我有:
<input type="file" id="f" name="f" onchange="c()" multiple />
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每次用户选择文件时,我都会通过将每个元素压f.files入数组来构建所有选定文件的列表:
var Files = [];
function c() {
for (var i=0;i<this.files.length;i++) {
Files.push(this.files[i]);
};
}
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在提交表单时,f.files仅包含上次选择操作中的项目,因此我将需要更新自己累积f.files的FileList项目列表:
const upload=document.getElementById("f");
upload.files = files;
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但是第二行给出:
未捕获到的TypeError:无法在'HTMLInputElement'上设置'files'属性:提供的值不是'FileList'类型的。
我为它分配一个数组并不高兴。如何FileList从FileList先前收集的元素列表中构造一个对象?
附带问题:我认为Javascript使用动态类型。为什么在这里抱怨类型错误?
End*_*ess 14
就像你说的
您只能使用FileList设置文件,不幸的是FileList是不可构造的或不可更改的,但是有一种方法可以使文件列表全面
// Used for creating a new FileList in a round-about way
function FileListItem(a) {
a = [].slice.call(Array.isArray(a) ? a : arguments)
for (var c, b = c = a.length, d = !0; b-- && d;) d = a[b] instanceof File
if (!d) throw new TypeError("expected argument to FileList is File or array of File objects")
for (b = (new ClipboardEvent("")).clipboardData || new DataTransfer; c--;) b.items.add(a[c])
return b.files
}
var files = [
new File(['content'], 'sample1.txt'),
new File(['abc'], 'sample2.txt')
];
fileInput.files = new FileListItem(files)Run Code Online (Sandbox Code Playgroud)
<input type="file" id="fileInput" multiple />Run Code Online (Sandbox Code Playgroud)
这样做将触发更改事件,因此您可能需要打开和关闭更改事件侦听器
sup*_*ary 10
You can't modify a Filelist, but you can create a new one using a DataTransfer object, and if you wish you can copy your data into it to create a duplicate with the specific change you want to make.
let list = new DataTransfer();
let file = new File(["content"], "filename.jpg");
list.items.add(file);
let myFileList = list.files;
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You can then set it as the file attribute of the DOM node:
fileInput.files = myFileList;
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If you wished, you could iterate over your old FileList, copying files to the new one.
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