Question

我是cakephp的新手并尝试实施AJAX。我有一个视图add.ctp，其中我写了以下几行：

$('#office_type').change(function(){ 
    var office_id = $('#office_type').val();
    if(office_id > 0) {
    var data = office_id;
    var url_to_call = "http://localhost/testpage/officenames/get_office_names_by_catagory/";
    $.ajax({
        type: "GET",
        url: url_to_call,
        data = data,
        //dataType: "json",
        success: function(msg){
            alert(msg);
        }
    }); 
    }
 });

get_office_names_by_catagory()中的OfficenamesController.php函数是：

public function get_office_name_by_catagory($type = '') {  
    Configure::write("debug",0); 
    if(isset($_GET['type']) && trim($_GET['type']) != ''){
        $type = $_GET['type'];
        $conditions = array("Officename.office_type"=> $type);
        $recursive = -1;
        $office_names = $this->Officename->find("all",array("conditions"=>$conditions,"recursive"=>$recursive));
    }
    $this->layout = 'ajax';
    //return json_encode($office_names);
    return 'Hello !';    
}

但不幸的是，它没有提醒任何事情！怎么了？

Answer 1

可能由两个问题引起：

1）在你的js片段中，你正在查询

http://localhost/testpage/officenames/get_office_names_by_catagory/。

请注意get_office_names_by_category中的复数“姓名”。在PHP代码段中，您已定义了一个操作get_office_name_by_catagory。请注意单数“名称”。

2）您可能需要适当设置标题，以便整个页面不会在AJAX请求中呈现：请参阅此link。

Answer 2

我认为，您指定的数据格式错误：

$.ajax({
    type: "GET",
    url: url_to_call,
    data = data,             // i guess, here is the problem
    //dataType: "json",
    success: function(msg){
        alert(msg);
    }
});

要

$.ajax({
    type: "GET",
    url: url_to_call,
    data: { name: "John", location: "Boston" }, //example
    success: function(msg){
        alert(msg);
    }
});

您应该以key：value格式指定数据。

Answer 3

$('#office_type').change(function(){ 
    var office_id = $('#office_type').val();
    if(office_id > 0) {
    var data = office_id;
    var url_to_call = "http://localhost/testpage/officenames/get_office_name_by_catagory/"+office_id;
    $.ajax({
        type: "GET",
        url: url_to_call,
        success: function(msg){
            alert(msg);
        }
    }); 
    }
 });

在你的行动中

public function get_office_name_by_catagory($type = '') {  
    $this->autoRender = false;
    Configure::write("debug",0); 
    if(!empty($type)){

        $conditions = array("Officename.office_type"=> $type);
        $recursive = -1;
        $office_names = $this->Officename->find("all",array("conditions"=>$conditions,"recursive"=>$recursive));
    }
    $this->layout = 'ajax';
    //return json_encode($office_names);
    echo 'Hello !'; 
    exit;   
}

看看我做了什么，我已经改变了你的函数get_office_name_by_catagory的请求，因为函数中已经定义了一个参数$ $，所以如果我有/ get_office_name_by_catagory / 2的请求那么你会在$中找到值在行动中输入。

所以不需要使用$ _GET并且休息一切都很好！

Answer 4

试试这个，从ajax中删除类型并尝试。

$('#office_type').change(function(){ 
    var office_id = $('#office_type').val();
    if(office_id > 0) {
    var data = office_id;
    var url_to_call = yourlink +office_id;
    **$.ajax({
        url: url_to_call,
        success: function(msg){
            alert(msg);
        }
    });** 
    }
 });

在你的行动中

public function get_office_name_by_catagory($type = '') {  
    $this->autoRender = false;
    Configure::write("debug",0); 
    if(!empty($type)){

        $conditions = array("Officename.office_type"=> $type);
        $recursive = -1;
        $office_names = $this->Officename->find("all",array("conditions"=>$conditions,"recursive"=>$recursive));
    }
    $this->layout = 'ajax';
    //return json_encode($office_names);
    echo 'Hello !'; 
}

cakephp 2.x中的简单ajax函数无法正常工作

4 个答案: