Conditional replacement of string fragment with sed (one-liner!)

Question

I am trying to process the result of diff operation with sed. This is my diff output, which I pipe into sed

3d2
< 12-03-22_JET_D_CL_UR_l4053_0061 True_Warning All 9 149261 
62a62
> 13-01-29_VUE_EPM3_v37_CSAV2_0370 True_Warning All 13 22125 
68c68
< 13-05-14_Regular_Front_0062 True_Warning All 13 123383 
---
> 13-05-14_Regular_Front_0062 True_Warning All 21 123383 
119c119
< CADS4_PMP363_20130202_DPH_069 True_Warning All 13 233405 
---
> CADS4_PMP363_20130202_DPH_069 True_Warning All 9 233409 
149c149
< CADS4_PMP363_20130315_Fujifilm_UK_186 True_Warning All 21 18611 
---
> CADS4_PMP363_20130315_Fujifilm_UK_186 True_Warning All 17 18615

I need to sort out the difference string and prepend the 3rd word in the strings with either "Old" or "New" - depending on the first character. My best effort so far is

diff new_jumps/true.jump old_jumps/true.jump | sed -n "/^[<>]/ s:\(.\) \(\S\+\) \(.\+\):\2 \1,\3: p" | replace ">" Old | replace "<" New

Which give me this result (exactly what I wanted).

12-03-22_JET_D_CL_UR_l4053_0061 New,True_Warning All 9 149261 
13-01-29_VUE_EPM3_v37_CSAV2_0370 Old,True_Warning All 13 22125 
13-05-14_Regular_Front_0062 New,True_Warning All 13 123383 
13-05-14_Regular_Front_0062 Old,True_Warning All 21 123383 
CADS4_PMP363_20130202_DPH_069 New,True_Warning All 13 233405 
CADS4_PMP363_20130202_DPH_069 Old,True_Warning All 9 233409 
CADS4_PMP363_20130315_Fujifilm_UK_186 New,True_Warning All 21 18611 
CADS4_PMP363_20130315_Fujifilm_UK_186 Old,True_Warning All 17 18615

My question is - how can I change conditional expression within sed one-liner that will eliminate the need to use replace afterwards? (I assume that it is possible) Thanks in advance

EDIT:

I know, I missed the option to chain sed expressions, but what I had in mind - is it possible to do it within one substitute operation?

score 2 · Answer 1 · answered Jul 03 '14 at 08:45

2

By adding more commands to sed using semicolon (;), like this:

diff new_jumps/true.jump old_jumps/true.jump | sed -n "/^[<>]/ s:\(.\) \(\S\+\) \(.\+\):\2 \1,\3:; s/</New/gp; s/>/Old/gp"

answered Jul 03 '14 at 08:45

Arnestig

2,285
18
30

condorwasabi · Accepted Answer · 2014-07-03T10:08:44.483

1

With awk I get a faster response. Try this:

diff new_jumps/true.jump old_jumps/true.jump | awk '{ if($1=="<" || $1==">"){($1=="<")?temp="New,":temp="Old,";print $2,temp$3,$4,$5}}'

Here's another solution suggested by Jidder:

awk '/^</{i="old,"}/^>/{i="new,"}i{$2=$2" "i;print;i=0}'

edited Jul 03 '14 at 10:08

answered Jul 03 '14 at 08:51

condorwasabi

616
6
19

Better solution inserted – condorwasabi Jul 03 '14 at 09:00
@volcano This leaves the `<` `>` on the front is that what you wanted ? – Jul 03 '14 at 09:26
Thanks, I will probably use it – volcano Jul 03 '14 at 09:27
@Jidder, well that is not a rocket science :) – volcano Jul 03 '14 at 09:27
@Volcano heres another way then :) `awk '/^{i="old,"}/^>/{i="new,"}i{$2=$2" "i;print;i=0}'` – Jul 03 '14 at 09:28
@Jidder, why don't you put it into an answer? – volcano Jul 03 '14 at 09:44
Sorry, I forgot to leave the "<" and ">" out of the output. I edited the answer. – condorwasabi Jul 03 '14 at 09:53
@Volcano, no point as its effectively the same as the above answer just a different way of doing it.condorwasabi can add it to his answer though if he/she wants :) – Jul 03 '14 at 10:01
@condorwasabi i suggested it :( – Jul 03 '14 at 10:09
One last request - how do I remove $1? setting it to null value does not remove the leading space – volcano Jul 03 '14 at 10:32
Do you want it to remove it from the output? If so, I already edited the answer. Now the output doesn't show "<" or ">" anymore (and there are no leading spaces in the output). – condorwasabi Jul 03 '14 at 10:37

score 0 · Answer 3 · edited May 23 '17 at 10:26

@volcano: here is a one-liner solution in sed, but relies in the interaction with the shell. IMHO if you want to have only one sed substitution command, you cannot avoid that behavior: you have to output to the shell the information of which first character has been seen on the line, the shell somewhat does the mapping to "Old" or "New" strings, and gives the result back to sed.

So the one-liner is not exactly a one-liner because we have to define things in the shell... ;)

replace() { if [ "$1" == ">" ] ; then echo -n "Old"; else echo -n "New" ; fi }
export -f replace
sed -n '/^[<>]/ s:\(.\) \(\S\+\) \(.\+\):echo "\2 $(replace \\\1),\3";:ep' yourfile

Please note that the e flag to the substitution command is a GNU sed extension, we use it here to avoid calling the shell explicitly. If you don't use GNU sed, you can simply replace the last line above by the following:

sed -n '/^[<>]/ s:\(.\) \(\S\+\) \(.\+\):echo "\2 $(replace \\\1),\3";:p' yourfile | bash

The solution I am giving here has been inspired by that other one.

Please also note that all this gymnastics is avoidable if you accept to replace your three-letter tokens "Old" and "New" by their initials, because then we can neatly use the y command to first act in a tr fashion, likewise:

sed -n '/^[<>]/ y/<>/ON/; s:\(.\) \(\S\+\) \(.\+\):\2 \1,\3:p' yourfile

Conditional replacement of string fragment with sed (one-liner!)

3 Answers3