如何优化（最小化）jQuery的AJAX调用 [英] How to optimize (minimize) jQuery AJAX calls

问题描述

我有超过50 AJAX从我的code不同的功能要求。所有这些调用有不同的数据/地址/回调PARAMS类似的结构：

I have over 50 AJAX calls from different functions of my code. All these calls have a similar structure with different data/url/callback params:

var jqXHR = $.post('/dba/port.php', {
        mode: "del_wallfunds",
        pdata: cdata,
        wname: wName
    },
    function (data) {}, "json")
    .done(function (data) {
        var msg = data.msg;
        if (msg.indexOf("Error") == -1) {
            alertify.success(msg);
            delSelected(selGroup);
        } else {
            alertify.error(msg);
        }
    })
    .fail(function () {
        alertify.error("Error .....");
    });

我在想怎么写，将返回VAR jqXHR减少的code总规模的函数。这是不是一个问题，通过所有的静态变量，如URL，错误字符串等的但问题是，在.done所有的回调函数都不同，我不知道如何通过这些回调作为变量。

I am thinking how to write a function that would return that var jqXHR to minimize the total size of the code. It is not a problem to pass all static variables like URL, error strings etc. But the problem is that all callback functions on ".done" are different and I don't know how to pass these callback functions as variables.

一种方法是调用.done一个通用的功能，并通过一个开关变量的函数，但它似乎并没有成为一个优雅的解决方案。

One way would be to call a single "universal" function on .done and pass a "switch" variable to that function, but it doesn't seem to be an elegant solution.

在一些优雅的方式如何将它有什么建议？

Any suggestions how to it in some elegant way?

感谢

推荐答案

无论哪种呼唤你的功能，当通过完成回调函数作为参数：

Either pass the done callback function as an argument when calling your function:

function ajaxCall(url, data, doneCallback) {
    return $.post(url, data, doneCallback, "json").fail(...);
    // or
    return $.post(url, data, function() {}, "json").done(doneCallback).fail(...);
}

var jqXhr = ajaxCall('yoururl.php', {key: 'value'}, function(data) {
    // do something
});

或者从函数返回的jqXhr对象，并指定完成回调，则：

function ajaxCall(url, data) {
    return $.post(url, data, function() {}, "json").fail(...);
}

var jqXhr = ajaxCall('yoururl.php', {key: 'value'});
jqXhr.done(function(data) {
    // do something
});

或者切换到使用 jQuery.ajax（）代替，并通过在整个方案中的对象：

Alternatively switch to using jQuery.ajax() instead, and pass the entire options object in:

function ajaxCall(options) {
    return $.ajax(options).fail(...);
}

var jqXhr = ajaxCall({
    url: 'yoururl.php',
    data: {key: 'value'},
    dataType: 'json'
});
jqXhr.done(function(data) {
    // do something
});

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