Question

获取到mysql的Json数据。问题。在JSON数据中有opt1，opt2，opt3 ...... opt7。

下面我通过for循环获得opt1..7值，但我得到每个opt1 ... 7值undefined。为什么呢？

JSON数据：

{"qz_id":"15","qzn_id":"14","ins_user_id":"1","qnumber":"8","quiz_question":"jdfkjf dkfj dkjfkd kid jfkjdk fdkjfk djfkdjfkd kfkd fk djfkd jfk d",
            "opt1":"jkj","opt2":"kjkj","opt3":"kjkj","opt4":"kjkjkjk","opt5":"","opt6":"","opt7":"","right_opt":"opt1","qns_explanation":"",
            "Answeroption":"0","q_mark":"0","qns_neq_mark":"0",
            "qns_DifficultyLevel":"1","cdate":"2017","stt":"In-active","del_stt":"No","quiz_name":"ssdsds demo","exam_id":"5","sec_id":"6"}

AJAX代码：

$.ajax({
 url: "quiz_process.php",
 type: "POST",
 data: dataString,
 cache: false,
 success: function(data) {  
   var obj=JSON.parse(data);
   /*opt a*/
   var optv="";
   var optionv="";
   for(var i=1;i<=7;i++) {
     optv='opt'+i;
     optionv = optv+'::'+obj.optv;
     console.log(optionv);
   }

 }
});

Answer 1

如果要通过变量访问对象属性，则必须稍微更改语法：

optionv = optv+'::'+obj[optv];

否则，当您执行obj.optv时，您尝试访问对象中不存在的optv属性。

如何获得JSON obj的动态值？

1 个答案: