jQuery ajax 返回值 [英] jQuery ajax return value

问题描述

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

<前>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({类型:POST"，网址:data.php"，数据:request_type=generator"，异步:假，成功:功能(味精){var pinNumber = msg;返回 pinNumber;//引脚号应该返回}});jQuery('.pin_generated_table').append(cardNumber+' = '+pinNumber+'
');//变量 pinNumber 应该可以到这里}

不明白的可以问我..^^谢谢

解决方案

AJAX 默认是异步的，如果不进行同步调用，你就不能从回调中返回一个值，你几乎肯定不想这样做.

>

您应该为 success: 处理程序提供一个真正的回调函数，并将您的程序逻辑放在那里.

How can I return the value "pinNumber" from jquery ajax so I can append it outside of the ajax. Here is my code

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

    }

Please ask me if you don't understand.. ^^ thanks

解决方案

AJAX is asynchronous by default, you cannot return a value from the callback without making a synchronous call, which you almost certainly don't want to do.

You should supply a real callback function to the success: handler, and put your program logic there.

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