如何将排序结果通过管道传递给grep？

Question

$ grep HxH 20170213.csv | awk -F',' '{print $13}' | cut -b 25-27 | sort -u
868
881
896
904
913
914
918
919
920

Question> How to pipe the sorted results and feed into grep ?

Now I have to do the following command manually.

grep 868 /tmp/aaa/*.csv
grep 881 /tmp/aaa/*.csv
...
grep 920 /tmp/aaa/*.csv

Answer 1

Since your output is numeric (output lines do not contain spaces), you can use a for loop with command substitution:

for id in $(grep HxH 20170213.csv | awk -F',' '{print $13}' \
            | cut -b 25-27 | sort -u); do
    grep $id /tmp/aaa/*.csv
done

Another option is to use xargs :

grep HxH 20170213.csv | awk -F',' '{print $13}' | cut -b 25-27 | sort -u \
    | xargs -n1 grep /tmp/aaa/*.csv -e

The xargs variant requires one to jump through a couple hoops to get right:

• by default xargs would stick more than one pattern to the same grep , which is prevented using -n1 ;
• xargs specifies the stdin contents as the last argument in the command line, which is a problem because grep expects pattern then file name. Fortunately, grep PATTERN FILES... can be spelled as grep FILES... -e PATTERN , which is why grep must be followed by -e .