Re: Regex to find out length - Regular Expressions Discussion Boards

Re: remove parenthese from function like string

Michael Eckstein18-Sep-22 5:55

18-Sep-22 5:55

By whatever1 and whatever2 I meant that these strings are arbitrary, not just exactly whatever1(2).
So it should work both for

__X(" (Joe)") John __X("Silver") Doe

with the result

" (Joe)" John "Silver" Doe

and

__X(" (Jane)") Michael J. __X("Frank Sinatra") Fox

with the result

" (Jane)" Michael J. "Frank Sinatra" Fox

I'd be happy even with replacing

__X(" (Joe)") John was there
for
" (Joe)" John was there

Re: remove parenthese from function like string

moxol18-Sep-22 8:28

moxol

18-Sep-22 8:28

Regex that I gave you will work whatever the string inside parentheses is.
It focuses on removing the __X(" and ").

You can use this one for any __X

string newExp = Regex.Replace(Regex.Replace(oldExp, @"__\w+?\(""", @""""), @"""\)", @"""");

modified 18-Sep-22 13:41pm.

Re: remove parenthese from function like string

Michael Eckstein18-Sep-22 9:01

Michael Eckstein

18-Sep-22 9:01

Is this possible to use in MS Visual Studio using 'Edit->Find and Replace->Replace in files' and checking 'Use regular expressions' and if so, then what should be in boxes 'Find what' and 'Replace with:' ?

In fact, it is a revert replace from

Find what: \("([^"]*)"(.*)
Replace with: (__X("$1")$2

modified 18-Sep-22 16:54pm.

Re: remove parenthese from function like string

moxol18-Sep-22 10:19

moxol

18-Sep-22 10:19

Yes, it can be done from Find and Replace in two steps.

Find: __\w+?$"
Replace: "

Find: "$
Replace: "

Re: remove parenthese from function like string

Michael Eckstein18-Sep-22 21:22

Michael Eckstein

18-Sep-22 21:22

Hi moxol, thank you for your idea using two steps! I used this:

Find: __X\("([^"]*)"(.*)
Replace: "$1"q!@tr@$2

repeated n-times until no other replacement was made

Then simply remove all q!@tr@) strings without regexp to remove the original closing parenthese, where q!@tr@ is any improbable string used as an identifier

Regex to find out length

RanCohen16-Aug-22 0:50

RanCohen

16-Aug-22 0:50

Lets say I want to "encrypt" the ID number 123456789 - add '*' to all the digits up to the last four - *****6789.
A starting point is to replace .*(....) with *$1. But it would produce *6789, which is too short.
How can I know how many chars the first .* matched?

Re: Regex to find out length

Richard Deeming16-Aug-22 1:14

Richard Deeming

16-Aug-22 1:14

It depends which language and framework you're using. But a regex is almost certainly the wrong tool for this job.

For example, in C# you might use something like this:

string input = "1234567890";

string output = string.Create(input.Length, input, static (buffer, input) =>
{
	int pivot = buffer.Length - 4;
	buffer[0..pivot].Fill('*');
	input.AsSpan()[pivot..].CopyTo(buffer[pivot..]);
});

"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer

Re: Regex to find out length

Graham Breach16-Aug-22 3:12

Graham Breach

16-Aug-22 3:12

You could use a zero-length lookahead.

JavaScript

"1234345242342356789".replaceAll(/\d(?=\d{4})/g, "*")

Non Zero Number

cad-one26-Jul-22 9:11

cad-one

26-Jul-22 9:11

Hi am looking for a RegEx that accepts a non-zero number only

accepted examples : -10, -12.50, 5, 200, 103.05 etc
not accepted : 0

Can you please help

Re: Non Zero Number

k505426-Jul-22 11:46

k5054

26-Jul-22 11:46

It depends to a certain extent on what "non zero numbers" means, exactly. Does it mean 0 only and exactly, or does it have to not match 0. 0.0 0.00 etc.

For the simple case of a just excluding a zero we can use the following perl regex:

^[-+]?([1-9]\d+|[1-9]+.|\d*.\d+|0?.\d+)$

I've anchored this on line start and end ^ ... $, but you might have other requirements. This will match e.g 1.2 -15 0.03 .12 15. and -0.3, and does not accept any of ., 0, 0., or 1.2.3. However it does accept 0.0, 0.00, etc, and I have not yet found a way to eliminate them from the pattern. There might be a way using lookahead/lookbehind matches Lookahead and Lookbehind Tutorial—Tips &Tricks, but so far I haven't been able to figure out how to tell the regex engine that the lookahead/behind should match the entire pattern. For example we can eliminate 0.0 by adding a negative lookbehind regex of (?<!0.0), but this also elminates 0.03, which is not what is wanted. Maybe someone else knows how to resolve this.

But maybe regex isn't the way to go. If you're working in Java or C# or C++ or ... , it's probably better to just read some input, check that the entire input parses as a number, convert to a number and then omit or complain if the value is zero - depending on what your use case is.

Keep Calm and Carry On

Re: Non Zero Number

cad-one26-Jul-22 12:09

cad-one

26-Jul-22 12:09

unfortunately this code is accepting 00, -0, 0.0,

I agree. I will look into handling it in JS.

Thanks for trying. Appreciate it.

Regex before match

moxol20-Jul-22 23:09

moxol

20-Jul-22 23:09

If I have string

1213 494 4 toobad boys 1234 4444

I want to match "bad boys" and everything before and after bad boys until word boundary \b.
Something like this

"\\b.+?bad boys.+?\\b"

This returns

1213 494 4 toobad boys

and I want it to match "toobad boys"
I know that the problem is "\\b.+?bad" which matches everything up to "bad boys", but instead it has to match everything before "bad boys" until word boundary \b.

Help?

Re: Regex before match

Peter_in_278021-Jul-22 2:31

Peter_in_2780

21-Jul-22 2:31

The problem is that your first \b is matching too early.
Try something like [A-Za-z]*bad boys[A-Za-z]*
In words, that will match "none or some alphabetic characters (no spaces!)" then "bad boys" then "none or some alphabetic characters"

Software rusts. Simon Stephenson, ca 1994. So does this signature. me, 2012

Re: Regex before match

moxol21-Jul-22 3:00

moxol

21-Jul-22 3:00

I can't place [A-Za-z] since there will be also a non-Latin characters.
It has to be .+

Re: Regex before match

jsc4221-Jul-22 3:56

jsc42

21-Jul-22 3:56

Does \w* or \pL* or \p{L}* or (\p{L}|\p{M})* or \p{Word}* do what you want? Different RegExps support different features.

See, for example, Regex Tutorial - Unicode Characters and Properties[^]

Re: Regex before match

moxol21-Jul-22 4:55

moxol

21-Jul-22 4:55

Yes, I think

\w*bad boys\w*

will do it.
It's MySQL Regex expression, and it seems that it works for both Latin and non-Latin characters.

Perl regex anchor merely matches at the position of the end of previous match period

marbooth18-Jul-22 7:23

marbooth

18-Jul-22 7:23

What is Perl regex anchor to merely match at the position where the previous match ended, as anchor \G matches at the position where the previous match ended, and also at the first match attempt the \G matches at the start of the string in the way \A does.

Re: Perl regex anchor merely matches at the position of the end of previous match period

marbooth18-Jul-22 18:18

marbooth

18-Jul-22 18:18

Grateful to God who guide me to the best path

Perl's way

PERL

(?!^)\G

PERL

(?!\A)\G

Re: Perl regex anchor merely matches at the position of the end of previous match period

RedDk20-Jul-22 7:23

RedDk

20-Jul-22 7:23

Sometimes things at CP are kinda hinky. Then SO usually leaves less stones unturned:
.net - In Regex, what is the difference between (?!^)\G and \G(?!^)? - Stack Overflow[^]
Stones unturned

Isn't ?= needless?

Member 1570882016-Jul-22 23:38

Member 15708820

16-Jul-22 23:38

Hello,

I am learning Regex.
I use Regex for a file-searching programme in Linux. (FSearch)

I do not understand, why we need ?= in Regex.
It looks for me like ?= is needless.

Why do I think so?

Here is an example:

abc(?=.*xyz)

But isn't this exactly the same like:

abc.*xyz

So actually my question is:
Can you give me an example, where ?= is really needed, because I can not get the same result without ?= ?

Would appreciate some answers.

Thank you.

modified 17-Jul-22 7:00am.

Re: Isn't ?= needless?

Dave Kreskowiak17-Jul-22 7:28

Dave Kreskowiak

17-Jul-22 7:28

No, they are not the same. You proved it yourself with your own experiment.

Is it useful? In your example, that depends on the context in which the expression is being used.

There's a good read on the "look ahead asseertion" at Regex Tutorial - Lookahead and Lookbehind Zero-Length Assertions[^].

Asking questions is a skill
CodeProject Forum Guidelines
Google: C# How to debug code
Seriously, go read these articles.

Dave Kreskowiak

Re: Isn't ?= needless?

Member 1570882019-Jul-22 22:48

Member 15708820

19-Jul-22 22:48

Thank you for the link.

In the meantime I think, I have found out, what the difference is between
abc(?=.*xyz) and abc.*xyz

We have 4 elements:
1. The regex search command string
2. The strings (e.g. the names of the files).
3. The strings that had a result (e.g. the file names that have a result).
4. The result for each string (The match within each string)

2-3 are the same for both abc(?=.*xyz) and abc.*xyz.
But 4. is different.

Re: Isn't ?= needless?

Dave Kreskowiak20-Jul-22 4:41

Dave Kreskowiak

20-Jul-22 4:41

If you want to experiment with your expression, Google for "Expresso 3". It's free and an indispensable tool for understanding regular expressions.

Asking questions is a skill
CodeProject Forum Guidelines
Google: C# How to debug code
Seriously, go read these articles.

Dave Kreskowiak

regex code assistance for entering 600 in a input field

Member 1568365422-Jun-22 5:13

Member 15683654

22-Jun-22 5:13

Hello I was trying to understand how to work with a regex expression for when a User enters starts typing in 600 in the input field, then populate the error message. What would be the regex code for that?
Currently I have `^[\w'\-,.][^0-9_!¡?÷?¿\/\\+=@#$%ˆ&*(){}|~<>;:[\]]{2,}$`

Re: regex code assistance for entering 600 in a input field

Richard MacCutchan22-Jun-22 6:39

Richard MacCutchan

22-Jun-22 6:39

Do you really need all that just to capture "600"?

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