1

Below Json data which fetch to mysql . Problem. In JSON data have opt1,opt2,opt3......opt7.

Below I am gettings opt1..7 values by for loop but I got each opt1...7 value undefined. Why?

JSON data:

{"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 code:

$.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);
   }

 }
}); 
mplungjan
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shiv pratap
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1 Answers1

1

When you want to access an objects property via a variable, you'll have to change your syntax a bit:

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

Otherwise, when you do obj.optv you're trying to access an optv property of the object which does not exist in your case.