将 curl 命令转换为 java http 客户端代码以获取数据
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【中文标题】将 curl 命令转换为 java http 客户端代码以获取数据【英文标题】:Convert a curl command to java http client code to get data 【发布时间】:2017-09-23 17:27:08 【问题描述】:获取授权码后,我正在使用下面的 curl 命令从 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"
卷曲网址:
https://api.instagram.com/oauth/access_token?client_id=cf07d1a2c69940e59420b6db4c936f4a&client_secret=fb0a975ca2024a1592459308df5ead47&grant_type=authorization_code&redirect_uri=http://localhost:8080/Insta_SMI_M1/auth&code=2c5d97c6d6454b8592816d7d39efb935
上面的网址在邮递员或浏览器中使用时既不给出任何错误也不显示输出(其空白行)。
下面是相同的代码。
@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();
任何机构可以帮助我解决这个问题吗? 谢谢
【问题讨论】:
如果 URL 在 Postman 中无法正常工作,则表明您的 URL 有问题。 【参考方案1】:curl 正在执行 POST 请求,而您正在执行 java 的 GET 请求。按照此示例了解如何制作POST request using java(使用http-client)。您可以考虑使用以下代码来设置参数:
params.add(new BasicNameValuePair("client_id", "cf07d1a2c69940e59420b6db4c936f4a"));
params.add(new BasicNameValuePair("client_secret", "fb0a975ca2024a1592459308df5ead47"));
【讨论】:
感谢您的提示! (我非常需要它,因为我是 java spring web rest api 交互的新手)【参考方案2】:由于我在互联网上找不到从 instagram api 获取访问令牌的完整代码,我将把对我有用的代码放在下面。
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": ""
【讨论】:
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