【问题标题】:php can not parse json data to ajaxphp无法将json数据解析为ajax
【发布时间】:2018-10-23 13:40:47
【问题描述】:

这个问题在stackoverflow中很常见。对于像我这样的问题,我去过这个论坛提供的太多解决方案。但它没有帮助。所以我发布它。请好心回答我。

问题: 我正在尝试从 cakephp 控制器函数发送一个整数来查看有 ajax 调用。 对于我从这里和其他论坛尝试的所有解决方案,我不断收到此错误。

“SyntaxError: JSON 中位置 0 的意外标记 a”

请告诉我查看我的代码有什么问题: 我可以清楚地看到我在 php 中获取 json 格式的数据,而我在回显但不是在主 Ajax 调用中。

Console says :the result is {"data":1}

控制器:

public function addit()
    {
        $mycount      = 1;
        $responseJSON = array(
            'data' => $mycount
        );
        //$responseJSON = array('status' =>'true', 'result'=>$arr);
        header('content-type:application/json');
        $response = json_encode($responseJSON);
        echo $response;
    }

ajaxcall 在视图中:

$(document).ready(function(){
        $('#bn_cart').click(function(event){
          //alert('clicked');
          var form_data = $(this).serialize();

          var id = $('#id').val();
          alert("your item id is "+ id);
          var csrfToken = <?php echo(json_encode($this->request->getParam('_csrfToken'))) ?>;
               //alert("your form data "+csrfToken);
              event.preventDefault();
              $.ajax({
                headers: {
                      'X-CSRF-Token': csrfToken
                    },
                  url:'../addit',
                  type:'POST',

                  data: { id : id },
                  dataType:'json',

              success:function(xhr, response){
                    var respons = response;
                    console.log("conosle success says "+ (respons.result));
                    alert("success"+respons.result);

                },
              error:function(xhr, e,etype,response){
                    //alert("<br>error<br>"+ error.responseText.message);
                    alert("response = "+ response +"xhr = "+ xhr + "  e = " + e + "  etype = "+ etype);
                    console.log(" response =" + response + "error ="+ e +"xhr = "+ xhr + "  etype = "+ etype );
                    //  $("#result").html(error.Message);
                    // alert('error ='+(error.Message));
                  }
          });

        });
    });


Network>Header:
Request URL: http://localhost/shoppingCart/products/addit
Request Method: POST
Status Code: 200 OK
Remote Address: [::1]:80
Referrer Policy: no-referrer-when-downgrade
Cache-Control: no-store, no-cache, must-revalidate
Connection: Keep-Alive
Content-Length: 583
Content-Type: text/html; charset=UTF-8
Date: Tue, 23 Oct 2018 14:10:08 GMT
Expires: Thu, 19 Nov 1981 08:52:00 GMT
Keep-Alive: timeout=5, max=99
Pragma: no-cache
Server: Apache/2.4.29 (Win32) OpenSSL/1.0.2n PHP/7.1.15
X-DEBUGKIT-ID: 182187f0-546d-48d4-9e5a-6746a40dba64
X-Powered-By: PHP/7.1.15
Accept: application/json, text/javascript, */*; q=0.01
Accept-Encoding: gzip, deflate, br
Accept-Language: en-US,en;q=0.9
Connection: keep-alive
Content-Length: 4
Content-Type: application/x-www-form-urlencoded; charset=UTF-8
Cookie: csrfToken=0a8a1f6e98fe8274e80f9bdcb3ba5df66a5af4296126302d3e79bf44e856ed720438947bb93f041f772ac1e39d083aa2d88c5159697c9843a8b04eace893260b; CAKEPHP=mhphclr8cuvacrlotbit45dd3l; csrftoken=t0p47S5P7NBcwGGQ9sfuNGLi5JJDkll8ifuCWhG3W6MRSIewe9GtRNjanPUqms54
Host: localhost
Origin: http://localhost
Referer: http://localhost/shoppingCart/products/view/2
User-Agent: Mozilla/5.0 (Windows NT 6.1; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/69.0.3497.100 Safari/537.36
X-CSRF-Token: 0a8a1f6e98fe8274e80f9bdcb3ba5df66a5af4296126302d3e79bf44e856ed720438947bb93f041f772ac1e39d083aa2d88c5159697c9843a8b04eace893260b
X-Requested-With: XMLHttpRequest


Network>>Response : 
{"data":1}

【问题讨论】:

  • echo "&lt;br&gt;the result is ". $response; 将使响应无效 JSON,因为它不是有效的 JSON。只需回显$response
  • 从您的回声中删除“
  • @JonStirling ..感谢您的回复。不幸的是,它还没有帮助..同样的错误:response = undefined xhr = [object Object] e = parsererror etype = SyntaxError: Unexpected token a in JSON at position 0
  • @MasivuyeCokile 感谢您的回复......但它还不能帮助我
  • @vins 尝试在您的 ajax 设置中添加 encode:true 并硬刷新页面

标签: javascript php json ajax cakephp


【解决方案1】:

您需要返回以 json 编码的响应。您正在尝试回显无效的非 json 编码响应。

public function addit()
{
    $mycount      = 1;
    $responseJSON = array(
        'data' => $mycount
    );
    //$responseJSON = array('status' =>'true', 'result'=>$arr);
    header('content-type:application/json');
    return json_encode($responseJSON);
}

Ajax 调用:

success:function(data, status, jqxhr)
            var respons = data;
            console.log("conosle success says "+ (respons.result));
            alert("success"+respons.result);

        },

【讨论】:

  • 你需要向 OP 解释你做了什么而他不做
  • @Bernard Pagoaga 感谢您的回复。我已经尝试了很多次,它给出了不同的错误:response = undefinedxhr = [object Object] e = error etype = Internal Server Error(我还可以在控制台中看到数据)
  • 成功回调也将数据作为第一个参数,而不是 xhr 对象:api.jquery.com/jquery.ajax.
猜你喜欢
  • 2015-03-15
  • 2017-08-21
  • 1970-01-01
  • 1970-01-01
  • 1970-01-01
  • 1970-01-01
  • 2011-11-30
  • 1970-01-01
  • 1970-01-01
相关资源
最近更新 更多