Drupal $不返回完整的HTML

Question

我寫了一個Drupal模塊，我試圖通過$ get調用.module文件。它正在運行，但它正在返回啓用HTML頁面。我嘗試使用drupal_json_output，因爲我正在使用drupal -7。我也嘗試使用getJSON，但隨後它不再給我提供任何響應。請幫助

這裏是我的.module文件

<?php 

/** 
* Implementation of hook_init(). 
*/ 
function ajax_privacy_init() { 
    drupal_add_js(drupal_get_path('module','ajax_privacy').'/userprivacy.js'); 
} 

/** 
* Implementation of hook_menu(). 
*/ 
function ajax_privacy_menu() { 

    $items = array(); 

    $items['user/%'] = array(
    'title' => 'menu privacy', 
    'page callback' => 'ajax_privacy_get_html', // Render HTML 
    'page arguments' => array(2), 
    'type' => MENU_CALLBACK, 
    'access arguments' => array('access content'), 
); 
    return $items; 
} 
/** 
* Callback to return JSON encoded data. 
*/ 
function ajax_privacy_get_html($arg) { 

$array['text'] = $arg; 
die(drupal_json_output($array)); 

} 

，這裏是我的.js的文件真正

// Jquery wrapper for drupal to avoid conflicts between libraries. 
(function ($) { 
    // Jquery onload function. 

    Drupal.behaviors.ajax_privacy = { 
    attach: function (context, settings) { 
    // Your JS code. 

    user_id_ajax=urldetect(); 

    if(user_id_ajax!=-1) 
    ajax_request='user/'+user_id_ajax; 

    //alert(ajax_request); 
    $.get('',{q:ajax_request},ajaxAction); 

    return false; 
    } 
    }; 

    ajaxAction=function(response){ 
    alert(response); 
    }; 

    urldetect=function() 
    { 
    current_url=window.location.href; 
    //alert(current_url); 

    user_id_temp=-1; 
    slash_temp=-1; 

    if(current_url.search("user/")!=-1) 
    { 
    user_id_temp=current_url.substring(42); 
    slash_temp=user_id_temp.search("/"); 
    if(slash_temp!=-1) 
    { 
    user_id_temp=user_id_temp.substring(0,slash_temp); 
    } 

    } 

    return(user_id_temp); 
    }; 

})(jQuery); 

Answer 1

// Jquery wrapper for drupal to avoid conflicts between libraries. 
(function ($) { 
    // Jquery onload function. 

    Drupal.behaviors.ajax_privacy = { 
    attach: function (context, settings) { 
    // Your JS code. 

    user_id_ajax=urldetect(); 

    if(user_id_ajax!=-1) 
    ajax_request='user/'+user_id_ajax; 

    //alert(ajax_request); 
    //try adding 'json' in your $.get call 
    $.get('',{q:ajax_request},ajaxAction,'json'); 

    return false; 
    } 
    }; 

    ajaxAction=function(response){ 
    alert(response); 
    }; 

    urldetect=function() 
    { 
    current_url=window.location.href; 
    //alert(current_url); 

    user_id_temp=-1; 
    slash_temp=-1; 

    if(current_url.search("user/")!=-1) 
    { 
    user_id_temp=current_url.substring(42); 
    slash_temp=user_id_temp.search("/"); 
    if(slash_temp!=-1) 
    { 
    user_id_temp=user_id_temp.substring(0,slash_temp); 
    } 

    } 

    return(user_id_temp); 
    }; 

})(jQuery); 

Answer 2

添加訪問回調：

$items['user/%'] = array(
    'title' => 'menu privacy', 
    'page callback' => 'ajax_privacy_get_html', // Render HTML 
    'page arguments' => array(2), 
    'type' => MENU_CALLBACK, 
    'access arguments' => array('access content'), 
    'access callback' => TRUE // add this line 
); 

嘗試這個：

function ajax_privacy_get_html($arg) { 

$array['text'] = $arg; 
print json_encode($array); 
exit(); 
}