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json_encode sparse PHP array as JSON array, not JSON object
I have the following array in PHP:
4 Answers
4
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Determining Referer in PHP
...nd(), TRUE);
$_SESSION['token'] = $token;
$url = "http://example.com/index.php?token={$token}";
Then the index.php will look like this:
if(empty($_GET['token']) || $_GET['token'] !== $_SESSION['token'])
{
show_404();
}
//Continue with the rest of code
I do know of secure sites that do the...
PHP - find entry by object property from an array of objects
...tion and subsequent answers for more information on the latter - Reference PHP array by multiple indexes
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Are “elseif” and “else if” completely synonymous?
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From the PHP manual:
In PHP, you can also write 'else if' (in two words) and the behavior would be identical to the one of 'elseif' (in a single word). The syntactic meaning is slightly different (if you're familiar with C, this i...
Escape single quote character for use in an SQLite query
...at way), so it would be :
INSERT INTO table_name (field1, field2) VALUES (123, 'Hello there''s');
Relevant quote from the documentation:
A string constant is formed by enclosing the string in single quotes ('). A single quote within the string can be encoded by putting two single quotes in a ...
Parsing domain from a URL
...
Or simply: print parse_url($url, PHP_URL_HOST)) if you don't need the $parse array for anything else.
– rybo111
Aug 24 '16 at 12:03
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PHP json_decode() returns NULL with valid JSON?
... On my server, it doesn't. And I can't call json_last_error() because it's PHP 5.2.9. That function appears on PHP 5.3.0.
– Joel A. Villarreal Bertoldi
Mar 9 '10 at 15:54
1
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Laravel Controller Subfolder routing
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For Laravel 5.3 above:
php artisan make:controller test/TestController
This will create the test folder if it does not exist, then creates TestController inside.
TestController will look like this:
<?php
namespace App\Http\Controllers\test;...
Difference between if () { } and if () : endif;
...n my .phtml files (Zend Framework) I will write something like this:
<?php if($this->value): ?>
Hello
<?php elseif($this->asd): ?>
Your name is: <?= $this->name ?>
<?php else: ?>
You don't have a name.
<?php endif; ?>
...
JSON encode MySQL results
... $rows[] = $r;
}
print json_encode($rows);
The function json_encode needs PHP >= 5.2 and the php-json package - as mentioned here
NOTE: mysql is deprecated as of PHP 5.5.0, use mysqli extension instead http://php.net/manual/en/migration55.deprecated.php.
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