Question

我尝试使用json对象从数据库中检索数据。但是当我调用servlet时，Jquery将返回SyntaxError: JSON.parse: unexpected non-whitespace character after JSON data此错误仅在响应包含更多数据时显示。

我的脚本是：

    $.ajax({
       type: "GET",
        url: "VComment",
        data:'comm='+encodeURIComponent(comm)+'&'+'data-id='+encodeURIComponent(dataid)+'&'+'data-alid='+encodeURIComponent(dataalid),
        dataType: "json",
        success: function( data, textStatus, jqXHR) 
        {
            if(data.success)
            {
                    var newcommhtml = '<div id="c0'+thecid+'" class="cnew clearfix"> <section class="c-author">';
                    newcommhtml = newcommhtml + '<h3>Anonymous</h3>';
                    newcommhtml = newcommhtml + '<span class="pubdate">'+month+' '+day+', '+year+'</span> </section>';
                    newcommhtml = newcommhtml + '<section class="c-content">';
                    newcommhtml = newcommhtml + '<img src="images/green-avatar.png" alt="avatar" width="80" height="80" class="ava">';
                    newcommhtml = newcommhtml + '<p>'+nl2br(data.commentInfo.comment)+'</p> </section></div>';

                    var thelm = "#c0"+thecid;
                    commwrap.append(newcommhtml);
                    $(thelm).hide().fadeIn('slow');

                    setTimeout(function() { $(thelm).addClass('green'); }, 800);

                    $("#comm").val("");
                    thecid++;

                    if(errorspan.html() != null) {
                        errorspan.remove();
                    }
            }

          },
     error: function(jqXHR, textStatus, errorThrown)
      {
         alert("error"+errorThrown);
         console.log("Something really bad happened " + textStatus);
      },
});

收到回复..

    {"success":true,"commentInfo":{"uname":"shyam","comment":"rreter","itemId":0}}
    {"success":true,"commentInfo":{"uname":"shyam","comment":"dfdsfdd","itemId":0}}
    {"success":true,"commentInfo":{"uname":"shyam","comment":"xzdfdsfdd","itemId":0}}
    {"success":true,"commentInfo":{"uname":"shyam","comment":"sdfsd fsdfs","itemId":0}}
    {"success":true,"commentInfo":{"uname":"shyam","comment":"sdsd","itemId":0}}
    {"success":true,"commentInfo":{"uname":"shyam","comment":"dd","itemId":0}}
    {"success":true,"commentInfo":{"uname":"shyam","comment":"dddf","itemId":0}}

Servlet代码：

      while(rs.next()){
            Commenter comment = new Commenter();
            comment.setUname(rs.getString("uname").trim());
            comment.setComment(rs.getString("comments").trim());
            commentObj=gson.toJsonTree(comment);
            myObj.add("commentInfo", commentObj);
            out.println(myObj.toString());
            }

请有人告诉我如何解决这个问题......谢谢....

Answer 1

响应中有几个独立的JSON对象。将它们包裹在一个阵列中你会更好

[{"success":true,"commentInfo":{"uname":"shyam","comment":"rreter","itemId":0}},
{"success":true,"commentInfo":{"uname":"shyam","comment":"dfdsfdd","itemId":0}},
{"success":true,"commentInfo":{"uname":"shyam","comment":"xzdfdsfdd","itemId":0}},
{"success":true,"commentInfo":{"uname":"shyam","comment":"sdfsd fsdfs","itemId":0}},
{"success":true,"commentInfo":{"uname":"shyam","comment":"sdsd","itemId":0}},
{"success":true,"commentInfo":{"uname":"shyam","comment":"dd","itemId":0}},
{"success":true,"commentInfo":{"uname":"shyam","comment":"dddf","itemId":0}}]

并将功能更改为：

success: function( data, textStatus, jqXHR) 
    {
       for(var i = 0,len=data.length;i<len;i += 1){
        if(data[i].success)
        {
          //code
        }
       }
    }

服务器端的

只需更改为

out.println("[");
Boolean first = true
while(rs.next()){
        Commenter comment = new Commenter();
        comment.setUname(rs.getString("uname").trim());
        comment.setComment(rs.getString("comments").trim());
        commentObj=gson.toJsonTree(comment);
        myObj.add("commentInfo", commentObj);
        if(!first){
          out.print(",");
        } else {
          first = false;
        }
        out.println(myObj.toString());
        }   
out.println("]");

Answer 2

尝试此代码我认为它可以解决您的问题：

        ArrayList<JSONObject> CommArray=new ArrayList<JSONObject>();

         while(rs.next()){
            JSONObject Obj = new JSONObject();
            Obj.put("uname",rs.getString("uname").trim());     //Adds your uname to Object
            Obj.put("comment",rs.getString("comments").trim());//Adds your comment to Object
            CommArray.add(Obj);                                //Inserts your Object to ArrayList
            System.out.println(rs.getString("comments").trim());
            }    
         JSONArray arrayObj=JSONArray.fromObject(CommArray);//Converts the Array List to JSONArray
         commentObj=gson.toJsonTree(arrayObj);   //Converts the JSONArray to Jsontree
         myObj.add("commentInfo", commentObj);   //Adds the Tree to JsonObject as commentInfo Array
         out.println(myObj.toString());        //Prints the result
         rs.close();                                                              
         stmt.close();                                                            
         stmt = null;                                                             
         conn.close();                                                            
         conn = null;                                                  

     }                                                              
     catch(Exception e){}

我希望这能解决你的问题。

Answer 3

尝试暂时更改dataType并提出相同的错误请求，输出回复让我们看看有什么不对。

JSON.parse：JSON数据之后的意外非空白字符

3 个答案: