Getting a value from function [closed] - php

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I'm finishing a script that converts a string and pass it as a link, but besides that, it also shortens the URL with an API. The problem is that I can not think how to get only the URL instead of the entire chain.
The function:
function findAndShort($string) {
$text = preg_replace("/(https?|ftps?|mailto):\/\/([-\w\p{L}\.]+)+(:\d+)?(\/([\w\p{L}\/_\.#]*(\?\S+)?)?)?/u", '$0', $string);
return $text;
}
Example:
$chk = findAndShort("Blahblah http://domain.tld");
echo $chk;
In this case, only need the http://domain.tld, i try with $chk[0], but ofcourse, print the first character on the line..

Just add ^.* at the begining of your regex and use $1 instead of $0:
$text = preg_replace("/^.*((https?|ftps?|mailto):\/\/([-\w\p{L}\.]+)+(:\d+)?(\/([\w\p{L}\/_\.#,]*(\?\S+)?)?)?)/u", '$1', $string);
You can also simplified a bit,
~^.*((https?|ftps?|mailto)://[-\p{L}\p{N}_.]+(:\d+)?(/([\p{L}\p{N}/_.#,]*(\?\S+)?)?)?)~u

The thing with functions is you can only return a single value that is under the return in the function, so in your case, after testing your function, the return value is the original input value.
If you want a function to return only the URL component I would use explode to separate the string then run preg_match on each part of the resulting array from the explode function looking for the https, ftp or mailto, once I know what part of the array I'm working with I would stop the process, define the var and as I believe you want to return code for a link, I would do is then do something like
return ''.$url.'';
Note that the var $url would be created from the above process.
I hope this helps.

Related

PHP array_search not working (returns empty string) [closed]

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I'm trying to check if a filename of an image contains "cover". Somehow this stopped working for me (Pretty sure it worked already). I copied the part of my function not working.
$name=array("_IMG8555.jpg", "_IMG7769.jpg", "_IMG8458.jpg", "Cover.jpg", "_IMG7184.jpg");
$cov=array("Cover.png","Cover.jpg","Cover.jpeg", "cover.png","cover.jpg","cover.jpeg");
This does not work for me:
print_r(array_search($cov, $name)); //Returns empty String
print_r($name[array_search($cov, $name)]); //Returns first element of the name Array
Also I added a test Strings to make sure this is not result the searched string is the same as the search value.
print_r($name[3]===$cov[1]); //Returns true(1)
Can anyone help? Why does this simple script not work?
I also tried using in_array() but this is not working either.
The array_search() function search an array for a value and returns the key
array_search(key_value,array)
Loop your $cov array and get one key at a time and check with $name array
foreach($cov as $i => $cov_s){
if(in_array($cov_s, $name)){
return $name[array_search($cov_s, $name)];
}
}
return $name[0];
Try this code.
$name=array("_IMG8555.jpg", "_IMG7769.jpg", "_IMG8458.jpg", "Cover.jpg", "_IMG7184.jpg");
$cov=array("Cover.png","Cover.jpg","Cover.jpeg", "cover.png","cover.jpg","cover.jpeg");
foreach($cov as $c){
if(array_search($c,$name)){
//Do your success function
return true;
}
else
return false;
}

url encoding issue. instead %50 its passing space..how to pass %50 through urlencoded url [closed]

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I have tried to pass a variable trough url... its %50.
I am doing urlencoding to pass other languages through the url.
At that time %50 also been converted to a space or something else.
Can anyone help me to find out a way to send %50 as a variable through urlencoded link(url).
<?php
$string = '%50';
echo $encoded = urlencode($string);
// returns %2550
echo urldecode($encoded);
// returns %50
?>
So if you want to pass $string to a url you write something like:
http://yoursite.com/script.php?string=$encoded
To get your original string value you can just use $_GET in your script.php:
echo $_GET["string"];
// returns %50

How to acces the second value [closed]

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I was wondering if you could give two values ​​to a class and then acces to the second one by POST, something like this (part of the code):
echo "<select name='selecttsk' id='selecttsk'>";
while($w = $bd->obtener_fila($tasker, 0)){
echo "<option class='opcion1' value ='".$w[1]/$w[2]."'>".$w[0]."</option>";
}
echo "</select>";
?>
and then i need to do something like this in other file
$var = $_POST[selecttsk];
but i need $w[2]
thanks
I suppose your $_POST['selecttsk'] does have the values in the following format:
"foo/bar"
You could use "explode" in PHP to get the second part (bar), for example:
$postvar = $_POST['selecttsk'];
$vars = explode("/", $postvar);
// Then
$var = $vars[1]; // Will be the $w[2] from the form
Look into: http://php.net/explode
Beware that if $w[1] or $w[2] ever contains a "/" you might get unexpected results, you could use the limit function of explode to mitigate that issue.
However - I would generally not recommend this workflow.
Why do you need to send two variables with one select?
(could you show us som example of what $w contains)

Regular Expression Help for Date Validation - mm-dd-yyyy - PHP [closed]

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if(preg_match('/^[0-9]{1,2}\-[0-9]{1,2}\-[0-9]{4}$/', '10-30-2013')){
echo 'true';
}
else {
echo 'false';
}
This not give me true. I think I'm wrong with regex. please tell how to correct this regex
I suggest not using regex at all for this -- date validation with regex is a surprisingly difficult thing to get right, due to the variations possible in the number of days in any given month.
Far better to simply use PHP's DateTime class to actually parse it into a date object. This will also validate it; if it doesn't meet the specified format, then it won't parse.
$dateObj = DateTime::CreateFromFormat('m-d-Y',$inputString);
See PHP manual page for CreateFromFormat().
You should use dedicate functions to parse date ie.:
if (strptime ('10-30-2013' , 'm-d-Y') !== false) {
echo 'true'.PHP_EOL;
} else {
echo 'true'.PHP_EOL;
}
Actually as the others confirmed, your regex works fine and returns true.
But as you made your code shrunken, I think the input string you're trying to show us isn't exactly just a date and it's within a string or maybe has trailing spaces!
So using ^ and $ at your regex will result in failure.

call fo a file from PHP [closed]

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i writing a PHP script that call a file in the server
by using :
<?php
$ret=system("command");
?>
the problem is when the file need some parameters
i can't find a way of doing that
because when using
<?php
$ret=system("command");
?>
the PHP skips that part of asking for variables
and assign to id a random one
and i can't pass theme at the start like
$ret = system("command argument1 argument2 argument3...");
beause the nmber of parametres depend on the user
i mean he keep entring data to a dynamic array entill he enter"end"
$ret = system("command argument1 argument2 argument3...");
Just load the arguments on, just like you were calling the program from a command line.
$cmd = "cmd param1 param2";
system($cmd,$return_value);
($return_value == 0) or die("returned an error: $cmd");
If what you mean is that you have an array that can have any number of parameters use this:
$commandParameters = implode(" ",$dynamicArray);
$ret = system("command ".escapeshellarg($commandParameters));
Get the parameters from the user via. HTML form and then, when you know what (and how many) the parameters are - you can use "system()" like everyone suggested!

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