我们正在尝试创建一个We服务,以便使用node.js服务将文件上传到Azure文件存储。
下面是node.js服务器代码。
exports.post = function(request, response){
var shareName = request.headers.sharename;
var dirPath = request.headers.directorypath;
var fileName = request.headers.filename;
var body;
var length;
request.on("data", function(chunk){
body += chunk;
console.log("Get data");
});
request.on("end", function(){
try{
console.log("end");
var data = body;
length = data.length;
console.log(body); // This giving the result as undefined
console.log(length);
fileService.createFileFromStream(shareName, dirPath, fileName, body, length, function(error, result, resp) {
if (!error) {
// file uploaded
response.send(statusCodes.OK, "File Uploaded");
}else{
response.send(statusCodes.OK, "Error!");
}
});
}catch (er) {
response.statusCode = 400;
return res.end('error: ' + er.message);
}
});
}
下面是我们上传文件的客户端。
private static void sendPOST() throws IOException {
URL obj = new URL("https://crowdtest-fileservice.azure-mobile.net/api/files_stage/");
HttpURLConnection con = (HttpURLConnection) obj.openConnection();
con.setRequestMethod("POST");
con.setRequestProperty("sharename", "newamactashare");
con.setRequestProperty("directorypath", "MaheshApp/TestLibrary/");
con.setRequestProperty("filename", "temp.txt");
Path path = Paths.get("C:/Users/uma.maheshwaran/Desktop/Temp.txt");
byte[] data = Files.readAllBytes(path);
// For POST only - START
con.setDoOutput(true);
OutputStream os = con.getOutputStream();
os.write(data);
os.flush();
os.close();
// For POST only - END
int responseCode = con.getResponseCode();
System.out.println("POST Response Code :: " + responseCode);
if (responseCode == HttpURLConnection.HTTP_OK) { // success
BufferedReader in = new BufferedReader(new InputStreamReader(con.getInputStream()));
String inputLine;
StringBuffer response = new StringBuffer();
while ((inputLine = in.readLine()) != null) {
response.append(inputLine);
System.out.println(inputLine);
}
in.close();
// print result
System.out.println(response.toString());
} else {
BufferedReader br = new BufferedReader(new InputStreamReader(con.getErrorStream()));
String line = "";
while ((line = br.readLine()) != null) {
System.out.println(line);
}
System.out.println("POST request not worked");
}
}
它正在显示错误
请求'POST /api/files_stage/‘已超时。这可能是因为脚本无法写入响应,或者无法及时从异步调用中返回。
更新:
我也尝试过下面的代码。
var body = new Object();
body = request.body;
var length = body.length;
console.log(request.body);
console.log(body);
console.log(length);
try {
fileService.createFileFromStream(shareName, dirPath, fileName, body, length, function(error, result, resp) {
if (!error) {
// file uploaded
response.send(statusCodes.OK, "File Uploaded");
}else{
response.send(statusCodes.OK, "Error!");
}
});
} catch (ex) {
response.send(500, { error: ex.message });
}
但是面对这个问题
{“error”:“函数createFileFromStream的参数流应为对象”}
我是node.js的新手。请帮我解决这个问题。
发布于 2016-01-05 06:42:15
这里有几个问题。让我们一个接一个地检查它们。
在您的客户端中,您不能仅将二进制数据转储到Azure移动服务连接中。
这样做的原因是Azure移动服务有两个正文解析器,这两个解析器确保无论如何都会为您解析请求正文。因此,虽然您可以通过指定不常见的内容类型来遍历Express body解析器,但您仍然会遇到Azure body解析器,它会天真地假设它是UTF-8字符串,从而将您的数据流搞得一团糟。
因此,唯一的选择是通过指定它无法处理的内容类型来跳过Express解析器,然后通过使用Base64编码来编码您的二进制数据来与Azure解析器一起播放。
因此,在Java客户端替换
Path path = Paths.get("C:/Users/uma.maheshwaran/Desktop/Temp.txt");
byte[] data = Files.readAllBytes(path);
使用
con.setRequestProperty("content-type", "binary");
Path path = Paths.get("C:/Users/uma.maheshwaran/Desktop/Temp.txt");
byte[] data = Files.readAllBytes(path);
data = Base64.getEncoder().encode(data);
如果您使用的不是Java8,请用您有权访问的任何其他Base64编码器替换java.util.Base64编码器。
Azure 2.您尝试使用的 createFileFromStream
存储函数需要一个流。
同时,在手动解析请求正文时,您能得到的最好结果是一个字节数组。不幸的是,Azure移动服务使用的是NodeJS 0.8版本,这意味着没有简单的方法来从字节数组构造可读的流,并且您将不得不组装适合Azure存储api的自己的流。一些胶带和流@0.0.1应该可以做得很好。
var base64 = require('base64-js'),
Stream = require('stream'),
fileService = require('azure-storage')
.createFileService('yourStorageAccount', 'yourStoragePassword');
exports.post = function (req, res) {
var data = base64.toByteArray(req.body),
buffer = new Buffer(data),
stream = new Stream();
stream['_ended'] = false;
stream['pause'] = function() {
stream['_paused'] = true;
};
stream['resume'] = function() {
if(stream['_paused'] && !stream['_ended']) {
stream.emit('data', buffer);
stream['_ended'] = true;
stream.emit('end');
}
};
try {
fileService.createFileFromStream(req.headers.sharename, req.headers.directorypath,
req.headers.filename, stream, data.length, function (error, result, resp) {
res.statusCode = error ? 500 : 200;
res.end();
}
);
} catch (e) {
res.statusCode = 500;
res.end();
}
};
这些是此示例所需的依赖项。
"dependencies": {
"azure-storage": "^0.7.0",
"base64-js": "^0.0.8",
"stream": "0.0.1"
}
如果在服务的package.json中指定它们不起作用,您始终可以转到此link并通过控制台手动安装它们。
cd site\wwwroot
npm install azure-storage
npm install base64-js
npm install stream@0.0.1
3.要增加1Mb的默认上传限制,请为您的服务指定MS_MaxRequestBodySizeKB。
但请记住,由于您是以Base64编码的格式传输数据,因此必须考虑到此开销。因此,要支持上传20Mb大小的文件,您必须将MS_MaxRequestBodySizeKB
设置为大约20 * 1024 *4/3= 27307。
发布于 2015-12-18 02:54:16
当请求到达exports.post
中定义的函数时,整个请求已经存在,因此您不需要对其进行缓冲。您可以按照下面的代码行编写一些东西来简化它。
exports.post = function(request, response){
var shareName = request.headers.sharename;
var dirPath = request.headers.directorypath;
var fileName = request.headers.filename;
var body = request.body;
var length = body.length;
console.log(length);
try {
fileService.createFileFromText(shareName, dirPath, fileName, body, function(error, result, resp) {
if (!error) {
// file uploaded
response.send(statusCodes.OK, "File Uploaded");
} else {
response.send(statusCodes.OK, "Error!");
}
});
} catch (ex) {
response.send(500, { error: ex.message });
}
}
发布于 2015-12-18 05:53:52
我发现最简单的方法是使用pkgcloud,它抽象了云提供商之间的差异,并为上传和下载文件提供了一个干净的接口。它使用流,因此实现也是内存高效的。
var pkgcloud = require('pkgcloud')
var fs = require('fs')
var client = pkgcloud.storage.createClient({
provider: 'azure',
storageAccount: 'your-storage-account',
storageAccessKey: 'your-access-key'
});
var readStream = fs.createReadStream('a-file.txt');
var writeStream = client.upload({
container: 'your-storage-container',
remote: 'remote-file-name.txt'
});
writeStream.on('error', function (err) {
// handle your error case
});
writeStream.on('success', function (file) {
// success, file will be a File model
});
readStream.pipe(writeStream);
https://stackoverflow.com/questions/34331893
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