Question

获取空指针异常JSON AJAX / Servlet 调用以下servlet的Ajax post调用

 $.ajax({
            url: '/bin/commersenext/assetUpload',
            type: 'POST',
            contentType:'application/json',
            data: JSON.stringify(ob),
            dataType: 'json',
          success:function(msg){
            alert("data"+msg);
              console.log(msg);
            }

        });

Servlet代码 - 发生nullpointerexception的地方：

    JSONParser parser = new JSONParser();
            JSONArray jsonArray = null;
            String tabledata = request.getParameter("data");


                  try{
                     Object obj = parser.parse(tabledata);
                     JSONArray array = (JSONArray)obj;


                     System.out.println(array.get(1));

Answer 1

尝试将此条件放入您的代码中。

if(data.status == 0){
    var value = JSON.parse(data.data);
    console.log("data saved");
    return;
}

Answer 2

    BufferedReader br = new BufferedReader(new InputStreamReader(request.getInputStream()));
    String json = "";
    if(br != null){
        json = br.readLine();
        System.out.println(json);
    }

它现在正在工作.. !!!

进行Ajax调用以发送json。但是在servlet上它会变为空

2 个答案: