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 [...].
Related
For example, in Unix, a backslash (\) is a common escape character. So to escape a full stop (.) in a regular expression, one does this:
\.
But with % encoding URL parameters, we have an escape character, %, and a control code, so an ampersand (&) doesn't become:
%&
Instead, it becomes:
%26
Any reason why? Seems to just make things more complicated, on the face of it, when we could just have one escape character and a mechanism to escape itself where necessary:
%%
Then it'd be:
simpler to remember; we just need to know which characters to escape, not which to escape and what to escape them to
encoding-agnostic, as we wouldn't be sending an ASCII or Unicode representation explicitly, we'd just be sending them in the encoding the rest of the URL is going in
easy to write an encoder: s/[!\*'();:#&=+$,/?#\[\] "%-\.<>\\^_`{|}~]/%&/g (untested!)
better because we could switch to using \ as an escape character, and life would be simpler and it'd be summer all year long
I might be getting carried away now. Someone shoot me down? :)
EDIT: replaced two uses of "delimiter" with "escape character".
Percent encoding happens not only to escape delimiters, but also so that you can transport bytes that are not allowed inside URIs (such as control characters or non-ASCII characters).
I guess it's because the URL Specification and specifically the HTTP part of it, only allow certain characters so to escape those one must replace them with characters that are allowed.
Also some allowed characters have special meanings like & and ? etc
so replacing them with a control code seems the only way to solve it
If you find it hard to recognize them, bookmark this page
http://www.w3schools.com/tags/ref_urlencode.asp
I am having allot of trouble finding a regex that will allow me to throw a error if the user tries to submit special characters especially "/" "\". I have a expression already that helps with other special characters but not the forward and backward slash Bonus: I don't want these ether but its not likely they will be entered. ~,!,#,#,$,%,^,&,*,().
I am currently using ^[\w{./\\(),'}+:?®©-]+$
The regex you have specifically allows / and \ (as well as ( and )). Change it like so:
^[\w{.,'}+:?®©-]+$
and keep removing any other characters you don't want to allow either.
In case you're wondering, the construct [...] is called a character class.
You can also use a negated character class like ^[^/\\()~!##$%^&*]*$ to allow any characters except /\()~!##$%^&*.
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.
I am using BizTalk 2009 and I have a flat file that is similar to the following
"0162892172","TIM ","LastName ","760 "," ","COMANCHE ","LN "
"0143248282","GEORGE ","LastName ","625 "," ","ENID ","AVE "
When I parse it and start mapping it I need to get rid of the quotation marks. I have marked the Wrap Character attribute for the schema as a quotation mark but it doesn't remove it when BizTalk is parsing the file.
Is there an easy way to specify the removal of a wrap character or am I going to have to run it through a script functiod every time? Also I would like to be able to remove the trailing spaces as well, if at all possible.
If you're still seeing the quotes after parsing, it likely means you set the wrap character property incorrectly. Are you sure you also set Wrap Character Type == Character?
As for the extra spaces, those will be hard to get rid of during parsing, because the quotes would specifically tell bts that they were intentional, so yeah, your best bet is to probably remove those during mapping or whatever.
this page seems to suggest that removing the trailing spaces can be done with:
> Pad Character Type = Hexadecimal
> Pad Character = 0x20
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]/