1

我已经完成了一些代码来将多个文件从浏览器上传到服务器，同时也显示了进度条。XHR2 发送调用是异步的。问题是我想在所有 XHR2 发送调用完成后调用一些函数。

这是我的代码的一个非常简化的片段：

<input id="upload" multiple="multiple" />

<script>
var files = document.getElementById("upload").files;
for (var i = 0, length = files.length; i < length; i++) {
   var file = files[i];
   var uploadFunc = function (arg) {
      return function () {
         processXHR(arg);
      };
   }(file);      
   uploadFunc();
}

function processXHR(file) {
   var normalizedFileName = getNormalizedName(file.name);
   var url = 'upload_file/';
   var formData = new FormData();
   formData.append("docfile", file);
   var xhr = new XMLHttpRequest();
   var eventSource = xhr.upload;
   eventSource.addEventListener("progress", function (evt) {
      var position = evt.position || evt.loaded;
      var total = evt.totalSize || evt.total;
      console.log('progress_' + normalizedFileName);
      console.log(total + ":" + position);
      $('#progress_' + normalizedFileName).css('width', (position * 100 / total).toString() + '%');
   }, false);
   eventSource.addEventListener("loadstart", function (evt) {
      console.log('loadstart');
   }, false);
   eventSource.addEventListener("abort", function (evt) {
      console.log('abort');
   }, false);
   eventSource.addEventListener("error", function (evt) {
      console.log('error');
   }, false);
   eventSource.addEventListener("timeout", function (evt) {
      console.log('timeout');
   }, false);
   xhr.onreadystatechange = function (evt) {
      if (xhr.readyState == 4) {
         console.log('onreadystatechange: File uploaded.');
      }
   };
    xhr.open('post', url, true);
    xhr.send(formData);
}
</script>

每当“onreadystatechange：文件上传”。在控制台中打印，我知道其中一个文件已完成上传。但我无法编写任何会说“所有文件已上传”的代码。

谢谢你的帮助。 我也在使用 jquery，以防有人有使用 jquery 的解决方案。

4

2 回答 2

0

你知道你打了多少电话吗？如果是这样，至少ajax回调，你检查计数。就像是

if (ajaxCallCount === ajaxCallsLength) {
 // Do stuff involving post all ajax calls
} else {
 ajaxCallCount++;
}
于 2012-06-30T03:40:52.023 回答
0

You will need a counter and setInterval method in your case, try this:

<input id="upload" multiple="multiple" />

<script>
var files = document.getElementById("upload").files;
var counter = 0;

for (var i = 0, length = files.length; i < length; i++) {
   var file = files[i];
   var uploadFunc = function (arg) {
      return function () {
         processXHR(arg);
      };
   }(file);      
   uploadFunc();
}

var interval = setInterval(function() {
    if(counter == files.length) {
        clearInterval(interval);
        console.log("All Files uploaded!");
    }
}, 100);

function processXHR(file) {
   var normalizedFileName = getNormalizedName(file.name);
   var url = 'upload_file/';
   var formData = new FormData();
   formData.append("docfile", file);
   var xhr = new XMLHttpRequest();
   var eventSource = xhr.upload;
   eventSource.addEventListener("progress", function (evt) {
      var position = evt.position || evt.loaded;
      var total = evt.totalSize || evt.total;
      console.log('progress_' + normalizedFileName);
      console.log(total + ":" + position);
      $('#progress_' + normalizedFileName).css('width', (position * 100 / total).toString() + '%');
   }, false);
   eventSource.addEventListener("loadstart", function (evt) {
      console.log('loadstart');
   }, false);
   eventSource.addEventListener("abort", function (evt) {
      console.log('abort');
   }, false);
   eventSource.addEventListener("error", function (evt) {
      console.log('error');
   }, false);
   eventSource.addEventListener("timeout", function (evt) {
      console.log('timeout');
   }, false);
   xhr.onreadystatechange = function (evt) {
      if (xhr.readyState == 4) {
         counter += 1;
      }
   };
    xhr.open('post', url, true);
    xhr.send(formData);
}
</script>
于 2012-06-30T04:10:32.040 回答