Perhaps he hadn't. [Perhaps he didn't?]
Perhaps he didn't.
[Perhaps he hadn't?]
I want to match the bits in brackets; sentences that start with "Perhaps" and end with a question mark.
I thought this regex would work: Perhaps.*(?!=\.)\?
However, what I'm getting is this:
[Perhaps he hadn't. Perhaps he didn't?]
Perhaps he didn't.
[Perhaps he hadn't?]
Why is this? And how to fix it?
https://regexr.com/5dfhs
You may use a negated character class like this:
/Perhaps[^.?]*\?/g
RegEx Demo
To match complete word use:
/\bPerhaps[^.?]*\?/
And to avoid matching across lines use:
/\bPerhaps[^.?\r\n]*\?/
here [^.?] would match any character except . and ?
About your regex:
(?!=\.) is actually wrong syntax for a negative lookahead. It just means don't match if we have a literal = and . ahead.
Even if you correct it to use Perhaps.*(?!\.)\? it will still not work because (?!\.) will only be applied for matching ? and that will always succeed.
Though not recommended but if you really want to use a negative lookahead then use:
/Perhaps(?:(?!\.).)*\?/
Related
I am trying to combine:
^[a-zA-Z.][a-zA-Z'\\- .]*$
with
(\W|^)first\sname(\W|$)
which should check for the exact phrase, first name, if that is correct. It should match either the first regex OR the second exact match. I tried this, but appears invalid:
^(([a-zA-Z.][a-zA-Z'\\- .]*$)|((\W|^)first\sname(\W|$))
This is in javascript btw.
Combining regular expressions generally can be done simply in the following way:
Regex1 + Regex2 = (Regex1|Regex2)
^[a-zA-Z.][a-zA-Z'\\- .]*$
+ (\W|^)first\sname(\W|$) =
(^[a-zA-Z.][a-zA-Z'\\- .]*$|(\W|^)first\sname(\W|$))
Because some SO users have a hard time understand the math analogy, here's a full word explanation.
If you have a regex with content REGEX1 and a second regex with content REGEX2 and you want to combine them in the way that was described by OP in his question, a simple way to do this without optimization is the following.
(REGEX1|REGEX2)
Where you surround both regular expressions with parenthesis and divide the two with |.
Your regex would be the following:
(^[a-zA-Z.][a-zA-Z'\\- .]*$|(\W|^)first\sname(\W|$))
Your first regex has an error in it, though, that makes it invalid. Try this instead.
(^[a-zA-Z.][a-zA-Z'\- .]*$|(\W|^)first\sname(\W|$))
You had \\ in the second character class where you wanted \
The problem is that the first regex is messed up. You don't need to double escape characters. Therefore
\\-
Will match an ascii character between \(92) and (32). Remove one of the slashes.
Reference
While trying to submit a form a javascript regex validation always proves to be false for a string.
Regex:- ^(([a-zA-Z]:)|(\\\\{2}\\w+)\\$?)(\\\\(\\w[\\w].*))+(.jpeg|.JPEG|.jpg|.JPG)$
I have tried following strings against it
abc.jpg,
abc:.jpg,
a:.jpg,
a:asdas.jpg,
What string could possible match this regex ?
This regex won't match against anything because of that $? in the middle of the string.
Apparently using the optional modifier ? on the end string symbol $ is not correct (if you paste it on https://regex101.com/ it will give you an error indeed). If the javascript parser ignores the error and keeps the regex as it is this still means you are going to match an end string in the middle of a string which is supposed to continue.
Unescaped it was supposed to match a \$ (dollar symbol) but as it is written it won't work.
If you want your string to be accepted at any cost you can probably use Firebug or a similar developer tool and edit the string inside the javascript code (this, assuming there's no server side check too and assuming it's not wrong aswell). If you ignore the $? then a matching string will be \\\\w\\\\ww.jpg (but since the . is unescaped even \\\\w\\\\ww%jpg is a match)
Of course, I wrote this answer assuming the escaping is indeed the one you showed in the question. If you need to find a matching pattern for the correctly escaped one ^(([a-zA-Z]:)|(\\{2}\w+)\$?)(\\(\w[\w].*))+(\.jpeg|\.JPEG|\.jpg|\.JPG)$ then you can use this tool to find one http://fent.github.io/randexp.js/ (though it will find weird matches). A matching pattern is c:\zz.jpg
If you are just looking for a regular expression to match what you got there, go ahead and test this out:
(\w+:?\w*\.[jpe?gJPE?G]+,)
That should match exactly what you are looking for. Remove the optional comma at the end if you feel like it, of course.
If you remove escape level, the actual regex is
^(([a-zA-Z]:)|(\\{2}\w+)\$?)(\\(\w[\w].*))+(.jpeg|.JPEG|.jpg|.JPG)$
After ^start the first pipe (([a-zA-Z]:)|(\\{2}\w+)\$?) which matches an alpha followed by a colon or two backslashes followed by one or more word characters, followed by an optional literal $. There is some needless parenthesis used inside.
The second part (\\(\w[\w].*))+ matches a backslash, followed by two word characters \w[\w] which looks weird because it's equivalent to \w\w (don't need a character class for second \w). Followed by any amount of any character. This whole thing one or more times.
In the last part (.jpeg|.JPEG|.jpg|.JPG) one probably forgot to escape the dot for matching a literal. \. should be used. This part can be reduced to \.(JPE?G|jpe?g).
It would match something like
A:\12anything.JPEG
\\1$\anything.jpg
Play with it at regex101. A better readable could be
^([a-zA-Z]:|\\{2}\w+\$?)(\\\w{2}.*)+\.(jpe?g|JPE?G)$
Also read the explanation on regex101 to understand any pattern, it's helpful!
I am trying to build a RegEx in Javascript that does not match if a certain string appear. So I want to match this curly bracket { but only if in front of it is not the string else.
I am trying to do this ^ *[^else]* *{.*$ but in fact this doe not match if any character in elsestring appear, for example this does not match also this:
erai {
I want to match all the cases when { appear despite of this case else {.
Please could you help me. Here is my DEMO
You can use a negative lookahead. This is supported by JavaScript:
(?!\s*else).+ *({).*$|
DEMO
JavaScript RegEx doesn't support ifs but we can use a trick for it to work:
(?!RegExp)
That's the first part, if RegExp (which is a regex) doesn't appear, then we do the code after that:
.+ *({).*$
That's the RegEx we run. Broken does, it is:
.+ Match anything
* Until 0 - unlimited spaces
({) Capture the {
.*$ Match anything till the end
Now this won't work unless we add a | at the end, or an OR. This will trick it into working like an if statement
Debuggex Demo
What you are looking for is called negative lookbehind.
I have a regex, and a string that includes some matches for this regex. My regex handle all this matches like it is only one big match (of course I don't want such behaviour), let me show you an example:
My test string (sorry for scribble, but this doesn't matter):
sdfsd -dsf- sdfsdfssdfsfdsfsd -sdfsdf-
my regex in js code:
view.replace(/(\-(.+)\-)/g, '<span style="background-color:yellow">$1</span>');
my result:
sdfsd<span style="background-color:yellow">-dsf- sdfsdfssdfsfdsfsd -sdfsdf-</span>
As you can see each of this strings in the "-" must be enclosed in span, but there is only one span. How I can fix this? (honestly I don't want change my (.+) regex part, which I think might be a problem, but if there is no other way to do this, let me know).
In other words, result must be:
sdfsd<span style="background-color:yellow">-dsf-</span> sdfsdfssdfsfdsfsd <span style="background-color:yellow">-sdfsdf-</span>
Feel free to ask me in the comments, and thanks for your help.
honestly I don't want change my (.+) regex part, which I think might be a problem
Why not, it is actually the source of the problem, you can try the following regex which would work:
/(\-([^-]+)\-)/g
and if you think that dashes - can appear between - and - themselves then you can use the less efficient:
/(\-(.+?)\-)/g
+? causes a lazy match, or in other words after matching the initial -, then .+? matches a single character then it moves control to the following - which tries to match a dash, if it couldn't then .+? reads (consumes) another character from the input and so on until the following - is able to match.
You can try:
view.replace(/-([^-]+)-/g, '<span style="background-color:yellow">$1</span>');
I've been looking for this for hours, right now I've ended in a very ugly way (but working). I would like to find a reusable nice way to do this.
I've a string like this:
wanna[0].some[0].javascript
I would like to replace the last digit occurence between square brackets:
wanna[0].some[1].javascript
I've ended this (ugly) way:
myString.replace(/\d].javascript$/, "1].javascript")
which should be the best regex to match that?
myString.replace(/\d/, 1) // this should be for the first digit
myString.replace(/\d/g, 1) // this for every digit
I've read about negative look-ahead but I still didn't get if JS supports this.
Just use a negative lookahead to ascertain that there are no more brackets after the one you're matching:
var text = 'wanna[0].some[0].javascript';
text = text.replace(/\[\d](?!.*\[)/, '[1]');
Here's the fiddle: http://jsfiddle.net/bRkEP/
Not sure why you need lookahead. The answer you gave (though ugly) should work. As Crockford says of negative lookaheads in JavaScript: The Good Parts: 'This is not a good part.'
If the line you provided isn't working I imagine it's because you haven't escaped the square bracket. It should be:
myString.replace(/\d\].javascript$/, "1].javascript");
You could also do some capturing to make it easier on the eye.
I agree with the comment to your question that it seems like an odd problem to solve. Why are you changing the number in a string? This looks like something that would be better served with a number variable in the square brackets that you can then increment.