显示变量不工作在ajax响应

Question

success: function (response) { 
    var paid = "PURCHASED"; 
    var notpaid = "PREMIUM"; 
    $.each(response['courceResults'], function(k, cource) { 
courceResultsData +='<tr><td>' 
    if(cource.membership_chosen == 3){ 
    if ($.inArray(cource.id , mystr) != -1) { /*alert(paid);*/ paid } 

在上面的行中有一个错误，当我警告它的价值 正确;但是当输入变量或保持一个字符串“PURCHASED”if if 条件它不能正常工作我解决了这个连接..？

 else{ notpaid } 
    '</td></tr>'; 
    }); 

Answer 1

一些校正代码： -

success: function (response) { 
    var paid = "PURCHASED"; 
    var notpaid = "PREMIUM"; 
    $.each(response.courceResults, function(k, cource) { //i think it's response.courceResults not response['courceResults'] check and change accordingly 
     var courceResultsData ='<tr><td>'; // missed ; 
     if(cource.membership_chosen == 3){ 
      if ($.inArray(cource.id , mystr) != -1){ // from where the hell mystr is coming? check yourself 
       courceResultsData +=paid; // forgot concatenation 
      } else{ 
       courceResultsData +=notpaid ; // forgot concatenation and missed ; 
      } 
      courceResultsData +='</td></tr>';//forgot concatenation 
     } // missed 
    } // missed 
    console.log(courceResultsData); //check the final output 
} // missed