我使用下面的curl命令在获得授权码后从instagram api获取访问令牌。
curl \-F 'client_id=cf07d1a2c69940e59420b6db4c936f4a' \
-F 'client_secret=fb0a975ca2024a1592459308df5ead47' \
-F 'grant_type=authorization_code' \
-F 'redirect_uri=http://localhost:8080/Insta_SMI_M1/accessToken/' \
-F 'code=fcf66e5f09bf43a18ab15e5f1e0ae75f'
\https://api.instagram.com/oauth/access_token/
输出:
{"access_token": "5351945621.cf07d1a.1d35647e22f24ed0885f65545f3f1b0b", "user": {"id": "5351945621", "username": "abhaykumar", "profile_picture": "https://scontent-amt2-1.cdninstagram.com/t51.2885-19/11906329_960233084022564_1448528159_a.jpg", "full_name": "Quantum Four", "bio"}
Curl Url:
上述网址在邮递员或浏览器中使用时既没有出现任何错误也没有显示输出(空白行)。
以下是相同的代码。
@RequestMapping(value="/auth", method=RequestMethod.GET)
public String getAuthCode(HttpServletRequest request, HttpServletResponse response)
{
String code = request.getParameter("code");
System.out.println("code is: "+ code);
String url = "https://api.instagram.com/oauth/access_token?"
+ "client_id=" + Constants.CLIENT_ID
+ "&client_secret=" + Constants.CLIENT_SECRET
+ "&grant_type=authorization_code"
+ "&redirect_uri=" + Constants.REDIRECT_URI_AUTH
+ "&code="+code;
System.out.println("Access Token URL: "+ url);
StringBuffer result = null;
try {
System.out.println("1");
@SuppressWarnings({ "resource", "deprecation" })
HttpClient client = new DefaultHttpClient();
HttpGet request1 = new HttpGet(url);
System.out.println("2");
HttpResponse response1 = client.execute(request1);
BufferedReader rd = new BufferedReader(
new InputStreamReader(response1.getEntity().getContent()));
System.out.println("3");
result = new StringBuffer();
String line = "";
while ((line = rd.readLine()) != null) {
System.out.println("line " + line);
result.append(line);
System.out.println("3");
}
} catch (UnsupportedOperationException | IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println(result.toString());
return result.toString();
}
任何人都可以帮我吗? 感谢
答案 0 :(得分:1)
curl正在执行POST请求,您正在执行Java的GET请求。请按照此示例了解如何创建POST request using java(使用http-client)。您可以考虑以下代码来设置参数:
params.add(new BasicNameValuePair("client_id", "cf07d1a2c69940e59420b6db4c936f4a"));
params.add(new BasicNameValuePair("client_secret", "fb0a975ca2024a1592459308df5ead47"));
答案 1 :(得分:0)
由于我没有找到一个完整的代码来从互联网上的instagram api获取访问令牌,我会把下面的代码放到我身上。
public String accessTkn (String code)
{
try {
HttpClient httpclient = HttpClients.createDefault();
HttpPost httppost = new
HttpPost("https://api.instagram.com/oauth/access_token");
// Request parameters and other properties.
List<NameValuePair> params = new ArrayList<NameValuePair>(2);
params.add(new BasicNameValuePair("client_id", Constants.CLIENT_ID));
params.add(new BasicNameValuePair("client_secret", Constants.CLIENT_SECRET));
params.add(new BasicNameValuePair("grant_type", "authorization_code"));
params.add(new BasicNameValuePair("redirect_uri", Constants.REDIRECT_URI_AUTH));
params.add(new BasicNameValuePair("code", code));
httppost.setEntity(new UrlEncodedFormEntity(params, "UTF-8"));
//Execute and get the response.
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
System.out.println("entity "+ entity.getContent());
if (entity != null) {
InputStream instream = entity.getContent();
try {
return (getStringfromStream(instream));
// do something useful
} finally {
instream.close();
}
}
} catch (UnsupportedEncodingException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (ClientProtocolException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (UnsupportedOperationException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return "Abhay";
}
输出:
{"access_token": "5351945621.cf07d1a.1d35647e22f24ed0885f65545f3f1b0b", "user": {"id": "5351945621", "username": "abhaykumar", "profile_picture": "https://instagram.fsgn2-1.fna.fbcdn.net/t51.2885-19/11906329_960233084022564_1448528159_a.jpg", "full_name": "Abhay Kumar", "bio": "", "website": ""}}