Match string followed by any character followed by float number

Question

I am trying to match bla = 0.05 and replace the number with 1234 in a file. Expected output is bla = 1234

Code I tried:

sed -i '' "s/\(bla\)\(.*\)\([-+]?[0-9]*\.?[0-9]*\)/#\1\21234/g" foo

Also, why do I sometimes need '' and sometimes not to call sed ?

see https://stackoverflow.com/questions/5694228/sed-in-place-flag-that-works-both-on-mac-bsd-and-linux for usage of `-i` between different versions of sed... — Sundeep, Oct 22 '16 at 15:55
for the problem you are trying to solve, can you add more sample inputs with expected output for clarity? one issue apparent is that you need to use `\?` instead of `?` — Sundeep, Oct 22 '16 at 15:58
also, `.*` would consume too much... `sed -i '' 's/$bla$$[^0-9+-]*$$[-+]\?[0-9]*\.\?[0-9]*$/#\1\21234/g' foo` might help — Sundeep, Oct 22 '16 at 16:01

score 1 · Accepted Answer · edited May 23 '17 at 12:00

1

$ echo 'bla = 0.05' > foo
$ cat foo 
bla = 0.05
$ sed 's/\(bla[^0-9.+-]*\)\([-+]\?[0-9]*\.\?[0-9]*\)/\11234/g' foo
bla = 1234

If extended regex option is available, either -E or -r

$ sed -E 's/(bla[^0-9.+-]*)([-+]?[0-9]*\.?[0-9]*)/\11234/g' foo
bla = 1234

See sed in-place flag for requirements of using -i flag between different sed versions

edited May 23 '17 at 12:00

Community

answered Oct 22 '16 at 16:18

Sundeep

the -E command worked. THank you very much. Im trying to figure out why my version did not work – Tuni Oct 22 '16 at 16:40
1

because `.*` is [greedy](https://stackoverflow.com/documentation/regex/429/greedy-and-lazy-patterns#t=201610221644401303712) and every pattern after that can be `0` or more... for example, try this simpler command for experimenting... `echo 'bla = 0.05' | sed 's/$bla$$.*$[0-9]*/\2/'` – Sundeep Oct 22 '16 at 16:44
1

also `echo 'bla = 0.05' | sed 's/bla$.*$$[0-9]*$/\1 abc \2/'` – Sundeep Oct 22 '16 at 16:46
i understand. If i want to match 'foo is left' whilst the string left could also be right, can i do it like this ?: 'sed -E -i '' "s/(foo[^[left]|^[right]]*)([left]|[right])/\1$ContSide/g" bar' – Tuni Oct 22 '16 at 17:17
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@Thunfisch not sure about that, would highly recommend you to go through this: https://stackoverflow.com/questions/22937618/reference-what-does-this-regex-mean – Sundeep Oct 23 '16 at 02:50

1 Answers1