Bash Split strings on pattern match

Question

I have data such as

ID | Name | Grade
1|Robin Hood so hood | A
2|Gwen Stack DO Rick Stacy |C
3|Bling s/o Peter| A

where so is son of, DO is daughter of, S/O is son of.

I am gazed How I can arrive to

     ID | Name | Grade
    1|Robin Hood| A
    2|Gwen Stack  |C
    3|Bling| A

As at end I will be mailing these guys there grades where I am not willing to feed full string just Name

I was trying to use while loop, with awk which gives 2nd field into array then for loop to array which was uncessfull

Any SED or AWK method ??

If you just want ot remove so, DO, s/o you can do sed -r 's/ (so|DO|s\/o) / /'. If that's not what you want, please clarify your question — Alfwed
– Alfwed, Commented Nov 19, 2015 at 17:43
In the case where ID = 3, you have "Bling Peter", which are strings both before and after s/o. Do you really just want to have "Bling" there? — jas
– jas, Commented Nov 19, 2015 at 18:20

anubhava · Accepted Answer · 2015-11-19 18:16:24Z

4

You can use this sed command:

sed -i.bak 's~ [sSdD]/*[oO] [^|]*~~' file
ID | Name | Grade
1|Robin Hood|A
2|Gwen Stack|C
3|Bling|A

Or using awk:

awk -F ' [sSdD]/?[oO] [^|]*' '{print $1 $2}' file
ID | Name | Grade
1|Robin Hood|A
2|Gwen Stack|C
3|Bling|A

answered Nov 19, 2015 at 18:16

anubhava

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7 Comments

Ashish Over a year ago

I haven't understood [^|]*~~' this part of sed

anubhava Over a year ago

[^|]* matches 0 or more characters before a pipe

Ashish Over a year ago

I never used ~~ in regex expression [used only as condition but not expression. what is ~~

anubhava Over a year ago

~ is just an alternate sed delimiter. We could also use: 's# [sSdD]/*[oO] [^|]*##'

Ashish Over a year ago

+1 Upvote: I just realized few days ago if we accept answer in just few minutes others may not get time to comeup with there solution. So I am waiting for other potential ways of solving it

|

repzero · Accepted Answer · 2015-11-20 03:03:40Z

0

sed

sed -r 's/(so|DO|S[/]O)[^|]+([|])/\2/g

answered Nov 20, 2015 at 3:03

repzero

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Comments

Claes Wikner · Accepted Answer · 2015-11-20 21:13:21Z

0

awk '{sub(/s\/o Peter/,"")}{sub(/d \| A/,"\ |A")}NR>1{print $1,$2, $NF}NR==1' file
ID | Name | Grade
1|Robin Hood |A
2|Gwen Stack |C
3|Bling | A

answered Nov 20, 2015 at 21:13

Claes Wikner

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Comments

Michael Back · Accepted Answer · 2015-11-21 08:50:37Z

This answer should feel straightforward as it is using | as the field separator, and the sub is only working on the second field. It is also making sure that so, DO, and s/o are separate from names by making the regex in the sub take into account spaces on either side of the three:

awk 'BEGIN {OFS=FS="|"} sub(/ (s\/?o|DO) .*/, "", $2) + 1' file
ID | Name | Grade
1|Robin Hood| A
2|Gwen Stack|C
3|Bling| A

One other thing -- it looks like from the middle part of your question you may be wanting case insensitivity. If you have gawk, you can use IGNORECASE. Otherwise, here is the edited regex.

awk 'BEGIN {OFS=FS="|"} sub(/ ([Ss]\/?|[Dd])[Oo] .*/, "", $2) + 1' file

Here is the one if you want to accept case insensitive D/O as well.

awk 'BEGIN {OFS=FS="|"} sub(/ [SsDd]\/?[Oo] .*/, "", $2) + 1' file

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