Replace the first occurrence with sed

Question

As the example below, I want to keep only the word before the first 'John'.

However, the pattern I applied seems to replace John from the end to the head. So I need to call sed twice.

How could I find the correct way?

PATTERN="I am John, you are John also"
OUTPUT=$( echo "$PATTERN" | sed -r "s/(^.*)([ \t]*John[ ,\t]*)(.*$)/\1/" )
echo "$OUTPUT"
OUTPUT=$( echo "$OUTPUT" | sed -r "s/(^.*)([ \t]*John[ ,\t]*)(.*$)/\1/" )
echo "$OUTPUT"

My expectation is only call sed one time. Since if "John" appears several times it will be a trouble.

By the procedure above, it will generate output as:

Firstly it matches & trims the word after the final John; then the first John.

I am John, you are

I am

I want to execute one time and get

I am

Try [`awk -F'[[:blank:]]*John[[:blank:]]*' '{ print $1 }'`](http://rextester.com/GWJ61030) — Wiktor Stribiżew, May 11 '18 at 07:25
Don't use `ALL_UPPERCASE` variables in the shell. Those tend to be used for system stuff (e.g. `HOME`) or have a special meaning to the shell itself (e.g. `PATH`, `RANDOM`). — melpomene, May 11 '18 at 07:31
@chang jc, please mention always expected output in your post too. — RavinderSingh13, May 11 '18 at 07:38
@chang, I'd suggest to look up about greediness.. `.*` will try to match as much as possible and then backtrack from end if needed... https://stackoverflow.com/questions/5319840/greedy-vs-reluctant-vs-possessive-quantifiers might help.. — Sundeep, May 11 '18 at 07:54

RavinderSingh13 · Answer 1 · 2018-05-11T07:32:49.843

2

Following sed may help you on same.

echo "I am John, you are John also" | sed 's/ John.*//'

Or with variables.

pattern="I am John, you are John also"
output=$(echo "$pattern" | sed 's/John.*//')

edited May 11 '18 at 07:32

answered May 11 '18 at 07:28

RavinderSingh13

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1

Simple and accurate ++ – anubhava May 11 '18 at 07:30
2

`sed 's/ John.*//'` would be better – anubhava May 11 '18 at 07:32
It does not return only the first word – Arount May 11 '18 at 07:32
@RavinderSingh13 `echo "I am John, you are John also" | sed 's/ John.*//'` => `I am`, it's two words – Arount May 11 '18 at 07:34
@Sundeep "_ I want to keep only the word before the first 'John'._", well seems clear to me - edit, ok since he edited, it's not anymore :p – Arount May 11 '18 at 07:35
@Arount, I believe this one will work for OP as per OP's output. – RavinderSingh13 May 11 '18 at 07:37
why sed 's/ John.*//' match the first but sed -r "s/(^.*)([ \t]*John[ ,\t]*)(.*$)/\1/" ) matches the final? I cannot understand... – chang jc May 11 '18 at 07:40
1

@changjc, simple, if you DO NOT want the part in anywhere of your script after first occurrence of john then why to save it unnecessary in memory of `sed`. – RavinderSingh13 May 11 '18 at 07:42

score 1 · Answer 2 · edited May 11 '18 at 08:18

Another way of doing it is to use the grep command in Perl mode:

echo "I am John, you are John also" | grep -oP '^(?:(?!John).)*';
I am 
#there will be a space at the end
echo "I am John, you are John also" | grep -oP '^(?:(?!John).)*(?=\s)';
I am
#there is no space at the end

Regex explanations:

^(?:(?!John).)*

This will accept all characters from the beginning of the lines until it reaches the first John.

Regex demo

score 0 · Answer 3 · answered May 11 '18 at 07:38

0

Awk solution:

s="I am John, you are John also and there is another John there"
awk '{ sub(/[[:space:]]+John.*/, "") }1' <<<"$s"

The output:

I am

answered May 11 '18 at 07:38

RomanPerekhrest

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I paid attention to OP's input that has only a single space :) – anubhava May 11 '18 at 09:48
@anubhava, We professionals should tend to unified solutions to prevent that often-used comments like *"This won't work if .... or this won't work in case of"*. The difference is that the posters need the solution for theirselfs but we provide a solution for the community. That's all – RomanPerekhrest May 11 '18 at 09:53

Replace the first occurrence with sed

3 Answers3