本文介绍了显示变量在ajax响应中不起作用的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
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,当我警告即将到来的值时出现错误正确的 ;但是当类型变量或在以下情况中保留字符串"PURCHASED"时条件无法正常工作,我可以解决此串联问题..?
In the above line there is an error when i alert the value it is comingcorrect ; but when type variable or kept a string "PURCHASED" in ifcondition it is not working fine i resolve this concatenation..?
else{ notpaid }
'</td></tr>';
});
推荐答案
代码中的一些更正:-
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
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