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
}
发布于 2018-02-26 12:43:23
Ajax默认是异步的,如果不进行同步调用,就无法从回调中返回值。
应该向success:
处理程序,并将程序逻辑放在那里。
发布于 2018-02-26 14:16:03
var pinNumber = $.ajax({
type: "POST",
url: "data.php",
data: "request_type=generator",
async: false
}).responseText;
jQuery('.pin_generated_table').append(cardNumber+' = '+pinNumber+' ');
https://stackoverflow.com/questions/-100007434
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