I'm creating a javascript regex to match queries in a search engine string. I am having a problem with alternation. I have the following regex:
.*baidu.com.*[/?].*wd{1}=
I want to be able to match strings that have the string 'word' or 'qw' in addition to 'wd', but everything I try is unsuccessful. I thought I would be able to do something like the following:
.*baidu.com.*[/?].*[wd|word|qw]{1}=
but it does not seem to work.
replace [wd|word|qw] with (wd|word|qw) or (?:wd|word|qw).
[] denotes character sets, () denotes logical groupings.
Your expression:
.*baidu.com.*[/?].*[wd|word|qw]{1}=
does need a few changes, including [wd|word|qw] to (wd|word|qw) and getting rid of the redundant {1}, like so:
.*baidu.com.*[/?].*(wd|word|qw)=
But you also need to understand that the first part of your expression (.*baidu.com.*[/?].*) will match baidu.com hello what spelling/handle????????? or hbaidu-com/ or even something like lkas----jhdf lkja$##!3hdsfbaidugcomlaksjhdf.[($?lakshf, because the dot (.) matches any character except newlines... to match a literal dot, you have to escape it with a backslash (like \.)
There are several approaches you could take to match things in a URL, but we could help you more if you tell us what you are trying to do or accomplish - perhaps regex is not the best solution or (EDIT) only part of the best solution?
Related
Usually in my system i have the following string:
http://localhost/api/module
to find out the last part of the string (which is my route) ive been using the following:
/[^\/]+$/g
However there may be cases where my string looks abit different such as:
http://localhost/api/module/123
Using the above regex it would then return 123. When my String looks like this i know that the last part will always be a number. So my question is how do i make sure that i can always find the last string that does not contain a number?
This is what i came up with which really stricty matches only module for the following lines:
http://localhost/api/module
http://localhost/api/module/123
http://localhost/api/module/123a
http://localhost/api/module/a123
http://localhost/api/module/a123a
http://localhost/api/module/1a3
(?!\w*\d\w*)[^\/][a-zA-Z]+(?=\/\w*\d+\w*|$)
Explanation
I basically just extended your expression with negative lookahead and lookbehind which basically matches your expression given both of the following conditions is true:
(?!\w*\d\w*) May contain letters, but no digits
[a-zA-Z]+ Really, truly only consists of one or more letters (was needed)
(?=\/\d+|$)The match is either followed by a slash, followed by digits or the end of the line
See this in action in my sample at Regex101.
partYouWant = urlString.replace(/^.*\/([a-zA-Z]+)[\/0-9]*$/,'$1')
Here it is in action:
urlString="http://localhost/api/module/123"
urlString.replace(/^.*\/([a-zA-Z]+)[\/0-9]*$/,'$1')
-->"module"
urlString="http://localhost/api/module"
urlString.replace(/^.*\/([a-zA-Z]+)[\/0-9]*$/,'$1')
-->"module"
It just uses a capture expression to find the last non-numeric part.
It's going to do this too, not sure if this is what you want:
urlString="http://localhost/api/module/123/456"
urlString.replace(/^.*\/([a-zA-Z]+)[\/0-9]*$/,'$1')
-->"module"
/([0-9])\w+/g
That would select the numbers. You could use it remove that part from the url. What language are you using it for ?
I'm trying to write a regex to test for certain special characters, but I think I am overcomplicating things. The characters I need to check for are: &<>'"
My current regex looks like such:
/&<>'"/
Another I was trying is:
/\&\<\>\'\"/
Any tips for a beginner (in regards to regex)? Thanks!
You are looking for a character class:
/[&<>'"]/
In doing so, any of the characters in the square brackets will be matched.
The expression you were originally using, /&<>'"/, wasn't working as expected because it matches the characters in that sequential order. In other words, it would match a full string such as &<>'" but not &<.
I'm assuming that you want to be able to match all of the characters you listed, at one time.
If so, you should be able to combine a character set with the g (global-matching) flag, for your regex.
Here's what it could look like:
/[<>&'"]/g
Try /(\&|\<|>|\'|\")/
it depends on what regex system you use
While trying to submit a form a javascript regex validation always proves to be false for a string.
Regex:- ^(([a-zA-Z]:)|(\\\\{2}\\w+)\\$?)(\\\\(\\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 have a string of the format:
PATTERN or abcPATTERNdef or PATTERN or <<[TEST].[PATTERN]>>
I have created a regular expression (in JavaScript) as (^PATTERN)|([^\[]PATTERN) which is returning the first 3 occurrences while ignoring the last one, however I also seem to be getting the preceding character in the returned matches:
"PATTERN", "cPATTERN" and " PATTERN"
What I need are the matches without the preceding character.
I'm new to regular expressions and apologize if the question reflects that.
Any help would be greatly appreciated.
JavaScript doesn't have good support for lookbehinds, but if your format is going to remain the same, you can try something like this:
PATTERN(?!])
It matches all occurrences of "PATTERN" that are not followed by a ].
Let me know if this isn't the case, and I will update my answer to include the check for the opening [.
I'm a little confused as to what you're trying to do. Can you see if this works?
(^PATTERN)|[^\[](PATTERN)
Are you trying to find all the strings between the PATTERN's?
I'm trying to extract (potentially hyphenated) words from a string that have been marked with a '#'.
So for example from the string
var s = '#moo, #baa and #moo-baa are writing an email to a#bc.de'
I would like to return
['#moo', '#baa', '#moo-baa']
To make sure I don't capture the email address, I check that the group is preceded by a white-space character OR the beginning of the line:
s.match(/(^|\s)#(\w+[-\w+]*)/g)
This seems to do the trick, but it also captures the spaces, which I don't want:
["#moo", " #baa", " #moo-baa"]
Silencing the grouping like this
s.match(/(?:^|\s)#(\w+[-\w+]*)/g)
doesn't seem to work, it returns the same result as before. I also tried the opposite, and checked that there's no \w or \S in front of the group, but that also excludes the beginning of the line. I know I could simply trim the spaces off, but I'd really like to get this working with just a single 'match' call.
Anybody have a suggestion what I'm doing wrong? Thanks a lot in advance!!
[edit]
I also just noticed: Why is it returning the '#' symbols as well?! I mean, it's what I want, but why is it doing that? They're outside of the group, aren't they?
As far as I know, the whole match is returned from String.match when using the "g" modifier. Because, with the modifier you are telling the function to match the whole expression instead of creating numbered matches from sub-expressions (groups). A global match does not return groups, instead the groups are the matches themselves.
In your case, the regular expression you were looking for might be this:
'#moo, #baa and #moo-baa are writing an email to a#bc.de'.match(/(?!\b)(#[\w\-]+)/g);
You are looking for every "#" symbol that doesn't follow a word boundary. So there is no need for silent groups.
If you don't want to capture the space, don't put the \s inside of the parentheses. Anything inside the parentheses will be returned as part of the capture group.