将文件从动作脚本发送到servlet

Question

我不知道在Flash或Actionscript中进行编程。实际上，我是一名Java EE开发人员。

在Flash文件中，我有以下方法：

    private function recordComplete(e:Event):void
    {
        fileReference.save(recorder.output, "recording.wav");

    }

此方法会将录制的声音保存到我们将指定的文件夹中的“ recording.wav”中。

我想要做的是通过将录制的声音发送到Java Servlet来更改保存到磁盘的内容。

我找到了这段代码，但是我不知道如何在HTTP请求中发送的params中插入recorder.output：

var uploadRequest:URLRequest = new URLRequest("http://127.0.0.1:8080/uploading/upservlet");
        uploadRequest.method = URLRequestMethod.POST;
        uploadRequest.contentType = "multipart/form-data";
        uploadRequest.data = myByteArray;

        var uploader:URLLoader = new URLLoader;
        uploader.addEventListener(ProgressEvent.PROGRESS, onUploadProgress);
        uploader.addEventListener(Event.COMPLETE, onUploadComplete);
        uploader.dataFormat = URLLoaderDataFormat.BINARY;
        uploader.load(uploadRequest);

请帮忙。

Answer 1

默认情况下，flash无法multipart使用参数创建请求，您必须手动构造它。这是我在项目中使用的简单实用程序方法：

private static const BOUNDARY:String = "boundary";

/**
 * Create multipart request for URLLoader 
 * NOTE: Don't forget to set the URLLoader.dataFormat = URLLoaderDataFormat.BINARY;
 * @param url upload url
 * @param bytes bytes to upload
 */ 
public static function createMultiPartRequest(url:String, bytes:ByteArray, fileProp:String="file1", fileName:String="file1.png", params:Object=null):URLRequest
{
    var request:URLRequest = new URLRequest(url);

    var header1:String = "\r\n--" + BOUNDARY + "\r\n" + 
        "Content-Disposition: form-data; name=\""+fileProp+"\"; filename=\""+fileName+"\"\r\n" + 
        "Content-Type: image/png\r\n" + "\r\n";
    var headerBytes1:ByteArray = new ByteArray();
    headerBytes1.writeMultiByte(header1, "ascii");
    var postData:ByteArray = new ByteArray();
    postData.writeBytes(headerBytes1, 0, headerBytes1.length);

    if(bytes)
        postData.writeBytes(bytes, 0, bytes.length);

    if (!params)
        params = {};
    if (!params.Upload)
        params.Upload = "Submit Query";
    for (var prop:String in params) {
        var header:String = "--" + BOUNDARY + "\r\n" + "Content-Disposition: form-data; name=\""+prop+"\"\r\n" + "\r\n" + params[prop]+"\r\n" + "--" + BOUNDARY + "--";
        var headerBytes:ByteArray = new ByteArray();
        headerBytes.writeMultiByte(header, "ascii");
        postData.writeBytes(headerBytes, 0, headerBytes.length);
    }
    request.data = postData;
    request.method = URLRequestMethod.POST;
    request.contentType = "multipart/form-data; boundary=" + BOUNDARY;

    return request;
}

因此，您应该以以下方式修改代码：

    var uploadRequest:URLRequest = createMultiPartRequest("http://127.0.0.1:8080/uploading/upservlet", myByteArray, "file1", recorder.output, {param1:value1});

    var uploader:URLLoader = new URLLoader;
    uploader.addEventListener(ProgressEvent.PROGRESS, onUploadProgress);
    uploader.addEventListener(Event.COMPLETE, onUploadComplete);
    uploader.dataFormat = URLLoaderDataFormat.BINARY;
    uploader.load(uploadRequest);