I have a question. How should look a regex for the following words:
+23456745678 or +29845281058, (with comma)
I tried but no result:
if (!preg_match('#^\+[0-9]{11}$#',$string)) {
return ("Msisdn $string is incorrect!");
break;
}
Always I get this error.
Please help me. Thx in advance
The pattern should be:
'/\+[0-9]{11},?/'
The ^ in the beginning and $ at the end stand for beginning and end of line, which I am not sure if you really want there (my guess is not).
Test it here: http://www.phpliveregex.com/p/78v
Another option is to use \d for digit
'/\+\d{11},?/'
Related
I try to make system that can detect date in some string, here is the code :
$string = "02/04/16 10:08:42";
$pattern = "/\<(0?[1-9]|[12][0-9]|3[01])\/\.- \/\.- \d{2}\>/";
$found = preg_match($pattern, $string);
if ($found) {
echo ('The pattern matches the string');
} else {
echo ('No match');
}
The result i found is "No Match", i don't think that i used correct regex for the pattern. Can somebody tell me what i must to do to fix this code
First of all, remove all gibberish from the pattern. This is the part you'll need to work on:
(/0?[1-9]|[12][0-9]|3[01]/)
(As you said, you need the date only, not the datetime).
The main problem with the pattern, that you are using the logical OR operators (|) at the delimiters. If the delimiters are slashes, then you need to replace the tube characters with escaped slashes (/). Note that you need to escape them, because the parser will not take them as control characters. Like this: \/.
Now, you need to solve some logical tasks here, to match the numbers correctly and you're good to go.
(I'm not gonna solve the homework for you :) )
These articles will help you to solve the problem tough:
Character classes
Repetition opetors
Special characters
Pipe character (alternation operator)
Good luck!
In your comment you say you are looking for yyyy, but the example says yy.
I made a code for yy because that is what you gave us, you can easily change the 2 to a 4 and it's for yyyy.
preg_match("/((0|1|2|3)[0-9])\/\d{2}\/\d{2}/", $string, $output_array);
Echo $output_array[1]; // date
Edit:
If you use this pattern it will match the time too, thus make it harder to match wrong.
((0|1|2|3)[0-9])/\d{2}/\d{2}\s+\d{2}:\d{2}:\d{2}
http://www.phpliveregex.com/p/fjP
Edit2:
Also, you can skip one line of code.
You first preg_match to $found and then do an if $found.
This works too:
If(preg_match($pattern, $string, $found))}{
Echo $found[1];
}Else{
Echo "nothing found";
}
With pattern and string as refered to above.
As you can see the found variable is in the preg_match as the output, thus if there is a match the if will be true.
I actually used the expression from the solution given in Regular expression for Dutch phone number for my php code below, but this code is not working.
The code is simple but I don't see where I go wrong ?
define("REGEXP_PHONE_NL","(^\+[0-9]{2}|^\+[0-9]{2}\(0\)|^\(\+[0-9]{2}\)\(0\)|^00[0-9]{2}|^0)([0-9]{9}$|[0-9\-\s]{10}$)");
$string = "+31123456789"; //based on solution given in https://stackoverflow.com/questions/17949757/regular-expression-for-dutch-phone-number
echo(filter_var($string, FILTER_VALIDATE_REGEXP,array("options"=>array("regexp"=>REGEXP_PHONE_NL))));
Regex itself works, but you forgot to put same characters in the beginning and in the end of the pattern (delimeters).
<?php
define("REGEXP_PHONE_NL","/(^\+[0-9]{2}|^\+[0-9]{2}\(0\)|^\(\+[0-9]{2}\)\(0\)|^00[0-9]{2}|^0)([0-9]{9}$|[0-9\-\s]{10}$)/");
$string = "+316123456789"; //based on solution given in http://stackoverflow.com/questions/17949757/regular-expression-for-dutch-phone-number
var_dump(filter_var($string, FILTER_VALIDATE_REGEXP,array("options"=>array("regexp"=>REGEXP_PHONE_NL))));
See it working here.
You are missing to put regexp identifier character forward slash (/) in your regular expression pattern i add forward slash before and after,
Check this Demo code Viper
PHP
<?php
define("REGEXP_PHONE_NL","/(^\+[0-9]{2}|^\+[0-9]{2}\(0\)|^\(\+[0-9]{2}\)\(0\)|^00[0-9]{2}|^0)([0-9]{9}$|[0-9\-\s]{10}$)/");
$string = "+316123456789"; //based on solution given in http://stackoverflow.com/questions/17949757/regular-expression-for-dutch-phone-number
echo filter_var($string, FILTER_VALIDATE_REGEXP,array("options"=>array("regexp"=>REGEXP_PHONE_NL)));
?>
Result
+316123456789
I would like to get a string made of one word with a delimiter word before and after it
i tried but doen t work
$stringData2 = file_get_contents('testtext3.txt');
$regular2=('/(?<=first del)*MAIN WORD(?=last del)*\s');
preg_match_all($regular2,
$stringData2,
$out, PREG_PATTERN_ORDER);
thank you very much for any help
No quantifier needed, add delimeter at end, put \s inside lookahead.
'/(?<=first del)MAIN WORD(?=last del\s)/'
This regex
(?<=xx)[^\s]*(?=yy)
matches hello in:
xxhelloyy
but fails to match in:
xxhello worldyy
This is probably what you're looking for.
If you want the delimiter string included in the match, then you should not be using lookahead or look or look behind. It should be something rather basic, like this.
/\s?first del MAIN WORD last del\s?/
If you do want to return JUST the MAIN WORD part of the match, then this will work.
/(?<=\s?first del)MAIN WORD(?=last del\s?)/
Put a 'i' at the very end of that to make it case insensitive, if you want. I only mention this, because in the example you gave me above has different case between the example text and the desired response.
I searched everywhere but i couldn't find the right regex for my verificaiton
I have a $string, i want to make sure it contains at last one uppercase letter and one number. no other characters allowed just numbers and letter. is for a password require.
John8 = good
joHn8 = good
jo8hN = good
I will use preg_match function
The uppercase and letter can be everywhere in the word, not only at the begging or end
This should work, but is a bit of a mess. Consider using multiple checks for readability and maintainability...
preg_match('/^[A-Za-z0-9]*([A-Z][A-Za-z0-9]*\d|\d[A-Za-z0-9]*[A-Z])[A-Za-z0-9]*$/', $password);
Use lookahead:
preg_match('/^(?=.*[A-Z])(?=.*[0-9])[a-zA-Z0-9]+$/', $string);
Use this regex pattrn
^([A-Z]+([a-z0-9]+))$
Preg_match
preg_match('~^([A-Z]+([a-z0-9]+))$~',$str);
Demo
Your requisition need "precise syntax description", and a lot of examples for assert your description. Only 3 or 4 examples is not enough, is very open.
For last confirmed update:
preg_match('/^([a-z]*\d+[a-z]*[A-Z][a-z]*|[a-z]*[A-Z][a-z]*\d+[a-z]*)$/',$str)
History
first solution preg_match('/^[A-Z][a-z]+\d+$/',$str)
After your edit1: preg_match('/^[a-z]*[A-Z][a-z]*\d+$/',$str)
After your comment about utf8: hum... add at your question the valid language. Example: "José11" is a valid string?
After your edit2 ("jo8hN" is valid): and about number, can repeat? Well I suppose not. "8N" is valid? I suppose yes. preg_match('/^([a-z]*\d+[a-z]*[A-Z][a-z]*|[a-z]*[A-Z][a-z]*\d+[a-z]*)$/',$str) you can add more possibilities with "|" in this regex.
I want to find any pattern matching: ###-##-####
and replace the ###-##, with ***-**
but leave the -####
I tried this below, but nothing is being replaced at all.
preg_replace('/(^[\d]{3})(-)([\d]{2})(-[\d]{4}$)/','\2\4',$myText);
Any help is appreciated
Update, here is my entire code string as it currently stands, after trying a few of the suggestions below. I am comparing the second echo output to the first... and the social numbers all remain the same.
Also, as it was mentioned below, my string does contain more than just a social... it is thousands of characters long. which i think is my real issue. Sorry if i didnt clear that up in the beginning.
//Make the CSC credit report request.
$strCscResponse = $Csc->makeRequest($strFixedFormatRecord);
echo "<br/><br/><pre>" . $strCscResponse . "</pre><br/><br/>";
$strCscResponse = str_replace("!", " ", $strCscResponse);
$strCscResponse = preg_replace('/^\d{3}-\d{2}(-\d{4})$/','***-**$1',$strCscResponse);
echo "<br/><br/><pre>" . $strCscResponse . "</pre><br/><br/>";
update
I'd like to mark all the answers and "the answer" just because i didnt clarify the string has more than just a social in it. thank you for the help with this issue, embarrisingly enough it has been driving me wild for a couple days now.
There is one possible problem: you might not be matching the right string (if you are trying to find SSNs buried in a large block of text) - the ^ and $ anchors will only match beginning of string (or sometimes beginning of line) - if this is not what you want, but instead you want to find SSNs in a long string, you need to get rid of those anchors.
The other problem, potentially, is that you seem to want to replace things with asterisks, but you do not include asterisks in your replacement expression. you need to use a replacement expression like
`***-**\4`
Try this regex:
(\d{3})(-)(\d{2})(-\d{4})
Try this:
preg_replace('/^\d{3}-\d{2}(-\d{4})$/','***-**$1',$myText);
you have ^ and $ in your pattern, but I see no m modifier, so this
will only match if ###-##-#### is the entire string.
[\d] can be
shortened to \d
your \2\4 will leave --####, if you wanted *-####
you can simply have *\4