我正在尝试使用CrowdFlower的API上传数据,为此我正在编写JAVA Package 器。我正在使用HttpClient Apache。
CrowdFlower给予的cURL示例如下:curl -T 'sampledata.xlsx' -H 'Content-Type: application/vnd.ms-excel' https://api.crowdflower.com/v1/jobs/upload.json?key={api_key}
下面是我的代码:
public InputStream HTTPmethodPostUpload (String authKey, File file) throws ClientProtocolException, IOException{
HttpClient httpclient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost("https://api.crowdflower.com/v1/jobs/upload.json?key="+authKey);
MultipartEntity mpEntity = new MultipartEntity();
ContentBody cbody = new FileBody( file,"application/vnd.ms-excel");
mpEntity.addPart("sampledata.xlsx", cbody );
httpPost.setEntity(mpEntity);
HttpResponse response = httpclient.execute(httpPost);
HttpEntity entityResponse = response.getEntity();
return entityResponse.getContent(); }
字符串
但是,它返回一个错误,并显示以下消息:
- {不可接受的格式,Content-Type必须是“formats”中列出的格式之一,但您发送了“multipart/form-data; boundary= yTuwTm 4 hWmnasxIMB 9dC-sxdELIGoNJVudjJdCz”",“formats”:[“application/vnd.oasis.opendocument. spreadsheet”,“application/vnd.openxmlformats-officedocument.spreadsheetml.sheet”,“application/vnd.ms-excel”,“text/csv”,“text/plain”]}}*
我不太了解Apache HttpClient,所以我不明白我的代码中的问题在哪里。
1条答案
按热度按时间1wnzp6jl1#
在java中,你在哪里设置了header?
字符串
试试下面的:
型