Ah, OK ! So, if we consider that there were 2 separate tasks and starting with your exact input text, below :

ladgdfg_jfdggs@yertynno.com.vn:1234556:1654473467:53687091200
jegfa@nkcore.com:jeryr55673:1757810123:53687091200
ghyfm36@rocgfhmmail.com:mintgg18509:11767409268:53687091200
v7cifgthhy@yasbco.com.tw:512454215:380488:53687091200

The following regex S/R :

• SEARCH (?-s)^.+?:.+?\K:.+

• REPLACE Leave EMPTY

• Click on the Replace All button, exclusively

should leave you with your expected result :

ladgdfg_jfdggs@yertynno.com.vn:1234556
jegfa@nkcore.com:jeryr55673
ghyfm36@rocgfhmmail.com:mintgg18509
v7cifgthhy@yasbco.com.tw:512454215

Cheers,

guy038

for my question: this is ANOTHER txt, then different. but anyway you code work great, my intent was that! very very many thanks for your help!
Regards

Nice ! I even succeded to merge your two kinds of text, by adding a third alternative in the search regex !

So, starting with the input text, below, where the e-mail adresses may be :

• Preceded with any chars ( asda, 12345 …), followed with, at least, one space char OR beginning the line

• Followed with, at least, one field, determined by a colon separator char

bla bla blah
bla bla blah
bla bla blah
asda ladgdfg_jfdggs@yertynno.com.vn:1234556:1654473467:53687091200
bla bla blah
bla bla blah
jegfa@nkcore.com:jeryr55673:1757810123:53687091200
bla bla blah
bla bla blah
12345      ghyfm36@rocgfhmmail.com:mintgg18509:0000000000000:1111111111:55555555555555:777777777777777777:99999999999
bla bla blah
bla bla blah
bla bla blah
asda           v7cifgthhy@yasbco.com.tw:512454215:380488:53687091200

Use the following regex S/R :

• SEARCH (?-s)^(?!\h*[\w.\x20]+@).*\R?|^.*?\x20+|.+?:.+?\K:.+

• REPLACE Leave EMPTY

• Of course, the Regular expression search mode is selected and the Wrap around option is ticked

• You must use the Replace All button, exclusively ( Not the Replace button ! )

We get the following output text :

ladgdfg_jfdggs@yertynno.com.vn:1234556
jegfa@nkcore.com:jeryr55673
ghyfm36@rocgfhmmail.com:mintgg18509
v7cifgthhy@yasbco.com.tw:512454215

Notes :

This regular expression is composed of 3 alternatives :

• The first part (?-s)^(?!\h*[\w.\x20]+@).*\R?, which have been explained previouly. If deletes any entire line, which does not contain a @ char, after a non-null range of words chars and possible dot and/or space chars

• The second part ^.*?\x20+ which deletes all text between beginning of line and a non-null range of space characters, if that line contains an e-mail address, further on

• The third part .+?:.+?\K:.+ deletes, after the e-mail adress, from the second : char found till the last standard char of the current scanned line

• Note, in this last part, the use of the lazy quantifier +? to identify all standard chars between the first and the second : chars and the \K syntax before the second : char, which resets the overall search and just grabs all the remaining chars of current line, from the second :, included

Remark :

I strongly advice you to learn the regular expresions world ! Of course, you’ll have to study for 1/2 months, but it’s really worth it ;-)). To begin with, consult this FAQ :

https://community.notepad-plus-plus.org/topic/15765/faq-desk-where-to-find-regex-documentation

Best Regards,

guy038

What have you tried?
What did you get?
What didn’t work about it?

ladgdfg_jfdggs@yertynno.com.vn:1234556:1654473467:53687091200
jegfa@nkcore.com:jeryr55673:1757810123:53687091200
ghyfm36@rocgfhmmail.com:mintgg18509:11767409268:53687091200
v7cifgthhy@yasbco.com.tw:512454215:380488:53687091200

i need have like this:

code_text
ladgdfg_jfdggs@yertynno.com.vn:1234556
jegfa@nkcore.com:jeryr55673
ghyfm36@rocgfhmmail.com:mintgg18509
v7cifgthhy@yasbco.com.tw:512454215

i want delete all :number:number

No problem ! If we assume that there are no space char in e-mail addressses ( generally the case ! ) and further on, in current scanned line, we just need to add an alternative which, implicitly, looks for lines with an e-mail address and deletes from beginning of current line till a non-null range of space characters. Thus, the following regex S/R :

SEARCH (?-s)^(?!\h*[\w.\x20]+@).*\R?|^.*?\x20+

REPLACE Leave EMPTY

And you’ll obtain your expected result :

pamfhdjosa.mardedyly@ydefgo.com:cloggr4trr
cartgrtgrg@gfgbil.com:1r5yyr78115
jatrhtrh@grthil.com:mrthhrhmm17
lathmhjm@hhhjail.com:p456456inha

Best Regards,

guy038

Ok ! So, assuming your short example, below :

.: data :.
=================================
asda pamfhdjosa.mardedyly@ydefgo.com:cloggr4trr
=================================
 
ubba:/my-wallets.zip                               379912     2019-08-07T02:52:11+02:00
.: data :.
=================================
asda cartgrtgrg@gfgbil.com:1r5yyr78115
Pry Statinf: Kree
=================================
 
ubba:/crypto.zip                                   380547     2019-08-07T02:50:07+02:00
.: data :.
=================================
asda jatrhtrh@grthil.com:mrthhrhmm17
Pry Statinf: Kree
=================================
 
ubba:/wallets.zip                                  379807     2019-08-07T02:58:33+02:00
.: data :.
=================================
asda lathmhjm@hhhjail.com:p456456inha
Pry Statinf: Kree

The following regex S/R :

SEARCH (?-s)^(?!\h*[\w.\x20]+@).*\R?

REPLACE Leave EMPTY

gives this output text :

asda pamfhdjosa.mardedyly@ydefgo.com:cloggr4trr
asda cartgrtgrg@gfgbil.com:1r5yyr78115
asda jatrhtrh@grthil.com:mrthhrhmm17
asda lathmhjm@hhhjail.com:p456456inha

Are you expecting this result ?

I simply suppose that the non-null part of an e-mail address, located before the @ char, may contain :

• Any word character

• A dot symbol

• A space character

Best Regards,

guy038

.: data :.
=================================
asda pamfhdjosa.mardedyly@ydefgo.com:cloggr4trr
=================================
 
ubba:/my-wallets.zip                               379912     2019-08-07T02:52:11+02:00
.: data :.
=================================
asda cartgrtgrg@gfgbil.com:1r5yyr78115
Pry Statinf: Kree
=================================
 
ubba:/crypto.zip                                   380547     2019-08-07T02:50:07+02:00
.: data :.
=================================
asda jatrhtrh@grthil.com:mrthhrhmm17
Pry Statinf: Kree
=================================
 
ubba:/wallets.zip                                  379807     2019-08-07T02:58:33+02:00
.: data :.
=================================
asda lathmhjm@hhhjail.com:p456456inha
Pry Statinf: Kree

i want delete all text and line except
mail.address@mail.com:xxxxxxxx

with your string not work for me.
work with (?-s)^(?!\h*[\w.]+@).
but i have this result:

 pamfhdjosa.mardedyly@ydefgo.com:cloggr4trr





 cartgrtgrg@gfgbil.com:1r5yyr78115






 jatrhtrh@grthil.com:mrthhrhmm17






 lathmhjm@hhhjail.com:p456456inha

empty line between emails and also one space before any mail address.
I hope it helps me to complete the correct string!

Well, it’s just up to you ! All your previous e-mail addresses did not contain any dot before the @ char. Thus, I supposed that there were only word characters !

With , your last example name.gress@mail.com, of course, the regex S/R should become :

SEARCH (?-s)^(?!\h*[\w.]+@).+\R

REPLACE Leave EMPTY

Then, you said :

now i would like to add part of string to delete the rest of the text in my txt

I’m sorry but it’s really not clear to my mind !

Please, give us a sample of your input text, with various e-mail adresses and show us the output text you are expecting to, from your present input example !

Just a hint : when replying, type your input and output text as code, using the symbol </> , as below :

See you later,

Cheers,

guy038

now i would like to add part of string to delete the rest of the text in my txt…

bt some mail is of the type:
name.gress@mail.com
and the first part of the email is deleted (in this specific case the term “name”)…
how to leave full email address?

Hmm. This forum doesn’t exist for someone else to do your work for you. Some hints about how a feature of Notepad++ works, yes, but really, don’t expect to say “I need this very specific thing” and expect someone to hand it to you.

You’re expected to try things on your own, and present your (failed) results here and then you’ll probably receive help on just that aspect.

But wholesale problem statement and solution provided…nope.

second method it relates only to a specific email address, instead I have several, all different…

So, apparently, you want to keep all lines beginning with an e-mail address, don’t you ?

In that case, here is a regex S/R to delete any line, which does not begin with an e-mail address, with possible leading blank chars :

SEARCH (?-s)^(?!\h*\w+@).+\R

REPLACE Leave Empty

Notes :

• The part (?!\h*\w+@) is a negative look-ahead structure that search for the literal @ char, preceded by word chars and possible blank chars ( Space or tabulation chars ) This structure returns true and valids the overall regex only when an e-mail address cannot be found

• In that case, the part .+ grabs, from beginning of lines ( ^ ), all standard characters, of the current line scanned

• Finally, the part \R matches all types of line-breaks ( \r\n for Windows files, \n for Unix files and \r for Mac files )

• These entire lines are deleted, as the replace*ent zone is empty

Now, here is the method to select specific e-mail addresses, from the remaining lines and copy them in a new document.

Let’s suppose that you want to search for all the occurrences of the exact address ccc@cc.cns:mssgahsk :

• Open the Mark dialog ( Search > Mark... menu item )

• First, tick the option Bookmark line, if necessary

• Click on the Clear All marks button ( IMPORTANT )

• Un-tick the Purge for each search option, if necessary

• Select the Normal search mode ( IMPORTANT )

• Tick the Match case option, preferably

• Tick the Wrap around option

• SEARCH ccc@cc.cns:mssgahsk

• Click on the Mark All button

• You may change the Find what : contents and repeat the last 2 points, in order to bookmark other specific addresses, as well !

• Now, select the Search > Bookmark > Copy Bookmarked Lines menu option

• Finally, open a new document ( Ctrl + N )

• And paste all the bookmarked lines, in this new tab ( Ctrl + V )

Best Regards,

guy038

Ah, OK, I see !

Now, in your combined.txt file, if we consider the general case, below :

.....
.....
A aaaa@aa.aaa:aaaaaaa
.....
.....
B aaaa@aa.aaa:kkkkkkk
.....
.....
C aaaa@aa.aaa:zzzzzzz
.....
.....
D xxxx@xx.xxx:aaaaaaa
.....
.....
E xxxx@xx.xxx:kkkkkkk
.....
.....
F xxxx@xx.xxx:zzzzzzz
.....
.....
G zzzz@zz.zzz:aaaaaaa
.....
.....
H zzzz@zz.zzz:kkkkkkk
.....
.....
I zzzz@zz.zzz:zzzzzzz
.....
.....

• Firstly, which line, from A to I, do you want to keep ?

• Secondly, do you want to delete any text, of the remaining lines, after the second colon : included ?

Cheers,

guy038

this command join all txt content into new one txt. and this it’s ok.
now i want purge all content.
on my new combined.txt i want delete all content except a line that contains:
data: xxxx@xx.xxx:kkkkkkk
this value data:xxx… repeats itself and it’s always different.
how to purge combined.txt ??

Here is a solution, which involves regular expressions !

• First copy your input text ( Ctrl + A then Ctrl + C )

• Open a new tab ( Ctrl + N )

• Paste your text ( Ctrl + V )

• Open the Replace dialog ( Ctrl + H )

• SEARCH (?-s)^(\h*\w+@.+?:.+?):.+

• REPLACE \1

• Tick, preferably, the Wrap around option

• Select the Regular expression serach mode

• Click on the Replace All button

Et voilà !

Notes :

• First, the part (?-s) means that the regex dot symbol . will match only a single standard character( Not a line-break )

• Then the part ^ is the default zero-length assertion, matching any beginning of line location

• The part \h*\w+@ part looks for possible horizontal blank chars \h*, followed with any non-null range of word chars \w+, followed with the literal char @

• Then, the part .+?: finds the shortest non-null range of chars till a colon :

• Now, the part .+? matches the shortest non-null range of chars, till …

• The last part :.+, which represents the second : of each line, followed with all the remaining chars of current scanned line

• As the global part \h*\w+@.+?:.+? is embedded in parentheses, it’s stored as group 1, which will be re-written as \1, only, in the replacement zone

Best Regards,

guy038