I am using the following regex
/[a-zA-Z0-9]+/i.test(value)
If I enter a space in the word, it passes.
I don't see where spaces are aloud in the regex, why is it passing?
You need to set the beginning and end bounderies so that the entire string must match the regular expression, otherwise it'll look for any match (which in this case is one or more of the characters specified).
Try this:
/^[a-zA-Z0-9]+$/i.test(value)
Because you haven't anchored it.
For these sorts of tests, it's typically safer to make sure you don't have the negated character class:
/[^a-zA-Z0-9]/
Related
I want to introduce a backspace character at the beginning of the line where a particular pattern is not found. Please advise.
Thanks,
Sagar
If you mean that you want to "remove the first character" then you can do this:
1) Write your regex pattern of what you want to find. For example, if you want to match Remove me at the start of the line, use:
^R\(emove me\)
Here we use ^ to assert the position to the start of the string. We also capture everything apart from the string we wish to keep in a backreference so it can be used later.
2) Replace the matches we find with whatever we grabbed in our backreference, in this case emove me, in effect backspacing the first character.
3) Make sure regular expression is checked and the cursor is at the start of the file, and hit Replace All.
Before
After:
I'm trying to create a RegEx Validator that checks the file extension in the FileUpload input against a list of allowed extensions (which are user specified). The following is as far as I have got, but I'm struggling with the syntax of the backward slash (\) that appears in the file path. Obviously the below is incorrect because it just escapes the (]) which causes an error. I would be really grateful for any help here. There seems to be a lot of examples out there, but none seem to work when I try them.
[a-zA-Z_-s0-9:\]+(.pdf|.PDF)$
To include a backslash in a character class, you need to use a specific escape sequence (\b):
[a-zA-Z_\s0-9:\b]+(\.pdf|\.PDF)$
Note that this might be a bit confusing, because outside of character classes, \b represents a word boundary. I also assumed, that -s was a typo and should have represented a white space. (otherwise it shouldn't compile, I think)
EDIT: You also need to escape the dots. Otherwise they will be meta character for any character but line breaks.
another EDIT: If you actually DO want to allow hyphens in filenames, you need to put the hyphen at the end of the character class. Like this:
[a-zA-Z_\s0-9:\b-]+(\.pdf|\.PDF)$
You probably want to use something like
[a-zA-Z_0-9\s:\\-]+\.[pP][dD][fF]$
which is same as
[\w\s:\\-]+\.[pP][dD][fF]$
because \w = [a-zA-Z0-9_]
Be sure character - to put as very first or very last item in the [...] list, otherwise it has special meaning for range or characters, such as a-z.
Also \ character has to be escaped by another slash, even inside of [...].
Sample text =
legacycard.ashx?save=false&iNo=3&No=555
Sample pattern =
^legacycard.ashx(.*)No=(\d+)
Want to grab group #2 value of "555" (the value of "No=" in the sample text)
In Expresso, this works, but in ASP.NET UrlRewrite, it is not catching.
Am I missing something?
Thanks!
I would do something along these lines:
^legacycard.ashx\?(?:.+&)*No=(\d+)
The \? will escape the question mark that normally separates the URL and the parameters, then you make sure that it will capture every parameter key/value pair (anything that ends on &) before the parameter you actually care about. Using ?: lets you specify that the set of brackets is non capturing (I'm assuming you won't need any of the data, has the potential to slightly speeds up your regex) and leaves you just 555 captured. The added benefit of this approach is that it'll work regardless of parameter order.
Just use this regex:
^legacycard\.ashx\?save=(false|true)&iNo=(?<ino>\d+)&No=(?<no>\d+)
Then Regex Replace with
${no}
Looks fine to me, your regex should match the entire string
legacycard.ashx?save=false&iNo=3&No=555
not sure why you have groups, but groups should also return
?save=false&iNo=3&
and
555
For good measure you should know that the . in legacycard.ashx is also interpreted by regex and you would normally escape it, in this case it dosen't matter because a single dot matches everything, also a dot. :)
Try this
^legacycard.ashx(\?No=|.*?&No=)(\d+)
this should work.
What is the difference between below two regular expressions
(.|[\r\n]){1,1500}
^.{1,1500}$
The first matches up-to-1500 chars, and the second (assuming you haven't set certain regex options) matches a first single line of up-to-1500 chars, with no newlines.
. does not match new lines.
The second one matches the first 1500 characteres of a line IF the line contains 1500 characters or less
First expression matches some <= 1500 characters of the file(or other source).
Second expression matches a entire line with charsNumber <= 1500.
. matches any character except \n newline.
If it's for use in a RegularExpressionValidator, you probably want to use this regex:
^[\s\S]{1,1500}$
This is because the regex may be run on either the server (.NET) or the client (JavaScript). In .NET regexes you can use the RegexOptions.Singleline flag (or its inline equivalent, (?s)) to make the dot match newlines, but JavaScript has no such mechanism.
[\s\S] matches any whitespace character or anything that's not a whitespace character--in other words, anything. It's the most popular idiom for matching anything including a newline in JavaScript; it's much, much more efficient than alternation-based approaches like (.|\n).
Note that you'll still need to use a RequiredFieldValidator if you don't want the user to leave the textbox empty.
I'm trying to complete a regular expression that will pull out matches based on their opening and closing characters, the closest I've gotten is
^(\[\[)[a-zA-Z.-_]+(\]\])
Which will match a string such as "[[word1]]" and bring me back all the matches if there is more than one, The problem is I want it to pick up matchs where there may be a space in so for example "[[word1 word2]]", now this will work if I add a space into my pattern above however this pops up a problem that it will only get one match for my entire string so for example if I have a string
"Hi [[Title]] [[Name]] [[surname]], How are you"
then the match will be [[Title]] [[Name]] [[surname]] rather than 3 matches [[Title]], [[Name]], [[surname]]. I'm sure I'm just a char or two away in the Regex but I'm stuck, How can I make it return the 3 matches.
Thanks
You just need to make you regex non-greedy by using a ? like:
^(\[\[)[a-zA-Z.-_ ]+?(\]\])
Also there is a bug in your regex. You've included - in the char class thinking of it as a literal hyphen. But - in a char class is a meta char. So it effectively will match all char between . (period) and _ (underscore). So you need to escape it as:
^(\[\[)[a-zA-Z.\-_ ]+?(\]\])
or you can put is in some other place in the regex so that it will not have things on both sides of it as:
^(\[\[)[a-zA-Z._ -]+?(\]\])
or
^(\[\[)[-a-zA-Z._ ]+?(\]\])
You need to turn off greedy matching. See these examples for different languages:
asp.net
java
javascript
You should use +? instead of +.
The one without the question mark will try to match as much as possible, while the one with the question mark as little as possible.
Another approach would be to use [^\]] as your characters instead of [a-zA-Z.-_]. That way, a match will never extend over your closing brackets.