Find all . between the start of the line and the first occurence : and find all " " between the start of the line and the first occurence :

Long-Manith SREY

Hi,

Here is the data I currently have (“before” data):

tokenized files: "toto.properties"
spring.flyway.tata: "classpath:/xxxx/oracle"
spring.flyway.toto: "classpath:/xxxx/oracle"
spring.flyway.tata: "EN"
spring.flyway.schemas: "OWNEN"
spring.flyway.titi.aper: "APLEN"
spring.flyway.titi.aplr: "AP"
spring.flyway.titi.user: "APL"
db.url: "jdbc:oracle:thin:@.fr.pop.intra:12232/1"

Here is how I would like that data to look (“after” data):

I have two needs
1-> i would like to find all . between the start of the line and the first occurence ":"
2->i would like to find all " " between the start of the line and the first occurence ":"

So the result shoule be this way

To accomplish this, I have tried using the following Find/Replace expressions and settings
(?:^|\G(?!^))[^.]\K.(?=.?: )
The problem is notepad++ give me also this result

Could you please help me to find the right expression?

Thank in advance

guy038

Hello, @long-manith-srey and All,

From your post, we can merge your two needs, saying :

From start of any line, I would like to find all space and dot characters which come before the first occurrence of the : symbol

If so, I think that the following regex does the expected job !

Move to or open your file
If necessary, move to its very beginning with the Ctrl + Home shortcut ( IMPORTANT because of the negative look-ahead (?!\A) )
Open the Find or Mark dialog ( Ctrl + F / Ctrl + M )
SEARCH or MARK (?-s)(^|(?!\A)\G)[^:\r\n]+?\K[ .]
Untick all options
Select the Regular expression search mode
Click on the Find All in Current Document or on the Mark All button

Of course, we’ll be able, to tune this regex, next time, in order to prevent from some matches or to get some additional matches, if any !

On the other hand, the FR regex [ .] could be expressed differently if you need replacement and separate characters for the space and dot chars !

Best Regards

guy038

P.S. :

Using the notation from this generic regex, we have :

BSR = ^

ESR = [^:\r\n]. Note that we could have used the ((?!:).)+? syntax, instead, but the former syntax is more simple !

FR = [ .]

So the alternate regex is (?-s)(^|(?!\A)\G)((?!:).)+?\K[ .]

Long-Manith SREY

@guy038
Thank you so much*It’s perfect :)