What is RegEx for find quoted string having only close parenthesis at the end, like this :
"People)"
But not
"(People)"
Something like so: "[^(]+?\)" should fit the bill. You might also need to escape the quotation marks and the backslash as well, depending on what regex engine you are using.
Some details on how does this regex work are available here.
Can you try the following ?
String REGEX_TEST_STRING="\"People)\"";
System.out.println(REGEX_TEST_STRING.matches("\"P.*\)\""));
This code returns true for "People)" and false for "(People)"
HTH.
Related
I would like to append _OLD to the end of each strings that starts with SR_ but before the symbol ' or without it
For example my text is the following:
SR_Apple
When the 'SR_APPLE' rotten, we must discard it.
I would like the find and replace to do:
SR_Apple_OLD
When the 'SR_APPLE_OLD' rotten, we must discard it.
I have tried (SR_*)+$.*(?='\s) based on what i Learned but no luck so far. Please help. Thx in Adv
For simple cases you should be able to use
Find: (\bSR_[\w]+)
Replace: $1_OLD
(\bSR_.+?)('|$) and $1_OLD$2 could also work if the text after SR_ is more complex
The lookbehind you're using is only matching the string if it ends with a ' so it won't find the text not in quotes.
regex101 is a useful tool for debugging expressions
I have an example string:
*DataFromAdHoc(cbgv)
I would like to extract by RegEx:
DataFromAdHoc
So far I have figured something like that:
^[^#][^\(]+
But Unfortunately without positive result. Do you have maybe any idea why it's not working?
The regex you tried ^[^#][^\(]+ would match:
From the beginning of the string, it should not be a # ^[^#]
Then match until you encounter a parenthesis (I think you don't have to escape the parenthesis in a character class) [^\(]+
So this would match *DataFromAdHoc, including the *, because it is not a #.
What you could do, it capture this part [^\(]+ in a group like ([^(]+)
Then your regex would look like:
^[^#]([^(]+)
And the DataFromAdHoc would be in group 1.
Use ^\*(\w+)\(\w+\)$
It just gets everything between the * and the stuff in brackets.
Your answer may depend on which language you're running your regex in, please include that in your question.
This regex:
"REGION\\((.*?)\\)(.*?)END_REGION\\((.*?)\\)"
currently finds this info:
REGION(Test) my user typed this
END_REGION(Test)
I need it to instead find this info:
#region REGION my user typed this
#endregion END_REGION
I have tried:
"#region\\ (.*?)\\\n(.*?)#endregion\\ (.*?)\\\n"
It tells me that the pattern assignment has failed. Can someone please explain what I am doing wrong? I am new to Regex.
It seems the issue lies in the multiline \n. My recommendation is to use the modifier s to avoid multiline complexities like:
/#region\ \(.*?\)(.*?)\s#endregion\s\(.*?\)/s
Online Demo
s modifier "single line" makes the . to match all characters, including line breaks.
Try this:
#region(.*)?\n(.*)?#endregion(.*)?
This works for me when testing here: http://regexpal.com/
When using your original text and regex, the only thing that threw it off is that I did not have a new line at the end because your sample text didn't have one.
Constructing this regex doesn't fail using boost, even if you use the expanded modifier.
Your string to the compiler:
"#region\\ (.*?)\\\n(.*?)#endregion\\ (.*?)\\\n"
After parsed by compiler:
#region\ (.*?)\\n(.*?)#endregion\ (.*?)\\n
It looks like you have one too many escapes on the newline.
if you present the regex as expanded to boost, an un-escaped pound sign # is interpreted as a comment.
In that case, you need to escape the pound sign.
\#region\ (.*?)\\n(.*?)\#endregion\ (.*?)\\n
If you don't use the expanded modifier, then you don't need to escape the space characters.
Taking that tack, you can remove the escape on the space's, and fixing up the newline escapes, it looks like this raw (what gets passed to regex engine):
#region (.*?)\n(.*?)#endregion (.*?)\n
And like this as a source code string:
"#region (.*?)\\n(.*?)#endregion (.*?)\\n"
Your regular expression has an extra backslash when escaping the newline sequence \\\n, use \\s* instead. Also for the last capturing group you can use a greedy quantifier instead and remove the newline sequence.
#region\\ (.*?)\\s*(.*?)#endregion\\ (.*)
Compiled Demo
We are moving from phpbb to a simpler system and some of the bbcode needs converting, particularly the "quote" code. The current phpbb based quote code looks like this:
[quote="username":nw4lek0o]The quoted text[/quote:nw4lek0o]
and it needs to be simplified to this:
[quote=username]The quoted text[/quote]
So, basically two things: strip the double quotes from around the username, and strip the ID string from the opening and closing tag.
I'm not good at Regex. Help?
Use this regex:
\[quote="(.+?)":.+?\](.+?)\[/quote:.+?]
And replace it with:
[quote=$1]$2[/quote]
Demo: http://regex101.com/r/jL3xU2
find (?=\").|(?::\w+) replace with blank.
demo here : http://regex101.com/r/oN8fS2
I have data that looks like this:
[Shift]);[Ctrl][Ctrl+S][Left mouse-click][Backspace][Ctrl]
I want to find all [.*] tags that have the word mouse in them. Keeping in mind non-greedy specifiers, I tried this in Vim: \[.\{-}mouse.\{-}\], but this yielded this result,
[Shift]);[Ctrl][Ctrl+S][Left mouse-click]
Rather than just the desired,
[Left mouse-click]
Any ideas? Ultimately I need this pattern in Perl syntax as well, so if anyone has a solution in Perl that would also be appreciated.
\[[^]]*mouse[^[]*\]
That is, match a literal opening bracket, then any number of characters that aren't closing brackets, then "mouse," then any number of non-opening-brackets, and finally a literal closing bracket. Should be the same in Perl.
You can use the following regex:
\[[^\]]*mouse.*?\]