Question

我们正在尝试创建一个Web服务，使用node.js服务将文件上传到Azure文件存储。

以下是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);
}

});

}

以下是我们上传文件的客户。

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");
    }
}

显示错误

请求'POST / api / files_stage /'已超时。这可能是由无法写入响应的脚本或其他原因引起的无法及时从异步调用返回。

已更新：

我也试过下面的代码。

  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 });
    }

但面对问题

{“error”：“函数createFileFromStream的参数流应该是一个对象“}

我是node.js的新手。请帮我解决这个问题。

Answer 1

这里有几个问题。让我们一个接一个地看看它们。

<强> 1。在Java客户端中，您不能只将二进制数据转储到Azure移动服务连接中。

原因是Azure移动服务有两个主体解析器，无论如何都能确保为您解析请求主体。因此，虽然您可以通过指定不常见的内容类型来遍历Express正文解析器，但您仍然会通过天真地假设它是UTF-8字符串来点击Azure主体解析器，这会破坏您的数据流。

因此，唯一的选择是通过指定无法处理的内容类型跳过Express解析器，然后通过使用Base64编码对二进制数据进行编码来播放Azure解析器。

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);

与

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);

如果您不在Java 8上，请将java.util.Base64编码器替换为您有权访问的任何其他Base64编码器。

<强> 2。您尝试使用的createFileFromStream Azure存储api功能需要一个流。

同时，手动解析请求主体时可以获得的最佳结果是字节数组。不幸的是，Azure移动服务使用NodeJS版本0.8，这意味着没有简单的方法来构建来自字节数组的可读流，并且您必须组装适合Azure存储API的自己的流。一些胶带和stream@0.0.1应该没问题。

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，您可以随时转到此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

第3。要增加1Mb的默认上传限制，请为您的服务指定MS_MaxRequestBodySizeKB。

请记住，因为您要将数据作为Base64编码传输，因此您必须考虑此开销。因此，要支持上传大小不超过20Mb的文件，您必须将MS_MaxRequestBodySizeKB设置为大约20 * 1024 * 4/3 = 27307.

Answer 2

当请求到达exports.post中定义的函数时，整个请求已经存在，因此您不需要缓冲它。您可以通过编写下面代码的内容来简化它。

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 });
    }
}

Answer 3

我发现最简单的方法是使用pkgcloud来抽象云提供商之间的差异，并提供一个干净的界面来上传和下载文件。它使用流，因此实现也是内存效率。

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);

Answer 4

我们可以在SO How to send an image from Android client to Node.js server via HttpUrlConnection?上利用这个线程的答案，它创建一个自定义中间件来将上传文件内容放到缓冲区数组中，然后我们可以使用createFileFromText()将文件存储在Azure中存储

以下是代码段：

function rawBody(req, res, next) {
    var chunks = [];

    req.on('data', function (chunk) {
        chunks.push(chunk);
    });

    req.on('end', function () {
        var buffer = Buffer.concat(chunks);

        req.bodyLength = buffer.length;
        req.rawBody = buffer;
        next();
    });

    req.on('error', function (err) {
        console.log(err);
        res.status(500);
    });
}
router.post('/upload', rawBody,function (req, res){

    fileService.createShareIfNotExists('taskshare', function (error, result, response) {
        if (!error) {
            // if result = true, share was created.
            // if result = false, share already existed.
            fileService.createDirectoryIfNotExists('taskshare', 'taskdirectory', function (error, result, response) {
                if (!error) {
                    // if result = true, share was created.
                    // if result = false, share already existed.
                    try {
                        fileService.createFileFromText('taskshare', 'taskdirectory', 'test.txt', req.rawBody, function (error, result, resp) {
                            if (!error) {
                                // file uploaded
                                res.send(200, "File Uploaded");
                            } else {
                                res.send(200, "Error!");
                            }
                        });
                    } catch (ex) {
                        res.send(500, { error: ex.message });
                    }

                }
            });
        }
    });

})
router.get('/getfile', function (req, res){
    fileService.createReadStream('taskshare', 'taskdirectory', 'test.txt').pipe(res);
})

Answer 5

有几件事：

<强> 1。 createFileFromText可以使用纯文本。但是对于那些二进制内容来说它会失败，因为它使用UTF-8编码。

您可能希望在Saving blob (might be data!) returned by AJAX call to Azure Blob Storage creates corrupt image

中引用blob的类似问题

<强> 2。 createFileFromStream或createWriteStreamToExistingFile \ createWriteStreamToNewFile Azure存储API可能是有用的功能。

请注意，这些API是流的目标。您需要将请求正文中的缓冲区/字符串转换为流。您可以参考How to wrap a buffer as a stream2 Readable stream?

createFileFromStream：

fileService.createFileFromStream(req.headers.sharename, 
  req.headers.directorypath, 
  req.headers.filename, 
  requestStream, 
  data.length, 
  function (error, result, resp) {
    res.statusCode = error ? 500 : 200;
    res.end();
  }
);

createWriteStreamToNewFile：

var writeStream = fileService.createWriteStreamToNewFile(req.headers.sharename, 
  req.headers.directorypath, 
  req.headers.filename, 
  data.length);

requestStream.pipe(writeStream);

第3。您的代码中存在几个问题

console.log(body); // This giving the result as undefined

原因是您定义var body并且它是undefined。代码body += chunk仍会使body未定义。

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!");
  }
});

当createFileFromStream发生错误时，它也可能是网络传输中的错误，您可能还想返回错误代码而不是statusCodes.OK。

使用node.js在Azure文件存储中上载文件

5 个答案: