我有curl命令:
curl -k -f --header "Authorization: bearer cb084803-xxxx-xxxx-xxxx-2a275e112c38 "
--header "Content-Type:multipart/form-data"
--header "X-PartnerUserId:login01"
--form "title=fileName"
--form "tag=mytag"
--form "archive=@test.pdf;type=application/pdf"
-X POST https://api.EXAMPLEAPI.com/api/v2.0/doc
我想在Java API中创建HTTP请求,它将执行相同的操作 但我不知道怎样才能为参数做什么(-form ....)
Map<String, String> mapResponse = new HashMap<String, String>();
String resourceURL = config.getProperty(OAuthConstants.RESOURCE_SERVER_URL_UPLOAD);
OAuth2Details oauthDetails = createOAuthDetails(config);
HttpPost httpPost = new HttpPost(resourceURL);
httpPost.setHeader(OAuthConstants.AUTHORIZATION, getAuthorizationHeaderForAccessToken(oauthDetails.getAccessToken()));
httpPost.setHeader(OAuthConstants.CONTENTE_TYPE, OAuthConstants.ENCODED_CONTENT_DATA);
httpPost.setHeader(OAuthConstants.PARTNERUSERID, login);
//------------------------------------------------------------------------
httpPost.set... // add: --form "title=fileName"
httpPost.set... // add: --form "tag=mytag"
httpPost.set... // add: --form "archive=@test.pdf;type=application/pdf"
//------------------------------------------------------------------------
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpResponse response = null;
int code = -1;
try {
response = httpClient.execute(httpPost);
code = response.getStatusLine().getStatusCode();
任何帮助表示感谢。