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Implode an array with JavaScript?
Can I implode an array in jQuery like in PHP?
7 Answers
7
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How to Get the Current URL Inside @if Statement (Blade) in Laravel 4?
... Route::current()->getName() to check your route name.
Example: routes.php
Route::get('test', ['as'=>'testing', function() {
return View::make('test');
}]);
View:
@if(Route::current()->getName() == 'testing')
Hello This is testing
@endif
...
jQuery Ajax File Upload
...ed to use the XHR object but could not get results on the server side with PHP.
var formData = new FormData();
formData.append('file', $('#file')[0].files[0]);
$.ajax({
url : 'upload.php',
type : 'POST',
data : formData,
processData: false, // tell jQuery not to proces...
Getting raw SQL query string from PDO prepared statements
...O driver,
otherwise, it will be -1).
You can find more on the official php docs
Example:
<?php
/* Execute a prepared statement by binding PHP variables */
$calories = 150;
$colour = 'red';
$sth = $dbh->prepare('SELECT name, colour, calories
FROM fruit
WHERE calories < :calorie...
How to design RESTful search/filtering? [closed]
I'm currently designing and implementing a RESTful API in PHP. However, I have been unsuccessful implementing my initial design.
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Is there a query language for JSON?
...ons inside the delegate as you can imagine: simple comparison, startsWith, etc. I haven't tested but you could probably nest filters too for querying inner collections.
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Display JSON as HTML [closed]
...er, most browsers display the XML formatted (indented, proper line breaks, etc). I'd like the same end result for JSON.
12 ...
Do I have to guard against SQL injection if I used a dropdown?
... {
// Not Expected
}
Then use mysqli_* if you are using a version of php >= 5.3.0 which you should be, to save your result. If used correctly this will help with sql injection.
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How to convert JSON string to array
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@RahulMehta If you're using PHP's built-in json_decode() it will return NULL if your JSON is invalid (for example, no quoted keys). That's what the documentation says and that's what my PHP 5.2 installation returns. Are you using a function other than...
Create JSON-object the correct way
I am trying to create an JSON object out of a PHP array. The array looks like this:
5 Answers
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