I have the following possible string:
'', or '4.', or '*.4' or '4.35'
all the above format are valid, others are all invalid.
basically, if I don't care the digit or word character, this is what I used in PHP for the validation:
else if ( !ereg('^\*|.*\..*$',$bl_objver) )
Now, I would like to add some clientside validation, so I just translate it into javascript:
var ver_reg = new RegExp("^\*|.*\..*$");
if (ver_reg.test(obj_ver) == false)
but firebug always shows some error, like: "invalid quantifier |...*$" etc..
any suggestions?
(I'm not convinced your expression is correct, but for the moment just going with what you have.)
Using the RegExp object, you need to escape the slashes:
var ver_reg = new RegExp("^\\*|.*\\..*$");
Alternatively you can use regex literal notation:
var ver_reg = /^\*|.*\..*$/;
That answers your direct question, but...
As for the expression, well, what you definitely want to correct is the start/end anchors each applying to one side of the alternation.
i.e. you're saying <this>|<that> where <this> is ^\* and <that> is .*\..*$
What you want is ^(?:<this>|<that>)$ to ensure the start/end markers are not part of the alternatives (but using ?: since we're not capturing the group).
So /^(?:\*|.*\..*)$/ using the second example above - this fix would also need applying to the PHP version (which can use the same syntax).
I'd also question your use of . instead of \w or [^.] or similar, but without knowing what you're actually doing, I can't say for sure what makes most sense.
Hope this helps! :)
Related
I'm new to Regular expressions and can't seem to find out how I have to solve this:
I need a regular expressions that "allows" only numbers, letters and /. I wrote this:
/[^a-zA-Z0-9/]/g
I think it's possible to strip the first / off, but don't know how.
so #/register/step1 becomes register/step1
Who knows how I could get this result?
Thanks!
You can use a non-global match, if the pattern is contiguous in the string:
var rx=/(([a-zA-Z0-9]+\/*)+)/;
var s='#/register/step1';
var s1=(s.match(rx) || [])[0];
alert(s1)>>> returned value: (String) "register/step1"
"/register/step1".match(/[a-zA-Z0-9][a-zA-Z0-9/]*/); // ["register/step1"]
\w is Equivalent to [^A-Za-z0-9_], so:
"/register/step1".match(/\w[\w/]*/); // ["register/step1"]
edit: Don't know why i didn't suggest this first, but if you're simply enforcing the pattern rather than replacing, you could just replace that slash (if it exists) before checking the pattern, using strpos(), substr(), or something similar. If you are using a preg_replace() already, then you should look at the examples on the function docs, they are quite relevant
i have this regex code
/^(https?:\/\/+[\w\-]+\.[\w\-]+)/i
it works but there is a problem
you NEED http:// in the url for it to validate, and what i am making, the user will not want to add http:// to the url they want to just have example.com, if its possible i need it to work weather it has http:// or not
i don't know how to make my own regex, and ive searched but cannot find a one that does what i need, unless im just not looking in the right place. (Google :P)
Don't bother with regex. Use parse_url function.
You can just make it optional
/^((?:https?:\/\/+)?[\w\-]+\.[\w\-]+)/i
The (?:) around the part you don't want to have is a non capturing group, the ? afterwards makes it optional.
I'm not sure what the + after the second slash is good for, it says at least one of the preceding character. That means it allows also stuff like http://////////.
I hope you are aware, that this regex is far from matching valid URLs.
For example it will match stuff like
http://////////------------.-
or at least
http://N.O
^ after this position you can write what you want and it will match valid.
Here on Regexr you can see what your regex is matching.
See Purple Coder's answer for a probably better solution.
/^((https?:\/\/+)?[\w-]+.[\w-]+)/i
I'm using this :
// Validate that the string contains at least a dot .
var filterWebsite = /^([a-zA-Z0-9:_\.\-/])+\.([a-zA-Z0-9_\.\-/])+$/;
How can I match everything with a PHP regular expression? I tried: /[.\r\n]*/, but it isn't working. Any ideas? Thanks.
This is for a method I made for a PHP class to parse e-mails:
public function getHeader($headerName) {
preg_match('/[\r\n]' . $headerName . '[:][ ](.+)[\r\n][^ \t]/Uis', "\n" . ltrim($this->originalMessage), $matches); return preg_replace('/[\r\n]*/', '', $matches[1]);
}
/.*/s (see perl's docs). The s option means (quoting from that URL):
Treat string as single line. (Make . match a newline)
I assume, based on your inclusion of \n and \r above, that you want to match across multiple lines. In this case, use:
/.*/s
(note the explicit /s modifier, that is, change . to match any character whatsoever, even a newline, which it normally would not match.)
See http://www.perl.com/doc/manual/html/pod/perlre.html
Why do you want to match everything? There's no point in using it as a condition because it's always true. If you want to capture the text you don't need a regex to do it because you just use the entire string. If you're trying to get around taint-checking, then shame on you (and ask a separate question about doing that right).
Note that we have a bit of the XY Problem here. You have some task X in mind, and think Y is part of the solution. You ask about Y but never tell us X. It's hard to answer your real question when we don't know what you are trying to do. :)
What about /.*/s?
In a character class ( the [] ), . just means period.
Does /[\.\r\n]+/ do what you want?
This kludge has also worked for me before:
my $abstract_text = /Abstract:([\s\S]+?)\nReferences/m;
It's useful if you want to capture patterns with arbitrary text included or intervening between multiple captures.
im new to regular expressions in php.
I have some data in which some of the values are stored as zero(0).What i want to do is to replace them with '-'. I dont know which value will get zero as my database table gets updated daily thats why i have to place that replace thing on all the data.
$r_val=preg_replace('/(0)/','-',$r_val);
The code im using is replacing all the zeroes that it finds for eg. it is even replacing zero from 104.67,giving the output 1-4.56 which is wrong. i want that data where value is exact zero that must be replaced by '-' not every zero that it encounter.
Can anyone please help!!
Example of the values that $r_val is having :-
10.31,
391.05,
113393,
15.31,
1000 etc.
This depends alot on how your data is formatted inside $r_val, but a good place to start would be to try:
$r_val = preg_replace('/(?<!\.)\b0\b(?!\.)/', '-', $r_val);
Where \b is a 0-length character representing the start or end of a 'word'.
Strange as it may sound, but the Perl regex documentation is actually really good for explaining the regex part of the preg_* functions, since Perl is where the functionality is actually implemented.
Again, it would be more than helpful if you could supply an example of what the $r_val string really looks like.
Note that \b matches at word boundaries, which would also turn a string like "0.75" into "-.75". Not a desirable result, I guess.
Whilst the other answer does work, it seems overly complex to me. I think you need only to use the ^ and $ chars either side of 0.
$r_val = preg_replace('/^0+$/', '-', $r_val);
^ indicates the regex should match from the beginning of the line.
$ indicates the regex should match to the end of the line.
+ means match this pattern 1 or more times
I altered the minus sign to it's html code equivalent too. Paranoid, yes, but we are dealing with numbers after all, so I though throwing a raw minus sign in there might not be the best idea.
Why not just do this?
if ( $r_val == 0 )
$r_val = '-';
You do not need to use a regex for this. In fact, I'd advise against doing so for performance reasons. The operation above is approximately 20x faster than the regex solution.
Also, the PHP manual advises against using regexes for simple replacements:
If you don't need fancy replacing rules (like regular expressions), you should always use this function instead of ereg_replace() or preg_replace().
http://us.php.net/manual/en/function.str-replace.php
Hope that helps!
I wonder if you anyone can construct a regular expression that can detect if a person searches for something like "site:cnn.com" or "site:www.globe.com.ph/". I've been having the most difficult time figuring it out. Thanks a lot in advance!
Edit: Sorry forgot to mention my script is in PHP.
Ok, for input into an arbitary text field, something as simple as the following will work:
\bsite:(\S+)
where the parentheses will capture whatever site/domain they're trying to search. It won't verify it as valid, but validating urls/domains is complex and there are many easily googlable regexes for doing that, for instance, there's one here.
What are you matching against? A referer url?
Assuming you're matching against a referer url that looks like this:
http://www.google.com/search?client=safari&rls=en-us&q=whatever+site:foo.com&ie=UTF-8&oe=UTF-8
A regex like this should do the trick:
\bsite(?:\:|%3[aA])(?:(?!(?:%20|\+|&|$)).)+
Notes:
The colon after 'site' can either be unencoded or it can be percent encoded. Most user agents will leave it unencoded (which I believe is actually contrary to the standard), but this will handle both
I assumed the site:... url would be right-bounded by the equivalent of a space character, end of field (&) or end of string ($)
I didn't assume x-www-form-urlencoded encoding (spaces == '+') or spaces encoded with percent encoding (space == %20). This will handle both
The (?:...) is a non-capturing group. (?!...) is a negative lookahead.
no it's not for a referrer url. My php script basically spits out information about a domain (e.g. backlinks, pagerank etc) and I need that regex so it will know what the user is searching for. If the user enters something that doesn't match the regex, it does a regular web search instead.
If this is all you are trying to do, I guess I'd take the more simple approach and just do:
$entry = $_REQUEST['q'];
$tokens = split(':', trim($entry));
if (1 < count($tokens) && strtolower($tokens[0]) == 'site')
$site = $tokens[1];