3

我试图在 Android 上的 Google Drive 中下载文件，并且我正在关注https://developers.google.com/drive/manage-downloads#examples中的示例

if (entry.getDownloadUrl() != null && entry.getDownloadUrl().length() > 0) 
{
    try
    {
        LogUtils.xi(this, "start opening url:", entry.getDownloadUrl());
        /* HttpRequestFactory: Thread-safe light-weight HTTP request factory layer on top of the HTTP */
        HttpResponse resp = manager.getDrive().getRequestFactory()
                        .buildGetRequest(new GenericUrl(entry.getDownloadUrl()))
                        .execute();
        return resp.getContent();
    } 
    catch (IOException e) 
    {
        LogUtils.xe(this, e.toString());
        e.printStackTrace();
        throw ExceptionUtils.getIOException("Can't open inputstream.", e);
    }
}
else 
{
    LogUtils.xw(this, "The file doesn't have any content stored on Drive.");
    return null;
}

但是收到错误 302：有人知道这里发生了什么吗？

03-29 14:56:07.316 E/XXGoogleDrive(25539): [25566][DriveDLFutureTask] [getInputStream]com.google.api.client.http.HttpResponseException: **302 Moved Temporarily**
03-29 14:56:07.326 W/System.err(25539): com.google.api.client.http.HttpResponseException: 302 Moved Temporarily
03-29 14:56:07.326 W/System.err(25539):     at com.google.api.client.http.HttpRequest.execute(HttpRequest.java:1062)
4

1 回答 1

0

出现“临时移动”的原因是该页面正在尝试重定向到用户身份验证页面或在文件大小超过 25 MB 的情况下进行病毒扫描

我通过使用带有'alt = media'查询参数的文件获取请求来使其工作，工作代码如下

private static byte[] getBytes(Drive drive, String accessToken, String fileId, long position, long byteCount) {
        byte[] receivedByteArray = null;
        String downloadUrl = "https://www.googleapis.com/drive/v3/files/" + fileId + "?alt=media&access_token="
                + accessToken;
        if (downloadUrl != null && downloadUrl.length() > 0) {
            try {
                com.google.api.client.http.HttpRequest httpRequestGet = drive.getRequestFactory().buildGetRequest(new GenericUrl(downloadUrl));
                httpRequestGet.getHeaders().setRange("bytes=" + position + "-" + (position + byteCount - 1));
                com.google.api.client.http.HttpResponse response = httpRequestGet.execute();
                InputStream is = response.getContent();
                receivedByteArray = IOUtils.toByteArray(is);
                response.disconnect();
                System.out.println("google-http-client-1.18.0-rc response: [" + position + ", " + (position + receivedByteArray.length - 1) + "]");
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return receivedByteArray;
    }
于 2017-05-18T11:19:42.253 回答