Question

大家好，我从来没有遇到过谷歌无法解决的问题...... 所以这是代码：

function loadData(url, type){
        loading_show();  
        var quality = getCheckedRadio();
        $.ajax
        ({
            type: "POST",
            url: "http://.../core.php",
            data: "url="+url +"&quality="+quality +"&type="+type,
            success: function (msg)
            {
                $("#url").ajaxComplete(function(event, request, settings)
                {
                    //loading_hide();
                    //$("input[type=button]").removeAttr("disabled");         
                    $("#url").html(msg);
                    window.onload=$("#url").fadeIn('slow');
                });
            }
        });
        return false;
}
function getmp3(id){
        var quality = getCheckedRadio();
        $.ajax
            ({
                type: "POST",
                url: "http://.../core.php",
                data: "id="+id +"&quality="+quality,
                success: function (msgs)
                {
                    $("#dload").ajaxComplete(function(event, request, settings)
                    {
                        loading_hide();
                        $("input[type=button]").removeAttr("disabled");         
                        $("#dload").html(msgs);
                        window.onload=$("#dload").fadeIn('slow');
                    });
                }
            });
}

所以，第一个函数工作得很好，core.php返回一些html数据和一个javascript：“getmp3（'bla'）;”因此它会触发第二个函数。问题是'getmp3'函数中的$ .ajax一遍又一遍地重复，它会泛滥浏览器。它做了它所拥有的，但它不会一次又一次地停止呼叫。

Answer 1

您不需要ajaxComplete函数，可以将代码直接放在success函数中。

function loadData(url, type){
        loading_show();  
        var quality = getCheckedRadio();
        $.ajax
        ({
            type: "POST",
            url: "http://.../core.php",
            data: "url="+url +"&quality="+quality +"&type="+type,
            success: function (msg)
            {
                    //loading_hide();
                    //$("input[type=button]").removeAttr("disabled");         
                    $("#url").html(msg).fadeIn('slow');
            }
        });
        return false;
}
function getmp3(id){
        var quality = getCheckedRadio();
        $.ajax
            ({
                type: "POST",
                url: "http://.../core.php",
                data: "id="+id +"&quality="+quality,
                success: function (msgs)
                {
                        loading_hide();
                        $("input[type=button]").removeAttr("disabled");         
                        $("#dload").html(msgs).fadeIn('slow');
                }
            });
}

jquery ajax调用不会一次又一次地停止重复

1 个答案: