节点/多特 |无法根据字段名称获取 sperarte 文件名

Question

我创建了一个与 multer 结合使用的 Express 应用程序，以在我的 Node.js 应用程序中上传项目

我尝试做的是选择比方说：

Upload 1 - that has a fieldname of upfile1
Upload 2 - that has a fieldname of upfile2
Upload 3 - that has a fieldname of upfile3

基本上，我需要单独选择我的多上传应用程序的每个上传文件名项。对于不同的任务，需要在应用程序中以不同的方式处理每个上传。让我们使用 console.log 为例，我需要执行以下操作：

console.log(req.body.upfile1.filename);
console.log(req.body.upfile2.filename);
console.log(req.body.upfile3.filename);

以选择在应用程序中处理的不同项，这些字段名称使用 Name 属性在我的视图中定义。

下面是我的代码

视图 [索引.html]

<form id="app-form" method="POST" class="fileupload" method="post" action="app" enctype="multipart/form-data">
<h1>Multi File Uploads</h1>
<input type="file" name="upfile1" value="">
<input type="file" name="upfile2" value="">
<input type="file" name="upfile3" value="">
<input type="submit" />
</form>

NodeJS [app.js]

app.get("/", function(req, res) {
res.sendFile(__dirname + "/index.html");
});
app.post("/app", upload.any(), function(req, res) {
let files = req.files;
files.forEach(file => {
console.log(file.filename);
});
res.send(req.files);
res.end();
});

帮助将不胜感激，谢谢！

Answer 1

标记中的输入元素必须包装在表单中(它们可能已经包装在表单元素中，但未显示在您的问题中(。您还应该将表单的 enctype 属性设置为多部分/表单数据。

<form method="post" enctype="multipart/form-data" action="/upload">
<input type="file" name="upfile1">
<input type="file" name="upfile2">
<input type="file" name="upfile3">
<input type="submit" value="Submit">
</form>

完成后，您可以配置 multer 并创建一个路由来处理文件上传：

const upload = multer({
dest: path.join(__dirname, './upload') // You might want to change this according to your preferences
// Since you're using any(), you might want to set fileFilter to control which files should be uploaded. See: https://github.com/expressjs/multer#filefilter
});
const findFileByFieldname = (files, fieldname) => {
return files.find(file => file.fieldname === fieldname) || {};
}
app.post("/upload", upload.any(), (req, res) => {     
const upfile1Filename = findFileByFieldname(req.files, 'upfile1').filename;
const upfile2Filename = findFileByFieldname(req.files, 'upfile2').filename;
const upfile3Filename = findFileByFieldname(req.files, 'upfile3').filename;
res.json({
upfile1Filename,
upfile2Filename,
upfile3Filename,
});
});
// Example response (Node v8.11.4, Express v4.16.3, Multer v1.3.1)
// {"upfile1Filename":"360726b532a01b0e31832f067b5922c8","upfile2Filename":"144e1298437afb51f36eb37c77814650","upfile3Filename":"4c908da20e770130377e4006db945af6"}