能否请您告诉我如何在节点js中下载服务器文件
这是我的代码 Node js代码(请求代码)
var express = require('express');
var multer = require('multer');
var bodyParser = require('body-parser');
var cors = require('cors');
var app = express();
var path = require('path');
var PORT = process.env.PORT || 3000;
// use of body parser
app.use(bodyParser.urlencoded({
extended: true
}));
app.use(bodyParser.json());
app.use(cors());
app.get('/download', function(req, res){
console.log(__dirname);
var file = path.join(__dirname , '/uploads/file-1534458777206.xlsx');
console.log(file)
res.download(file); // Set disposition and send it.
});
app.listen(PORT, () => {
console.log(`App is listening to ${PORT}`);
})
我正在客户上这样要求
$('.download').on('click', function () {
$.ajax({
url: 'http://localhost:3000/download',
type: 'GET',
dataType: 'json',
success: function (data) {
console.log('data')
},
error: function (jqXHR, textStatus, errorThrown) {
// Handle errors here
console.log('ERRORS: ' + textStatus);
// STOP LOADING SPINNER
}
});
在控制台上出现错误
ERRORS: parsererror
3index.html?_ijt=9lu9erpan2oq6qf28851ngj0ra:32 ERRORS: parsererror
2index.html?_ijt=9lu9erpan2oq6qf28851ngj0ra:32 ERRORS: parsererror
服务器日志
C:\Users\B0207296\WebstormProjects\uploadFile\server
C:\Users\B0207296\WebstormProjects\uploadFile\server\uploads\file-1534458777206.xlsx
为什么文件没有下载到客户端,因为我提到文件存在于上面的URL中
答案 0 :(得分:0)
要通过ajax获取文件内容: 只需删除dataType,您的文件就会被解析为json,从而导致错误。
$('.download').on('click', function () {
$.ajax({
url: 'http://localhost:3000/download',
type: 'GET',
success: function (data) {
console.log(data); // File data
},
error: function (jqXHR, textStatus, errorThrown) {
// Handle errors here
console.log('ERRORS: ' + textStatus);
// STOP LOADING SPINNER
}
});
下载文件最好是用户虚拟链接
<!DOCTYPE html>
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
</head>
<body>
<b class="download">Download Here</b>
<script>
$('.download').on('click', function () {
var link = document.createElement('a');
link.href = 'http://localhost:3000/download';
var e = document.createEvent('MouseEvents');
e.initEvent('click', true, true);
link.dispatchEvent(e);
})
</script>
</body>
</html>