How to get value after space in string with sed or awk?

Question

I have string lists:

./index.blade.php core/resources/views/admin/pages/setting/index.blade.php
./country-flag vendor/country-flag

I want to fetch only string after space, how should i write it in sed or awk?

I tried this but it didn't work

echo "./country-flag vendor/country-flag"  | awk '/^[[blank]]/{p=1}p'

Expected output:

vendor/country-flag

Thanks for showing your efforts, could you please do mention your expected output in your question, thank you. — RavinderSingh13, Mar 24 '21 at 06:28
maybe this `echo "./country-flag vendor/country-flag" | sed 's/.* //'` — Vishal Singh, Mar 24 '21 at 06:39
Does this answer your question? [bash: shortest way to get n-th column of output](https://stackoverflow.com/questions/7315587/bash-shortest-way-to-get-n-th-column-of-output) — Sundeep, Mar 24 '21 at 07:00

score 3 · Answer 1 · answered Mar 24 '21 at 06:58

3

By default the field seperator in awk is space, so we can refer the second field after the space with the help of $2

echo "./country-flag vendor/country-flag"  |  awk '{print $2}'

vendor/country-flag

answered Mar 24 '21 at 06:58

Shubham Sharma

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score 0 · Answer 2 · answered Mar 24 '21 at 08:53

I want to explain why

echo "./country-flag vendor/country-flag"  | awk '/^[[blank]]/{p=1}p'

failed to yield desired result. Firstly bracket expressions in awk have form of [:name:] so it should be [[:blank:]] not [[blank]]. Secondly after changing your awk code to /^[[:blank:]]/{p=1}p it will be testing if line starts with blank (space or tab). If so set p to 1. As printing of line depends on value of p it will print first line starting with space or tab and all following lines, as p value is never changed to anything but 1.

dawg · Answer 3 · 2021-03-24T15:24:14.067

You can do:

$ echo "./country-flag vendor/country-flag"  | awk '/[[:blank:]]/{print $2}'
vendor/country-flag

More typical in awk (since it, by default splits on [[:space:]] already) is to use the NF variable that denotes the number of fields in the current record:

$ echo "./country-flag vendor/country-flag"  | awk 'NF>1{print $2}'
vendor/country-flag

With sed you just delete the first field:

echo "./country-flag vendor/country-flag"  | sed -E 's/^[^[:space:]]*[[:space:]]*//'

If you want to print a line with no spaces unmodified but only after the first space if there is one:

echo "./country-flag vendor/country-flag
./country-flagvendor/country-flag"  |
sed -E '/^[^[:space:]]*$/n; s/^[^[:space:]]*[[:space:]]*//'

Prints:

vendor/country-flag
./country-flagvendor/country-flag

score 0 · Accepted Answer · answered Mar 24 '21 at 22:49

As suggested by Vishal Singh, this sed solution is most terse

echo "./country-flag vendor/country-flag" | sed 's/.* //'

vendor/country-flag

You could also use grep

echo "./country-flag vendor/country-flag" | grep -Po '\b[^ ]+$'

vendor/country-flag

The grep options say: using perl -P print the matching part only o

The regular expression says: after word boundary \b match anything that isn't space [^ ] one or more times + before the end of the line $

How to get value after space in string with sed or awk?

4 Answers4