I need to check to see if a variable contains anything OTHER than 0-9 and the "-" and the "+" character and the " "(space).
The preg_match I have written does not work. Any help would be appreciated.
<?php
$var="+91 9766554433";
if(preg_match('/[0-9 +\-]/i', $var))
echo $var;
?>
You have to add a * as a quantifier to the whole character class and add anchors to the start and end of the regex: ^ and $ means to match only lines containing nothing but the inner regex from from start to end of line. Also, the i modifier is unnecessary since there is no need for case-insensitivity in this regex.
This should do the work.
if(!preg_match('/^[0-9 +-]*$/', $var)){
//variable contains char not allowed
}else{
//variable only contains allowed chars
}
Just negate the character class:
if ( preg_match('/[^0-9 +-]/', $var) )
echo $var;
or add anchors and quantifier:
if ( preg_match('/^[0-9 +-]+$/', $var) )
echo $var;
The case insensitive modifier is not mandatory in your case.
You can try regex101.com to test your regex to match your criteria and then on the left panel, you'll find code generator, which will generate code for PHP, Python, and Javascript.
$re = "/^[\\d\\s\\+\\-]+$/i";
$str = "+91 9766554433";
preg_match($re, $str, $matches);
You can take a look here.
Try see if this works. I haven't gotten around to test it beforehand, so I apologize if it doesn't work.
if(!preg_match('/^[0-9]+.-.+." ".*$/', $var)){
//variable contains char not allowed
}else{
//variable only contains allowed chars
}
Related
This is the code:
<?php
$pattern =' abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890';
$text = "kdaiuyq7e611422^^$^vbnvcn^vznbsjhf";
$text_split = str_split($text,1);
$data = '';
foreach($text_split as $value){
if (preg_match("/".$value."/", $pattern )){
$data = $data.$value;
}
if (!preg_match('/'.$value.'/', $pattern )){
break;
}
}
echo $data;
?>
Current output:
kdaiuyq7e611422^^$^vbnvcn^vznbsjhf
Expected output:
kdaiuyq7e611422
Please help me editing my code error. In pattern there is no ^ or $. But preg_match is showing matched which is doubtful.
You string $text have ^ which will match the begin of the string $pattern.
So the preg_match('/^/', $pattern) will return true, then the ^ will append to $data.
You should escape the ^ as a raw char, not a special char with preg_match('/\^/', $pattern) by the help of preg_quote() which will escape the special char.
There is no need to split your string up like that, the whole point of a regular expression is you can specify all the conditions within the expression. You can condense your entire code down to this:
$pattern = '/^[[:word:] ]+/';
$text = 'kdaiuyq7e611422^^$^vbnvcn^vznbsjhf';
preg_match($pattern, $text, $matches);
echo $matches[0];
Kris has accurately isolated that escaping in your method is the monkey wrench. This can be solved with preg_quote() or wrapping pattern characters in \Q ... \E (force characters to be interpreted literally).
Slapping that bandaid on your method (as you have done while answering your own question) doesn't help you to see what you should be doing.
I recommend that you do away with the character mask, the str_split(), and the looped calls of preg_match(). Your task can be accomplished far more briefly/efficiently/directly with a single preg_match() call. Here is the clean way that obeys your character mask fully:
Code: (Demo)
$text = "kdaiuyq7e611422^^$^vbnvcn^vznbsjhf";
echo preg_match('/^[a-z\d ]+/i',$text,$out)?$out[0]:'No Match';
Output:
kdaiuyq7e611422
miknik's method was close to this, but it did not maintain 100% accuracy given your question requirements. I'll explain:
[:word:] is a POSIX Character Class (functioning like \w) that represents letters(uppercase and lowercase), numbers, and an underscore. Unfortunately for miknik, the underscore is not in your list of wanted characters, so this renders the pattern slightly inaccurate and may be untrustworthy for your project.
i'm not very firm with regular Expressions, so i have to ask you:
How to find out with PHP if a string contains a word starting with # ??
e.g. i have a string like "This is for #codeworxx" ???
I'm so sorry, but i have NO starting point for that :(
Hope you can help.
Thanks,
Sascha
okay thanks for the results - but i did a mistake - how to implement in eregi_replace ???
$text = eregi_replace('/\B#[^\B]+/','\\1', $text);
does not work??!?
why? do i not have to enter the same expression as pattern?
Match anything with has some whitespace in front of a # followed by something else than whitespace:
$ cat 1812901.php
<?php
echo preg_match("/\B#[^\B]+/", "This should #match it");
echo preg_match("/\B#[^\B]+/", "This should not# match");
echo preg_match("/\B#[^\B]+/", "This should match nothing and return 0");
echo "\n";
?>
$ php 1812901.php
100
break your string up like this:
$string = 'simple sentence with five words';
$words = explode(' ', $string );
Then you can loop trough the array and check if the first character of each word equals "#":
if ($stringInTheArray[0] == "#")
Assuming you define a word a sequence of letters with no white spaces between them, then this should be a good starting point for you:
$subject = "This is for #codeworxx";
$pattern = '/\s*#(.+?)\s/';
preg_match($pattern, $subject, $matches);
print_r($matches);
Explanation:
\s*#(.+?)\s - look for anything starting with #, group all the following letters, numbers, and anything which is not a whitespace (space, tab, newline), till the closest whitespace.
See the output of the $matches array for accessing the inner groups and the regex results.
#OP, no need regex. Just PHP string methods
$mystr='This is for #codeworxx';
$str = explode(" ",$mystr);
foreach($str as $k=>$word){
if(substr($word,0,1)=="#"){
print $word;
}
}
Just incase this is helpful to someone in the future
/((?<!\S)#\w+(?!\S))/
This will match any word containing alphanumeric characters, starting with "#." It will not match words with "#" anywhere but the start of the word.
Matching cases:
#username
foo #username bar
foo #username1 bar #username2
Failing cases:
foo#username
#username$
##username
I tried to search around but couldn't find anything useful. I need to trim special characters from beginning and end of a string and identify if the remaining portion is a number.
For example
(5)
[[12]]
{3}
#!8(#
!255=
/879/
I need a preg_match expression for it. The regular expression should ignore the string if any alphabets come in between.
$string="yourstring";
$new_string=preg_replace('/[^A-Za-z0-9]/', '', $string);
if(is_numeric($new_string){
echo "number";
} else {
echo "string";
}
^(?!.*[a-zA-Z])\W*(\d+)\W*$
You can use this.Lookahead will validate if only numbers are there.Replace by $1.See demo.
https://regex101.com/r/cT0hV4/2
for an url routing I have
Patern :
/^\/stuff\/other-stuff\/(?:([^\/]\+?))$/i
Subject :
/stuff/other-stuff/foo-AB123456.html
why $num_matches is equal to 0 ??
$num_matches = preg_match_all($patern, $subject, $matches);
Help should be greatly appreciated :)
because of this:
[^\/]\+?
firstly there is no slash after other-stuff so you cannot find the sentence with a negated / secondly the + must not be escaped if you are doing this kind of match . + must only be escaped when you are doing a literal match.
the corrected regex should be :
^\/stuff\/other-stuff\/(?:(.+?))$
demo here : http://regex101.com/r/aV9cR0
will match foo-AB123456.html in the first capture
$patern= "#^/stuff/other-stuff/([^/]+)$#i";
$subject = "/stuff/other-stuff/foo-AB123456.html";
preg_match_all($patern, $subject, $matches);
print_r($matches[1]);
It looks to me like your regex could be simplified to something like:
(?i)^/stuff/other-stuff/[\w-.]+$
It would work like this:
<?php
$regex="~(?i)^/stuff/other-stuff/([\w-./]+)$~";
$string = "/stuff/other-stuff/foo-AB123456.html";
$hit = preg_match($regex,$string,$m);
echo $m[0]."<br />";
echo $m[1]."<br />";
?>
Output:
/stuff/other-stuff/foo-AB123456.html
foo-AB123456.html
Note that this could be done in a number of different ways.
Here are some details about the regex.
The ~ delimiter is nicer than the original / because you don't have to escape the slashes.
The parentheses in ([\w-.]+) capture the end of the url into Group 1. This is why $m[1] yields foo-AB123456.html
After the final slash, [\w-./]+ matches any number of letters or digits, underscores, dashes, dots and forward slashes. This is a "mini-spec" for what characters we expect there. If you want to allow anything at all, you could go with a simple dot.
i'm not very firm with regular Expressions, so i have to ask you:
How to find out with PHP if a string contains a word starting with # ??
e.g. i have a string like "This is for #codeworxx" ???
I'm so sorry, but i have NO starting point for that :(
Hope you can help.
Thanks,
Sascha
okay thanks for the results - but i did a mistake - how to implement in eregi_replace ???
$text = eregi_replace('/\B#[^\B]+/','\\1', $text);
does not work??!?
why? do i not have to enter the same expression as pattern?
Match anything with has some whitespace in front of a # followed by something else than whitespace:
$ cat 1812901.php
<?php
echo preg_match("/\B#[^\B]+/", "This should #match it");
echo preg_match("/\B#[^\B]+/", "This should not# match");
echo preg_match("/\B#[^\B]+/", "This should match nothing and return 0");
echo "\n";
?>
$ php 1812901.php
100
break your string up like this:
$string = 'simple sentence with five words';
$words = explode(' ', $string );
Then you can loop trough the array and check if the first character of each word equals "#":
if ($stringInTheArray[0] == "#")
Assuming you define a word a sequence of letters with no white spaces between them, then this should be a good starting point for you:
$subject = "This is for #codeworxx";
$pattern = '/\s*#(.+?)\s/';
preg_match($pattern, $subject, $matches);
print_r($matches);
Explanation:
\s*#(.+?)\s - look for anything starting with #, group all the following letters, numbers, and anything which is not a whitespace (space, tab, newline), till the closest whitespace.
See the output of the $matches array for accessing the inner groups and the regex results.
#OP, no need regex. Just PHP string methods
$mystr='This is for #codeworxx';
$str = explode(" ",$mystr);
foreach($str as $k=>$word){
if(substr($word,0,1)=="#"){
print $word;
}
}
Just incase this is helpful to someone in the future
/((?<!\S)#\w+(?!\S))/
This will match any word containing alphanumeric characters, starting with "#." It will not match words with "#" anywhere but the start of the word.
Matching cases:
#username
foo #username bar
foo #username1 bar #username2
Failing cases:
foo#username
#username$
##username