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Php 如何将数组传递给Drupal菜单回调_Php_Javascript_Ajax_Drupal_Callback - Fatal编程技术网

Php 如何将数组传递给Drupal菜单回调

Php 如何将数组传递给Drupal菜单回调,php,javascript,ajax,drupal,callback,Php,Javascript,Ajax,Drupal,Callback,我有两个JavaScript数组: var xcoord=[]和var ycoord=[] 经过一些处理后,每个数组包含15个数值。我想将这些数据发送到Drupal中的菜单回调 我的AJAX代码: $.post('http://mysite.com/?q=menu_example/my_page', {'ycoord[]': ycoord, 'xcoord[]': xcoord } ); Drupal PHP: $items['menu_example/my_page/%/%'] =

我有两个JavaScript数组:
var xcoord=[]
var ycoord=[]
经过一些处理后,每个数组包含15个数值。我想将这些数据发送到Drupal中的菜单回调

我的AJAX代码:

$.post('http://mysite.com/?q=menu_example/my_page', 
    {'ycoord[]': ycoord, 'xcoord[]': xcoord }
);
Drupal PHP:

$items['menu_example/my_page/%/%'] = array(
    'title' => 'My Page', 
    'description' => 'i hope this works', 
    'page callback' => '_graphael', 
    'page arguments' => array(2, 3), // fill this, 
    'access callback' => true,
    'type' => MENU_CALLBACK,
);

在我的控制台中,我看到
xcoord
ycoord
的值被正确地传输,但是回调不太有效。我的问题:如何将数组传递给菜单回调?我是否仍应在
$items
键中使用
%
占位符?

传递给
$的第二个参数。post()
未附加到作为第一个参数传递的URL;它是在
$\u POST
中传递给PHP的数据

菜单回调的正确定义应如下所示

$items['menu_example/my_page'] = array(
  'title' => 'My Page', 
  'description' => 'I hope this works', 
  'page callback' => '_graphael', 
  'access callback' => TRUE,
  'type' => MENU_CALLBACK,
);
然后,页面回调应该在
$\u POST
中找到从jQuery传递的数据