Split the line into next line after a matching pattern

Question

In the input file shown below, I am trying to find the string job_type and move the it to the next line if it matches job_type.

I tried this, but it is not working:

sed "s/[A-Z][a-z]*job_type:/\njob_type:/g"

Input:

    insert_job: VAU_vaultnotification_ertgvfg_job job_type: xxx 
    insert_job: VAU_vaultnotification_ertgvfg_frd job_type: yyy 
    insert_job: VAU_vaultnotification_ertgvfg_erb job_type: SXC 
     job_type: CMD
    insert_job: VAU_vaultnotification_ertgvfg_frd job_type: YUI 

Expected output:

insert_job: VAU_vaultnotification_ertgvfg_job 
job_type: xxx 
insert_job: VAU_vaultnotification_ertgvfg_frd 
job_type: yyy 
insert_job: VAU_vaultnotification_ertgvfg_erb 
job_type: SXC 
 job_type: CMD
insert_job: VAU_vaultnotification_ertgvfg_frd 
job_type: YUI 

Answer 1

Your requirements are not clear to me, but maybe you want:

sed 's/\([[:upper:]][[:alpha:]_]*\)[[:space:]]\{1,\}\(job_type:\)/\1\
\2/g'

Or:

perl -Mopen=locale -pe 's/\b\p{Lu}\w*\K\s+(job_type:)/\n$1/g'

Answer 2

I would do something similar to this, using GNU sed:

sed '/insert_job/s/job_type/\njob_type/'

The first part /insert_job/ matches that on a line before doing the substitution. If "insert_job" (or whatever regex you prefer) is not found, it won't do the the substitution.

Answer 3

Using GNU sed

$ sed '/ \+\(insert_job.*\)\(job_type:.*\)/s//\1\n\2/' input_file
insert_job: VAU_vaultnotification_ertgvfg_job
job_type: xxx
insert_job: VAU_vaultnotification_ertgvfg_frd
job_type: yyy
insert_job: VAU_vaultnotification_ertgvfg_erb
job_type: SXC
     job_type: CMD
insert_job: VAU_vaultnotification_ertgvfg_frd
job_type: YUI

Answer 4

Using Raku (formerly known as Perl_6)

Assuming lines-to-be-modified all contain at least 4 whitespace-separated "columns" (lines with fewer than 4 "columns" are returned unmodified):

raku -ne 'my @a = .words; @a.elems >= 4 ?? put @a[0..1], "\n", @a[2..*] !! put @a;' 

Above assumes no string pattern in the input, only the appropriate number of whitespace-separated words (i.e. columns). For matching lines, the first 2 columns are returned, then a newline, then the remaining columns (the OP provides Sample Input wherein the string job_type begins the 3rd column).

However, if you want to use a regex and search for string patterns, Raku has your back (using the subst operator). Below, Raku searches for 8 regex atoms between / ^ ... $ /, between the ^ start-of-string and $ end-of-string zero-width assertions (anchors). Once matched, Raku's <( capture marker is used to drop everything before job_type from the match. Then in the subst replacement half, a \n newline is ~ concatenated to the $/ match-variable and returned.

raku -ne '.subst(/^ insert_job\: \s .+? \s <(job_type\: \s \w* \s*? $/, {"\n"~$/} ).put;'

Sample Input:

insert_job: VAU_vaultnotification_ertgvfg_job job_type: xxx 
insert_job: VAU_vaultnotification_ertgvfg_frd job_type: yyy 
insert_job: VAU_vaultnotification_ertgvfg_erb job_type: SXC 
 job_type: CMD
insert_job: VAU_vaultnotification_ertgvfg_frd job_type: YUI

Sample Output (first code example, above):

insert_job: VAU_vaultnotification_ertgvfg_job
job_type: xxx
insert_job: VAU_vaultnotification_ertgvfg_frd
job_type: yyy
insert_job: VAU_vaultnotification_ertgvfg_erb
job_type: SXC
job_type: CMD
insert_job: VAU_vaultnotification_ertgvfg_frd
job_type: YUI

Note, the OP hasn't made clear if all output should be left-justified or not (zero leading-whitespace characters). The first answer above is left-justified, however it's easy to trim or even indent all lines that start with job_type in the output (for example, in the second answer using {"\n "~$/} in the replacement will indent).

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