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Javascript ajax JSONP始终返回未捕获的语法错误:意外标记:_Javascript_Jquery_Json_Ajax - Fatal编程技术网

Javascript ajax JSONP始终返回未捕获的语法错误:意外标记:

Javascript ajax JSONP始终返回未捕获的语法错误:意外标记:,javascript,jquery,json,ajax,Javascript,Jquery,Json,Ajax,我正在向远程api发送一个ajax请求 这是我在浏览URL时收到的响应: { "organizations":[], "displayName":"Asaf Nevo", "roles":[ { "name":"provider", "id":"106" } ], "app_id":"app_id", "email":"email", "id":

我正在向远程api发送一个ajax请求

这是我在浏览URL时收到的响应:

{ 
  "organizations":[],
  "displayName":"Asaf Nevo",
  "roles":[
            {
              "name":"provider",
              "id":"106"
            }
           ],
  "app_id":"app_id",
  "email":"email",
  "id":"id"
}
现在api位于不同的来源,因此我需要在调用中使用JSONP:

$.ajax(
    {
        url: KeyrockConfig.defaultInstanceUrl + uri,
        type: method,
        dataType: "jsonp",
        data: params,
        success: function (result, status, xhr) {
            callback(result, status, xhr);
        },
        //TODO consider having a different callback for error
        error: function (xhr, status, error) {
            callback(error, status, xhr);
        }
    }
);
但我不断得到
未捕获的语法错误:意外标记:

我试着做:

$.ajax(
    {
        url: KeyrockConfig.defaultInstanceUrl + uri,
        type: method,
        dataType: "jsonp",
        jsonp:false,
        jsonpCallback: function (response) {
            alert(response);
        },

        data: params,
        success: function (result, status, xhr) {
            callback(result, status, xhr);
        },
        //TODO consider having a different callback for error
        error: function (xhr, status, error) {
            callback(error, status, xhr);
        }
    }
);

但警报从未发出过。我做错了什么?

我认为您需要将json格式化为字符串或数组。要测试这一点,请尝试这样的警报
alert(JSON.stringify(your_JSON_obj))我认为您需要将json格式化为字符串或数组。要测试这一点,请尝试这样的警报
alert(JSON.stringify(your_JSON_obj))