[SOLVED] Delete everything until specific symbol

pedropt · 06-02-2018, 03:54 PM

Hi everyone , when it turns to sed and to delete some symbols i have a lot of difficulties to find a proper solution .

Here it is what i have :

Quote:

erjreglkrglkregerlkghreldsoigslçkgje[34423dfewkfhw]

and this is what i need as output :

Quote:

34423dfewkfhw

How can i do it with sed ?

RandomTroll · 06-02-2018, 04:23 PM

I can't guess your general problem, but

Code:

 echo $string | cut -d\[ -f2 | cut -d\] -f1

will work on the example string.

scasey · 06-02-2018, 05:21 PM

What have you tried?

Because the special characters [ and ] have meaning to sed, they must be escaped with \ to be taken literally.

Code:

sed 's/.*\[//;s/\]//'

pedropt · 06-02-2018, 05:38 PM

Thanks Scasey , your sentence works like a charm .

However , what can i do to delete everything ahead the last symbol ] ?

in this case :

Quote:

erjreglkrglkregerlkghreldsoigslçkgje[34423dfewkfhw]:dwfwe982350

Sed is not easy to work on some regex because like you told before , it will interpet them as a command .

syg00 · 06-02-2018, 05:47 PM

Nobody can write matching regex if you keep changing the data and requirements. You need to fully define the situation before you can start with the regex. For simple regex to work it needs consistent data - all the records must conform to a pattern.

You still haven't shown any evidence you'd made any effort yourself.

scasey · 06-02-2018, 05:49 PM

Sorry, you're really going to have to show some work.
Do you understand what I said about escaping special characters?
Have you read

Code:

man sed
sed info
man perlre

?
Have you searched for "regex syntax" on the web?

pedropt · 06-02-2018, 06:12 PM

Sorry , this is a specific script to check up the server logs , however after your filter i get : next in the lines , and i already figure out how to do it .

Code:

sed 's/.*\[//;s/\]//' <file | sed 's/\:.*$//'

Thanks for your help on this one , and you were right .
And yes , after checking sed manua i get :

'/REGEXP/'
This will select any line which matches the regular expression
REGEXP. If REGEXP itself includes any '/' characters, each must be
escaped by a backslash ('\').

keefaz · 06-02-2018, 06:38 PM

or

Code:

sed 's/.*\[\(.*\)].*/\1/' file
grep -Po '\[\K.*(?=])' file

syg00 · 06-02-2018, 08:26 PM

Better hope there's only one set of square brackets. And matched.
Potential corner cases abound.

We would all hope logs are consistent, but ya never know ...

MadeInGermany · 06-03-2018, 04:18 AM

You can simply include all trailing characters in the search. What matches is substituted

Code:

sed 's/.*\[//;s/\].*//' file

Here sed opens file itself. While <file lets the shell open it.
The .* is greedy, so the rightmost [ is found.
Sometimes you have two [ and want the first [ then look for "not [" characters before the [

Code:

sed 's/[^[]*\[//;s/\].*//' file

The same for the variant with back-reference

Code:

sed 's/[^[]*\[\([^]]*\)].*/\1/' file

It looks for "not [" characters before the [ then "not ]" characters befor the ]