Question

我注意到请求的文件大小会影响ajax调用的响应时间。因此，如果我发出3个不同大小的文件的ajax GET请求，它们可能以任何顺序到达。我想要做的是保证在将文件附加到DOM时的排序。

如何设置队列系统，以便在我触发A1-＆gt; A2-＆gt; A3时。我可以保证按顺序将它们称为A1-> A2-> A3。

例如，假设A2在A1之前到达。我希望行动等到A1的到达和加载。

一个想法是使用定时回调创建状态检查器

// pseudo-code
function check(ready, fund) {
    // check ready some how
    if (ready) {
        func();
    } else {
        setTimeout(function () {
            check(ready, fund);
        }, 1); // check every msec
    }
}

但这似乎是一种资源繁重的方式，因为我每隔1毫秒触发相同的函数，直到资源被加载。

这是完成此问题的正确途径吗？

Answer 1

使用1毫秒定时回调的状态检查器 - 但这似乎是一种资源繁重的方式;这是完成这个问题的正确途径吗？

没有。你应该看一下Promises。这样，您可以轻松地将其表达为：

var a1 = getPromiseForAjaxResult(ressource1url);
var a2 = getPromiseForAjaxResult(ressource2url);
var a3 = getPromiseForAjaxResult(ressource3url);

a1.then(function(res) {
    append(res);
    return a2;
}).then(function(res) {
    append(res);
    return a3;
}).then(append);

例如，jQuery的.ajax函数实现了这一点。

Answer 2

您可以尝试这样的事情：

var resourceData = {};
var resourcesLoaded = 0;

function loadResource(resource, callback) {
    var xhr = new XMLHttpRequest();
    xhr.onload = function() {
        var state = this.readyState;
        var responseCode = request.status;

        if(state == this.DONE && responseCode == 200) {
            callback(resource, this.responseText);
        }
    };

    xhr.open("get", resource, true);
    xhr.send();
}

//Assuming that resources is an array of path names
function loadResources(resources) {
    for(var i = 0; i < resources.length; i++) {
        loadResource(resources[i], function(resource, responseText) {

            //Store the data of the resource in to the resourceData map,
            //using the resource name as the key. Then increment the
            //resource counter.
            resourceData[resource] = responseText;
            resourcesLoaded++;

            //If the number of resources that we have loaded is equal
            //to the total number of resources, it means that we have
            //all our resources.
            if(resourcesLoaded === resources.length) {
                //Manipulate the data in the order that you desire.
                //Everything you need is inside resourceData, keyed
                //by the resource url. 
                ...
                ...
            }                    
        });
    }
}

如果某些组件必须之前加载并执行（如某些JS文件）其他组件，您可以排队AJAX请求，如下所示：

function loadResource(resource, callback) {
    var xhr = new XMLHttpRequest();
    xhr.onload = function() {
        var state = this.readyState;
        var responseCode = request.status;

        if(state == this.DONE && responseCode == 200) {
            //Do whatever you need to do with this.responseText
            ...
            ...

            callback();
        }
    };

    xhr.open("get", resource, true);
    xhr.send();
}

function run() {
    var resources = [
        "path/to/some/resource.html",
        "path/to/some/other/resource.html",
        ...
        "http://example.org/path/to/remote/resource.html"
    ];

    //Function that sequentially loads the resources, so that the next resource 
    //will not be loaded until first one has finished loading. I accomplish
    //this by calling the function itself in the callback to the loadResource 
    //function. This function is not truly recursive since the callback 
    //invocation (even though it is the function itself) is an independent call 
    //and therefore will not be part of the original callstack.
    function load(i) {
        if (i < resources.length) {
            loadResource(resources[i], function () {
                load(++i);
            });
        }
    }
    load(0);
}

这样，在上一个文件加载完成之前，不会加载下一个文件。

如果您无法使用任何第三方库，则可以使用我的解决方案。但是，如果您执行Bergi suggested并使用Promises，您的生活可能会更容易。

Answer 3

无需每毫秒调用check()，只需在xhr onreadystatechange中运行即可。如果您提供更多代码，我可以进一步解释。

Answer 4

我会有一个要执行的函数队列，每个函数都会在执行之前检查先前的结果是否已完成。

var remoteResults[] 

function requestRemoteResouse(index, fetchFunction) {
  // the argument fetchFunction is a function that fetches the remote content
  // once the content is ready it call the passed in function with the result.
  fetchFunction(
    function(result) { 
      // add the remote result to the list of results
      remoteResults[index] = result
      // write as many results as ready.
      writeResultsWhenReady(index);
    });
}

function writeResults(index) {
  var i;
  // Execute all functions at least once
  for(i = 0; i < remoteResults.length; i++) {
    if(!remoteResults[i]) {
      return;
    }
    // Call the function that is the ith result
    // This will modify the dom.
    remoteResults[i]();
    // Blank the result to ensure we don't double execute
    // Store a function so we can do a simple boolean check.
    remoteResults[i] = function(){}; 
  }
}

requestRemoteResouse(0, [Function to fetch the first resouse]);
requestRemoteResouse(1, [Function to fetch the second resouse]);
requestRemoteResouse(2, [Function to fetch the thrid resouse]);

请注意，为了简单起见，目前这是O（n ^ 2），如果您在具有hasRendered属性的remoteResults的每个索引处存储了一个对象，它会变得更快但更复杂。然后你只会扫描回来，直到找到一个尚未发生的结果或一个已经渲染的结果。

概念 - 为异步资源设计可折叠队列

4 个答案: