<分区>
如何从 jquery ajax 返回值“pinNumber”,以便我可以将它附加到 ajax 之外。这是我的代码
var x = pinLast + 1; for(i=x;i<=pinMany;i++) { var i = x++; var cardNumber = i.toPrecision(8).split('.').reverse().join(''); var pinNumber = ''; jQuery.ajax({ type: "POST", url: "data.php", data: "request_type=generator", async: false, success: function(msg){ var pinNumber = msg; return pinNumber; //pin number should return } }); jQuery('.pin_generated_table').append(cardNumber+' = '+pinNumber+'
'); // the variable pinNumber should be able to go here }
有不懂的可以问我。。^^谢谢