Can't get result from grep in bash loop

Question

I'm trying to get info from log files using grep + tail -1:

#!/bin/bash
# declare an array called array and define 3 vales
array=( 00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 )
array2=( 00 30 )
for i in "${array[@]}"
do
        for j in "${array2[@]}"
        do
                 output=$(grep filtered /home/pavel/hasoffers_api/logs/tabatoo2_20180627$i$j* | tail -1)
                 echo /home/pavel/hasoffers_api/logs/tabatoo2_20180627$i$j*
                 echo "$output"
        done
done

this outputs just empty string, but I definitely know that phrase "filtered" is in the logs.

What can be the problem? I tried several variations of the same script but it just returns empty string.

Thanks!

I think the script looks ok. The only explanation I see is that the string "filtered" is not in the logs or there's no files named ".../tabatoo2_201806270000*". Without log files I/we can't reproduce the problem. — KamilCuk, Jun 29 '18 at 08:42
Also see [How to use Shellcheck](https://github.com/koalaman/shellcheck), [How to debug a bash script?](https://unix.stackexchange.com/q/155551/56041) (U&L.SE), [How to debug a bash script?](https://stackoverflow.com/q/951336/608639) (SO), [How to debug bash script?](https://askubuntu.com/q/21136) (AskU), [Debugging Bash scripts](http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_02_03.html), etc. — jww, Jun 29 '18 at 10:21
Thank you for comments, the script is actually working well, the phrase was different, my bad. Thanks! — Павел Иванов, Jun 29 '18 at 10:32

score 1 · Accepted Answer · answered Jun 29 '18 at 09:29

In this kind of assignment of a output value you should always redirect standard error (stderr) to standard out put (stdout). In your case if grep filtered /home/pavel/hasoffers_api/logs/tabatoo2_20180627$i$j* fails the error output will not be stored in variable output hence it will be difficult for you to debug the program. Change your script like below where error message will also be assigned to variable output.

output=$(grep filtered /home/pavel/hasoffers_api/logs/tabatoo2_20180627$i$j* 2>&1 | tail -1)

Now if search string is not available echo $output will print a empty string and if log file is not available it will print the error message

Hope this will help you.

score -1 · Answer 2 · answered Jun 29 '18 at 08:59

-1

For start maybe try to test if the file you get by building string "tabatoo2_20180627$i$j*" even exists.

answered Jun 29 '18 at 08:59

Jakub Pinkas

49
3

Can't get result from grep in bash loop

2 Answers2