Thanks for all the help. here is my final code:
res=python my.py -p 12345 | sed -n '5p' | xargs
Upvotes: 0
Views: 2483
Reputation: 5684
for finding a specific line in a file, have you considered using grep?
grep "thing I'm looking for" /path/to/my.file
this will output the lines that match the thing you're looking for. Moreover this can be piped to xargs as in your question.
If you need to look at a particularly numbered line of a file, consider using the head and tail commands (which can also be piped to grep).
cat /path/to/my.file | head -n5 | tail -n1 | grep "thing I'm looking for"
These commands take the first lines specified (in this case, 5 and 1 respectively) and only prints those out. Hopefully this will help you accomplish your task.
Happy coding! Leave a comment if you have any questions.
Upvotes: 1