Question

我有一个问题试图让Symfony和ajax相互配合得很好。我是Symfony和ajax整个世界的新手，所以对我来说很容易。我确信我的代码是一场噩梦，但我正在学习:)

我能够创建一个表单并将我的数据发布到没有Ajax的Symfony（通过URL发布）但是，当我尝试通过ajax（完全相同的形式）执行它时，我得到400的json响应。我知道它与Symfony无法理解进入它的帖子数据有关。在我的控制器操作中，它处理正常，直到它到达“if（$ form-＆gt; isValid（））”部分，这是Symfony不喜欢我的表单的地方。

这是我的jQuery ajax通过表单发送帖子数据：

 $(document).ready(function() {
        $("#myForm").submit(function(){

            // My form
            var $form = $(this).closest("#myForm");

            // If valid
            if($form){

                // The url where the form is being posted via ajax
                var url = $("#myForm").attr("action");

                // Post the data
                $.post(url,{
                    type: "POST",
                    data: $form.serialize(), // serializing the data being sent 
                    cache: false
                },function(data){

                    // If valid
                    if(data.responseCode == 200 ){
                        alert(data.responseCode);
                    }
                    else if(data.responseCode==400){
                        alert(data.responseCode);
                    }else{
                        alert("bad response all together...");
                    }
                }); 
            }
            return false;
        });
    });

这是我通过Symfony表单构建器创建的表单：

 <form novalidate class="form-horizontal" id="myForm" action="/symfonydev/web/app_dev.php/warehouse/ajax/insert/" method="POST" >
 <input type="text" id="warehouse_name" name="warehouse[name]" required="required" placeholder="Location Name" class="input-block-level" value="" />
 <input type="text" id="warehouse_address" name="warehouse[address]" required="required" placeholder="Address" class="input-block-level" value="" />
 <input type="text" id="warehouse_city" name="warehouse[city]" required="required" placeholder="City" class="input-block-level" value="" />
 <input type="text" id="warehouse_state" name="warehouse[state]" required="required" placeholder="State" class="input-block-level" value="" />
 <input type="text" id="warehouse_zip" name="warehouse[zip]" required="required" placeholder="Zip" class="input-block-level" value="" />
 <input type="text" id="warehouse_email" name="warehouse[email]" required="required" placeholder="Email Address" class="input-block-level" value="" />
 <input type="text" id="warehouse_phone" name="warehouse[phone]" required="required" placeholder="Phone" class="input-block-level" value="" />
 <input type="text" id="warehouse_fax" name="warehouse[fax]" required="required" placeholder="fax" class="input-block-level" value="" />
 <input type="hidden" id="warehouse__token" name="warehouse[_token]" value="5352017c711a3a9d87ca9158334b32ab9f1dd3af" />
 <input type="submit" value="Send" />
 </form>

也许是因为表单变量的序列化字符串是url编码的，Symfony是不是能够解析这些变量以将它们持久化到db？当ajax通过序列化字符串发布数据时，字段名称的URL编码如下。

 warehouse%5Bname%5D=Test&warehouse%5Baddress%5D=Test&warehouse%5Bcity%5D=Test&warehouse%5Bstate%5D=Test&warehouse%5Bzip%5D=Test&warehouse%5Bemail%5D=Test&warehouse%5Bphone%5D=Test&warehouse%5Bfax%5D=Test&warehouse%5B_token%5D=5352017c711a3a9d87ca9158334b32ab9f1dd3af

这是我的控制器动作：

 public function ajaxinsertAction(Request $request)
{

    // Get user's account
    $account = $this->getUser()->getAccount();

    // Warehouse form
    $warehouse = new Warehouse();
    $form = $this->createForm(new WarehouseType(), $warehouse);

    if ($request->isMethod('POST')) {

        // Get the Account of this user and set it on the warehouse being created.
        $account = $this->getUser()->getAccount();
        $warehouse->setAccount($account);
        $form->bind($request);

        if ($form->isValid()) {

            $em = $this->getDoctrine()->getManager();
            $em->persist($warehouse);
            $em->flush();

            $return = array("responseCode"=>200, "response"=>"Valid");
            $return = json_encode($return); // json encode the array
            return new Response($return,200,array('Content-Type'=>'application/json'));
            die;

        }else{

            $return = array("responseCode"=>400, "response"=>"Invalid");
            $return = json_encode($return); // json encode the array
            return new Response($return,400,array('Content-Type'=>'application/json'));
            die;
        }
    }
}

感谢您的帮助！

任何人都有关于Symfony和Ajax的好教程吗？我想了解更多关于这个框架的内容，除了Symfony的文档之外似乎只是一些稀疏的信息（虽然它们很少有关于ajax的文档，但它们相当不错）。

干杯！

Answer 1

对于在漫长的一天之后搜索类似问题的答案并且我被解雇的人。

我处于非常类似的情况。问题是你提交空数据......你提交表单......然后你拉到DOM。

var $form = $(this).closest("#myForm");

应该只是

var form=$(this); 

$.post(url, form.serialize(), function(data) {

});

Answer 2

您是否尝试过手动传递值？

 ...
               $.post(url,{
                    type: "POST",
                    "warehouse[city]": $('#warehouse_city').val(),
                    "warehouse[state]" : $('#warehouse_state').val(),
                    ...
                    cache: false
                }
                ....

Symfony2 Ajax jQuery

2 个答案: