How to replace characters over certain number?

[LordFader]

Hi,

I have the following problem and I can’t get any further:

I want to truncate text lines like this after 55 characters (with spaces) and replace them with " …".

STOXX Europe 600 Optimised Personal & Household Goods EUR (Acc)
MSCI EM Socially Responsible (Dist)
STOXX Europe 600 Optimised Industrial Goods & Services EUR (Acc)

So that the result looks as follows:
STOXX Europe 600 Optimised Personal & Household Goods E …
MSCI EM Socially Responsible (Dist)
STOXX Europe 600 Optimised Industrial Goods & Services …

I have tried the following, but it does not work:
Highlights only the first 5 characters

^.{5}

Finds “only” the last paragraph, no concrete character limit

\h+\S+$

What must the command look like that it works?
(gladly with a short explanation what each operator does, then I understand the principle better)

[guy038]

Hello, @till-ehrich and All,

You said :

I want to truncate text lines like this after 55 characters

OK. Now, what about the 56th character ?

If the 56th char is a space char, I suppose that you expect this replacement :

1234567890123456789012345678901234567890123456789012345 …

But, if the 56th char is not a space char, like below :

1234567890123456789012345678901234567890123456789012345ABDEFG HIJKL MNO

Do you expect this OUTPUT :

1234567890123456789012345678901234567890123456789012345 …

or this OUTPUT :

1234567890123456789012345678901234567890123456789012345ABCDEFG …

Best Regards,

guy038

[LordFader]

@guy038
Whether there is a space or a character after the 55 character is not important. Then only the rest should be replaced by " …".

ABCDE FGHIJ KLMNO
ABCDE FGHIJKLMNO
ABCDE FGHIJ  ...
ABCDE FGHIJ ...

If there is a blank character, one would have with a simple solution then " ..." at the end, thus with 2 blank characters behind each other. But I can change that in a second step simply with search and replace.

[guy038]

Hi, @till-ehrich and All,

Ah…, OK. In this case, use the following regex S/R :

• Open the Replace dialog ( Ctrl + H )

• SEARCH (?-s).{55}\K.+

• REPLACE \x20\x{2026}

• Check the Wrap around box option, preferably

• Select the Regular expression search mode

• Click on the Replace All button ( Do not use the Replace button ! )

BR

guy038

[Alan Kilborn]

OP probably wants ... rather than \x{2026} in the replacement…

[LordFader]

@guy038 said in How to replace characters over certain number?:

(?-s).{55}\K.+

Thank you very much, everything worked great. Tidy lists please the users :-)