How to remove all white spaces from stdout input via AWK in linux

Question

I want to remove all the whitespaces from the ouptut of $date by using $awk but somehow it does not work as intended, what am I missing here?

date | awk '{gsub(/ *\| */,",")}1'

Sat 28 Mar 23:13:22 CET 2020

`date | awk '{gsub(/[[:space:]]+/,",")}1'`? See https://ideone.com/OlrHKP — Wiktor Stribiżew, Mar 28 '20 at 22:19
$ date | awk '{gsub(/[[:space:]]+/,",")}1' Sat 28 Mar 23:21:36 CET 2020 — Ragnar Lothbrok, Mar 28 '20 at 22:22
Please add sample input (no descriptions, no images, no links) and your desired output for that sample input to your question (no comment). — Cyrus, Mar 28 '20 at 22:23
@RagnarLothbrok No, see https://ideone.com/OlrHKP, result is `Sat,28,Mar,2020,10:12:13,PM,UTC` — Wiktor Stribiżew, Mar 28 '20 at 22:23
Weird, it doesn't work for me. Tried to use fish and zsh instead of bash but it's the same. — Ragnar Lothbrok, Mar 28 '20 at 22:29
`$ date | awk '{gsub(/[[:space:]]+/,",")}1' Sat 28 Mar 23:25:35 CET 2020` — Ragnar Lothbrok, Mar 28 '20 at 22:29
`date | awk '{gsub(/ /,",")}1'` Works, tr also works as workaround, thanks. Take care. — Ragnar Lothbrok, Mar 28 '20 at 22:33
Does this answer your question? [How to remove all white spaces from a given text file](https://stackoverflow.com/questions/9953448/how-to-remove-all-white-spaces-from-a-given-text-file) — Wiktor Stribiżew, Mar 28 '20 at 22:37
Seem, like you've been mostly nudged in the needed direction. Note: `awk` (not sure if all implementations) defaults to extended regex, hence use `|` and not `\|` for or. Note2: `*` means any number of occurrences incl. none, you probably wanted to say `+` meaning one ore more. — Ondrej K., Mar 28 '20 at 22:41
@OndrejK. all awk implementations use EREs and it's not just a default - they ONLY use EREs. — Ed Morton, Mar 28 '20 at 22:58
@Ragnar - do you REALLY want to convert the date to `Sat28Mar23:13:22CET2020`? If so, why? — Ed Morton, Mar 28 '20 at 22:59

score 1 · Answer 1 · answered Mar 28 '20 at 23:01

1

for this sed might be easier

$ date | sed 's/ //g'

SatMar2819:00:12EDT2020

answered Mar 28 '20 at 23:01

karakfa

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score -1 · Answer 2 · edited Mar 28 '20 at 23:41

-1

Solution with the underscore set as the separator:

date | awk '{gsub(/ /,"_")}1'

Workaround without awk(no separator):

date | tr -d '[:space:]'

edited Mar 28 '20 at 23:41

Dharman

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answered Mar 28 '20 at 22:43

Ragnar Lothbrok

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That awk "solution" doesn't make sense. If you don't want an underscore then don't type an underscore. – Ed Morton Mar 28 '20 at 22:57

How to remove all white spaces from stdout input via AWK in linux

2 Answers2