I'm using ajax check function to check inserted category name which should be only alpha-numeric and also allowed space
I've used this function eregi_replace with the following regular expression [a-zA-Z0-9_]+
$check = eregi_replace('([a-zA-Z0-9_]+)', "", $catname);
But when i insert category name for example hello world it failed cause it does not accept space but if i write it as helloworld works so i understood that the error must be in the regular expression i'm using.
so what is the correct regular expression that filter any special characters and allow only for alpha-numeric and space.
Thanks a lot
A character class matching letters, numbers, the underscore and space would be
[\w ]
You should not be using any of the POSIX regular expression functions as they are now deprecated. Instead, use their superior counterparts from the PCRE suite.
Change your regular expression to:
([A-Za-z0-9_]+(?: +[A-Za-z0-9_]+)*)
I realize that it is not as straightforward as you might have hoped. Things to note:
The identifier must start with a non-space
If there are spaces, they should be between words and not matched at the end
?: is used to prevent an extra grouping in your expression, but is not required
The + after the space character allows multiple spaces between words. You can enforce a single space by removing it, but in some solutions, it is a better practice to normalize the space internally with a preg_split that matches on " +" (a space with a plus sign) and then use implode(" ", $array). But eh... if you are just validating, this should be fine.
you've got it nearly right, just add \s into your square brackets and "hello world" will pass.
([A-Za-z0-9_\s]+)
I've got some help by old friend and i've tested and works perfect - thank you all for answers and comments it was very helpful to me.
this works perfect
$check = eregi_replace('(^[a-zA-Z0-9 ]*$)', "", $catname);
Alphanumeric and white space regular expression
#Phil
yours works perfect but still will pass underscore ~ thanks
#Michael Hays
I do not know it didn't worked for whitespace , but your comments is very helpful ~ thanks
#kjetilh
I will read more about $preg ~ thanks
#Alastair
Works fine if i've replaced \s with just whitespace ! ~ thanks
eregi functions are deprecated as of php 5.3. Use preg instead.
Related
I have a string that contains normal characters, white charsets and newline characters between <div> and </div>.
This regular expression doesn't work: /<div>(.*)<\/div>. It is because .* doesn't match newline characters. How can I do this?
You need to use the DOTALL modifier (/s).
'/<div>(.*)<\/div>/s'
This might not give you exactly what you want because you are greedy matching. You might instead try a non-greedy match:
'/<div>(.*?)<\/div>/s'
You could also solve this by matching everything except '<' if there aren't other tags:
'/<div>([^<]*)<\/div>/'
Another observation is that you don't need to use / as your regular expression delimiters. Using another character means that you don't have to escape the / in </div>, improving readability. This applies to all the above regular expressions. Here's it would look if you use '#' instead of '/':
'#<div>([^<]*)</div>#'
However all these solutions can fail due to nested divs, extra whitespace, HTML comments and various other things. HTML is too complicated to parse with Regex, so you should consider using an HTML parser instead.
To match all characters, you can use this trick:
%\<div\>([\s\S]*)\</div\>%
You can also use the (?s) mode modifier. For example,
(?s)/<div>(.*?)<\/div>
There shouldn't be any problem with just doing:
(.|\n)
This matches either any character except newline or a newline, so every character. It solved it for me, at least.
An option would be:
'/<div>(\n*|.*)<\/div>/i'
Which would match either newline or the dot identifier matches.
There is usually a flag in the regular expression compiler to tell it that dot should match newline characters.
I have a string that contains normal characters, white charsets and newline characters between <div> and </div>.
This regular expression doesn't work: /<div>(.*)<\/div>. It is because .* doesn't match newline characters. How can I do this?
You need to use the DOTALL modifier (/s).
'/<div>(.*)<\/div>/s'
This might not give you exactly what you want because you are greedy matching. You might instead try a non-greedy match:
'/<div>(.*?)<\/div>/s'
You could also solve this by matching everything except '<' if there aren't other tags:
'/<div>([^<]*)<\/div>/'
Another observation is that you don't need to use / as your regular expression delimiters. Using another character means that you don't have to escape the / in </div>, improving readability. This applies to all the above regular expressions. Here's it would look if you use '#' instead of '/':
'#<div>([^<]*)</div>#'
However all these solutions can fail due to nested divs, extra whitespace, HTML comments and various other things. HTML is too complicated to parse with Regex, so you should consider using an HTML parser instead.
To match all characters, you can use this trick:
%\<div\>([\s\S]*)\</div\>%
You can also use the (?s) mode modifier. For example,
(?s)/<div>(.*?)<\/div>
There shouldn't be any problem with just doing:
(.|\n)
This matches either any character except newline or a newline, so every character. It solved it for me, at least.
An option would be:
'/<div>(\n*|.*)<\/div>/i'
Which would match either newline or the dot identifier matches.
There is usually a flag in the regular expression compiler to tell it that dot should match newline characters.
I have tried to work this out myself (even bought a Kindle book!), but I am struggling with backreferences in php.
What I want is like the following example:
var $html = "hello %world|/worldlink/% again";
output:
hello world again
I tried stuff like:
preg_replace('/%([a-z]+)|([a-z]+)%/', '\1', $html);
but with no joy.
Any ideas please? I am sure someone will post the exact answer but I would like an explanation as well please - so that I don't have to keep asking these questions :)
The slashes "/" are not included in your allowed range [a-z]. Instead use
preg_replace('/%([a-z]+)\|([a-z\/]+)%/', '\1', $html);
Your expression:
'/%([a-z]+)|([a-z]+)%/'
Is only capturing one thing. The | in the middle means "OR". You're trying to capture both, so you don't need an OR in there. You want a literal | symbol so you need to escape it:
'/%([a-z]+)\|([a-z\/]+)%/'
The / character also needs to be included in your char set, and escaped as above.
Your regex (/%([a-z]+)|([a-z]+)%/) reads this way:
Match % followed by + (= one or
more) a-z characters (and store this
into backreference #1).
Or (the |):
Match + (= one or more) a-z
characters (and store this into
backreference #2) followed by a
%.
What you are looking for is:
preg_replace('~%([a-z]+)[|]([a-z/]+)%~', '$1', $html);
Basically I just escaped the | regex meta character (you can do this by either surrounding it with [] like I did or just prepending a backwards slash \, personally I find the former easier to read), and added a / to the second capture group.
I also changed your delimiters from / to ~ because tildes are much more unlikely to appear in strings, if you want to keep using / as your delimiter you also have to escape their occurrences in your regex.
It's also recommended that you use the $ syntax instead of \ in your replacement backreferences:
$replacement may contain references
of the form \\n or (since PHP 4.0.4)
$n, with the latter form being the
preferred one.
Here is a version that works according to the OPs data/information provided (using a non-slash delimiter to avoid escaping slashes):
preg_replace('#%([a-z]+)\|([a-z/]+)%#', '\1', $html);
Using a non slash delimiter, would alleviate the need to escape slashes.
Outputs:
hello world again
The Explanation
Why yours did not work. First up the | is an OR operator, and, in your example, should be escaped. Second up, since you are using /'s or expect slashes it is better to use a non-slash delimiter, such as #. Third up, the slash needed to be added to list of allowed matches. As stated before you may want to include a bit more options, as any type of word with numbers underscores periods hyphens will fail / break the script. Hopefully that is the explanation you were looking for.
Here's what works for me:
preg_replace('/%([a-z]+)\|([a-z\/]+)%/', '\1', $html);
Your regular expression doesn't escape the |, and doesn't include the proper characters for the URL.
Here's a basic live example supporting only a-z and slashes:
preg_replace('/%([a-z]+)\|([a-z\/]+)%/', '\1', $html);
In reality, you're going to want to change those [a-z]+ blocks to something more expressive. Do some searches for URL-matching regular expressions, and pick one that fits what you want.
$html = "hello %world|/worldlink/% again";
echo preg_replace('/([A-ZA-z_ ]*)%(.+)\|(.+)%([A-ZA-z_ ]*)/', '$1$2$4', $html);
output:
hello world again
here is a working code : http://www.ideone.com/0qhZ8
^([a-zA-Z0-9!##$%^&*|()_\-+=\[\]{}:;\"',<.>?\/~`]{4,})$
Would this regular expression work for these rules?
Must be atleast 4 characters
Characters can be a mix of alphabet (capitalized/non-capitalized), numeric, and the following characters: ! # # $ % ^ & * ( ) _ - + = | [ { } ] ; : ' " , < . > ? /
It's intended to be a password validator. The language is PHP.
Yes?
Honestly, what are you asking for? Why don't you test it?
If, however, you want suggestions on improving it, some questions:
What is this regex checking for?
Why do you have such a large set of allowed characters?
Why don't you use /\w/ instead of /0-9a-zA-Z_/?
Why do you have the whole thing in ()s? You don't need to capture the whole thing, since you already have the whole thing, and they aren't needed to group anything.
What I would do is check the length separately, and then check against a regex to see if it has any bad characters. Your list of good characters seems to be sufficiently large that it might just be easier to do it that way. But it may depend on what you're doing it for.
EDIT: Now that I know this is PHP-centric, /\w/ is safe because PHP uses the PCRE library, which is not exactly Perl, and in PCRE, \w will not match Unicode word characters. Thus, why not check for length and ensure there are no invalid characters:
if(strlen($string) >= 4 && preg_match('[\s~\\]', $string) == 0) {
# valid password
}
Alternatively, use the little-used POSIX character class [[:graph:]]. It should work pretty much the same in PHP as it does in Perl. [[:graph:]] matches any alphanumeric or punctuation character, which sounds like what you want, and [[:^graph:]] should match the opposite. To test if all characters match graph:
preg('^[[:graph:]]+$', $string) == 1
To test if any characters don't match graph:
preg('[[:^graph:]]', $string) == 0
You forgot the comma (,) and full stop (.) and added the tilde (~) and grave accent (`) that were not part of your specification. Additionally just a few characters inside a character set declaration have to be escaped:
^([a-zA-Z0-9!##$%^&*()|_\-+=[\]{}:;"',<.>?/~`]{4,})$
And that as a PHP string declaration for preg_match:
'/^([a-zA-Z0-9!##$%^&*()|_\\-+=[\\]{}:;"\',<.>?\\/~`]{4,})$/'
I noticed that you essentially have all of ASCII, except for backslash, space and the control characters at the start, so what about this one, instead?
^([!-\[\]-~]{4,})$
You are extra escaping and aren't using some predefined character classes (such as \w, or at least \d).
Besides of that and that you are anchoring at the beginning and at the end, meaning that the regex will only match if the string starts and ends matching, it looks correct:
^([a-zA-Z\d\-!$##$%^&*()|_+=\[\]{};,."'<>?/~`]{4,})$
If you really mean to use this as a password validator, it reeks of insecurity:
Why are you allowing 4 chars passwords?
Why are you forbidding some characters? PHP can't handle some? Why would you care? Let the user enter the characters he pleases, after all you'll just end up storing a hash + salt of it.
No. That regular expression would not work for the rules you state, for the simple reason that $ by default matches before the final character if it is a newline. You are allowing password strings like "1234\n".
The solution is simple. Either use \z instead of $, or apply the D modifier to the regex.
I have a string that contains normal characters, white charsets and newline characters between <div> and </div>.
This regular expression doesn't work: /<div>(.*)<\/div>. It is because .* doesn't match newline characters. How can I do this?
You need to use the DOTALL modifier (/s).
'/<div>(.*)<\/div>/s'
This might not give you exactly what you want because you are greedy matching. You might instead try a non-greedy match:
'/<div>(.*?)<\/div>/s'
You could also solve this by matching everything except '<' if there aren't other tags:
'/<div>([^<]*)<\/div>/'
Another observation is that you don't need to use / as your regular expression delimiters. Using another character means that you don't have to escape the / in </div>, improving readability. This applies to all the above regular expressions. Here's it would look if you use '#' instead of '/':
'#<div>([^<]*)</div>#'
However all these solutions can fail due to nested divs, extra whitespace, HTML comments and various other things. HTML is too complicated to parse with Regex, so you should consider using an HTML parser instead.
To match all characters, you can use this trick:
%\<div\>([\s\S]*)\</div\>%
You can also use the (?s) mode modifier. For example,
(?s)/<div>(.*?)<\/div>
There shouldn't be any problem with just doing:
(.|\n)
This matches either any character except newline or a newline, so every character. It solved it for me, at least.
An option would be:
'/<div>(\n*|.*)<\/div>/i'
Which would match either newline or the dot identifier matches.
There is usually a flag in the regular expression compiler to tell it that dot should match newline characters.