在创建WCF REST服务时,我在json中接收数据并能够保存在数据库中。现在我需要选择上传文件(可选,图像或视频)与相同的服务。我尝试发送字节数组,但它可能因为序列化这么长的数组而给出错误的请求错误。我读到要上传大文件我需要使用流。发送其他参数时我该怎么做?我正在创建此服务以从移动设备接收数据。这是我的服务界面
[WebInvoke(UriTemplate = "SaveFBPost",
Method = "POST",
BodyStyle = WebMessageBodyStyle.Wrapped,
RequestFormat = WebMessageFormat.Json,
ResponseFormat = WebMessageFormat.Json)]
[OperationContract]
void SaveFacebookPost(FBPostData fbPostData);
公共类FBPostData:
[DataMember]
public string scheduleDate { get; set; }
[DataMember]
public string userId { get; set; }
[DataMember]
public string groupId { get; set; }
[DataMember]
public string postText { get; set; }
[DataMember]
public byte[] file { get; set; }
[DataMember]
public string fileType { get; set; }
[DataMember]
public string accessToken { get; set; }
答案 0 :(得分:1)
我首先要为web.config中的特定绑定增加maxBufferSize和maxArrayLength,看看是否能解决问题。
您应该能够提供有关错误的更多详细信息,这样您就可以确切了解为什么会收到400错误。
This blog article过去对我也很有用 - 底部也有一些很好的链接。
另请查看Stream课程。不确定“发送其他参数”是什么意思 - 如果你能澄清这一点,我们可以给你更多指导。
答案 1 :(得分:1)
我不得不使用来自android和multipart解析器的multipart上传来解决这个问题。我使用Apache mime库上传文件并发送如下参数:
HttpPost postRequest = new HttpPost(
context.getString(R.string.url_service_fbpost));
MultipartEntity reqEntity = new MultipartEntity(
HttpMultipartMode.BROWSER_COMPATIBLE);
if(postData.data != null && !"".equals(postData.fileName)){
ByteArrayBody bab = new ByteArrayBody(postData.data, postData.fileName);
reqEntity.addPart("uploaded", bab);
}
reqEntity.addPart("scheduleDate", new StringBody(postData.scheduleDate));
reqEntity.addPart("userId", new StringBody(postData.userId));
reqEntity.addPart("groupIds",
new StringBody(postData.groupIds.toString()));
reqEntity.addPart("postText", new StringBody(postData.postText));
reqEntity.addPart("postType", new StringBody(postData.postType));
reqEntity.addPart("accessToken", new StringBody(postData.accessToken));
if(postData.postId != null && postData.postId.length() > 0) {
reqEntity.addPart("postId", new StringBody(postData.postId));
}
postRequest.setEntity(reqEntity);
之后我使用c#multipart解析器来获取文件和参数。这是服务代码:
[OperationContract]
[WebInvoke(Method = "POST",
UriTemplate = "UploadFBPost",
BodyStyle = WebMessageBodyStyle.WrappedRequest)]
void UploadFBPost(Stream stream);
public void UploadFBPost(Stream stream)
{
MultipartParser parser = new MultipartParser(stream);
// Saves post data in database
if (parser.Success)
{
string fileName = null, userId = null, postText = null, postType = null, accessToken = null;
DateTime scheduleDate = DateTime.Now;
string[] groupIds = null;
int postId = 0;
// Other contents
foreach (MyContent content in parser.MyContents)
{
switch (content.PropertyName)
{
case "scheduleDate":
if (string.IsNullOrEmpty(content.StringData.Trim()))
scheduleDate = DateTime.Now;
else
scheduleDate = DateTime.ParseExact(content.StringData.Trim(), "M-d-yyyy H:m:s", CultureInfo.InvariantCulture);
break;
case "fileName":
fileName = content.StringData.Trim();
break;
case "userId":
userId = content.StringData.Trim();
break;
case "postText":
postText = content.StringData.Trim();
break;
case "accessToken":
accessToken = content.StringData.Trim();
break;
case "groupIds":
groupIds = content.StringData.Trim().Split(new char[] { ',' });
break;
case "postType":
postType = content.StringData.Trim();
break;
case "postId":
postId = Convert.ToInt32(content.StringData.Trim());
break;
}
}
string videoFile = null, imageFile = null;
if (parser.FileContents != null)
{
string filePath = GetUniqueUploadFileName(parser.Filename);
File.WriteAllBytes(filePath, parser.FileContents);
if (postType == "photo")
imageFile = Path.GetFileName(filePath);
else
videoFile = Path.GetFileName(filePath);
}
}
}
您需要根据要发送的数据修改多部分解析器。希望这会节省很少的时间。
感谢雷的帮助。
还有一件事。我不得不在网络配置中添加这些行:<httpRuntime maxRequestLength="2000000"/>
<bindings>
<webHttpBinding>
<binding maxBufferSize="65536"
maxReceivedMessageSize="2000000000"
transferMode="Streamed">
</binding>
</webHttpBinding>
</bindings>