regex to replace the entire character in string - regex

the string may have characters like "abc # xyz 1234-4321", i need regex to replace the characters inside the quoted string with the another string using text replace action in final builder.
Thanks in advance ,Any help would be appreciable.

".*?" will find anything between two ".
The dot matches everything, while the * tells it that you want somewhere between zero and many occurences. You can then just replace it with "yourstring".
If you don't want to write "yourstring" you could also use a capture group "(.*?)" but honestly it's overkill for this.

Related

Which pattern (regex) should I use for splitting a litte text using ':' character as separator?

I need to split a text onto two strings using RegEx (I can not use anyother method).
I have a text like:
This text should be first string: This text should be second string
I need to split the above text onto two string by character ':'. Trying achieve that I have use next pattern [^:]* but this is not working. I have tryed another patterns but nothing works.
Any idea which pattern should be the appropiate?
Thank you.
The straightforward regex is:
([^:]*):([^:]*)
Access the two parts / strings with \1 and \2.
Test it here.

How in notepad++ find/replace text between slashes?

How to find the text between the second and fourth slashes in a path like /folder/subfolder-1/subfolder-2/subfolder-3? I’m trying to replace this with something like /folder/new-folder/subfolder-3.
The most important for me is to be able to find the part after the n-th slash.
I tried the regex /((.*?)/){3}, but it doesn’t work.
Using Match resetter \K meta-character you are able to do it in a simpler way.
Find:
/.*?/\K(.*?/){2}
Replace with:
new-folder/
One way you could to it is by using this string in the pattern to replace
(/.+?)(/.+?){2}(/\S+)
And use this one in your pattern to replace it with
$1/new-folder$3
From your string:
/folder/subfolder-1/subfolder-2/subfolder-3
(/.+?) will match /folder as $1
(/.+?){2} will match /subfolder-1/subfolder-2 as $2 (not used)
(/\S+) will match everything that isn't a space, in this case/subfolder-3 as $3
Leaving you room to insert your new-folder in-between.
How can I just mark till the slash?
Find what: (/[^/]+/)[^/]+/[^/]+
Replace with: $1new-folder
To find text between second and forth slash you can use the regex ^(/[^/]*/)([^/]*/[^/]*) then you can reference to the text between slashes with \2 when replacing the text.
To keep the text before the slashes you can enter something like \1myNewTextBetweenSlashes2and4.
In notepad++ Find by this:
(/[^/]+)(?:/[^/]+/[^/]+/)(.*)
And Replace by this:
\1\/new-folder/\2
Make sure that: .matches newline is not checked
{2} indicates 2 levels after first level will be repalced by new-folder
Find:
(\/.*?\/)(.*?\/){2}(.*)
Replace:
$1new-folder/$3
Demo: https://regex101.com/r/XIA3IN/3

How can I replace this data in between certain delimiter with Notepad ++?

I have a list of data in this format
0000000000000000|000|000|00000|000000|CITY|GA|123456|8001234567
I need to replace the last piece of data with the word N/A so there is no phone number in the list.
0000000000000000|000|000|00000|000000|CITY|GA|123456|N/A
Thank you for the assistance, much appreciated.
The simplest and fastest solution for that would be to search for
[^|\r\n]+$
and replacing all with N/A.
Explanation:
[^|\r\n]+ matches one or more characters except | or newlines, and $ makes sure that the match only occurs at the end of a line.
Do a find/replace, with the mode set to "Regular expression".
Find:
(.*)\|[0-9]*
Replace:
\1|N/A
If your phone numbers contain any non-numeric characters (such as periods, hyphens, spaces, etc.), then I would recommend the following adjustment to the regex given by #Bitwise:
(.*)\|(.*)$
Also, in Notepad++, the backreference syntax is not
\1
but rather
$1
which means your replace string will actually be
$1|N/A
You can use
(?!.*\|)(.+)
to mark the end of the line.
In Notepad++ you can use the search and replace (regex) function.

Notepad++ Regex - Find multiple characters and replace them with one character

I have a file with +20K lines and some strings have this structure:
,"/d/s/aaa.jpg","/e/_/bbb.jpg","/_/2/bbb.jpg" ....
and I want to replace them with:
,"/aaa.jpg", "/bbb.jpg","/bbb.jpg" ...
Can some one provide me a regex expression that will find those 5 leading characters and replace them with "/"?
Thank You in advance.
Use the following:
Find what: /[^/]+/[^/]+(/[^/]+\.jpg)
Replace with: $1
Edited:
The following:
\"\/[^/]+\/[^/]+\/
will match the "/d/s/ or "/e/_/ part of a string. You can test it here: http://regexpal.com/
Make sure to replace it with:
"/
to do more you would need capture groups (i.e. capturing some parts of the regex to reuse that in the substitution or manipulate it somehow)

Boost:regex_search - match string between brackets

given this part of a LogString:
... HTTP/1.1" 206 391402 "-" "AppleCoreMedia/1.0.0.8B117 (iPod; U; CPU
OS 4_1 like Mac OS X; de_de)"
I need to match the last substring that is inside the brackets to get the Remote System Info from the string.
I'm not really into regular expressions but from my understanding I have to escape the brackets like "\(" and "\)".
Now to match the whole string I thought I'd have to do it like:
\(\.*\) with the dot matching anything and "*" for specifying a random amount of dots.
But it does't work. Neither does (\.*) or (.*) as you would expect.
There must be a problem with opening and closing the brackets because I get a Seg Fault when trying to match the string this way.
Can anyone help me please?
Ok sorry for bothering you. This did the trick:
boost::regex f_sourceDir(conf.pString("filter_SourceFiles")),
f_string(conf.pString("filter_String")),
m_first(conf.pString("field0")),
m_second("\\(([^)]+)\\)\"$",boost::regex::perl),
m_third(conf.pString("field2")),
m_fourth(conf.pString("field3"));
Seems to be a problem with parsing the configuration, it can't handle such strings. I think I have to change the Config class...
Thx anyway!
Try this
\(.*\)
See it here on Regexr
You need only to escape the brackets, but not the .. If you escape the . then you will match a literal dot and this will not be found.
If you not want to have the brackets then here is a solution using a capturing group
\((.*)\)
See it here on Regexr
You find the part inside the brackets stored in the capturing group 1.
Just in case, there are two ways of handling a string with a regex: matching and searching. Matching either matches the whole string or fails, searching does not have to match the whole string. What you are doing is searching.
The expression you need using Perl syntax (boost::regex let's you choose the syntax) is something like:
\(([^)]+)\)"$
That is: find a substring of one or more non-right-bracket symbols within brackets followed by the trailing " and the end of the string.
In C++ you need to quote this expression:
boost::regex re("\\(([^)]+)\\)\"$", boost::regex::perl);