Apache httpclient whith fastbill.api in java

Question

我是新来的java，需要这个项目。我必须结合使用Apache HttpClient和FastBill Api。

为FastBill阿比curl命令是

curl -v -X POST \ 
–u {E-Mail-Adresse}:{API-Key} \ 
-H 'Content-Type: application/json' \ 
-d '{json body}' \ 
https://my.fastbill.com/api/1.0/api.php 

我使用curl命令成功与此JSON文件

{ 
    "SERVICE":"customer.create", 
    "DATA": 
    { 
     "CUSTOMER_TYPE":"business", 
     "ORGANIZATION":"Musterfirma", 
     "LAST_NAME":"Mmann" 
    } 
} 

所以，我敢肯定，我的用户名，密码和JSON文件加工。 FastbillApi使用http基本身份验证。我在Java

public class Fastbill implements FastbillInterface { 

private static final String URL_SECURED_BY_BASIC_AUTHENTICATION = "https://my.fastbill.com/api/1.0/api.php"; 

public Customer createCustomer(String firstname, String lastname, CustomerType customertype, String organisation) { 

    CredentialsProvider provider = new BasicCredentialsProvider(); 
    UsernamePasswordCredentials credentials 
    = new UsernamePasswordCredentials("*****@****", "************"); //Api Username and API-Key 


    HttpClient client = HttpClientBuilder.create() 
     .setDefaultCredentialsProvider(provider) 
     .build(); 


    HttpPost httpPost = new HttpPost(URL_SECURED_BY_BASIC_AUTHENTICATION); 
    httpPost.setHeader("Content-Type", "application/json"); 
    String json = "{\"SERVICE\":\"customer.create\",\"DATA\":{\"CUSTOMER_TYPE\":\"business\",\"ORGANIZATION\":\"Musterfirma\",\"LAST_NAME\":\"Newmann\"}}"; 
    try { 
     HttpEntity entity = new ByteArrayEntity(json.getBytes("UTF-8")); 
     httpPost.setEntity(entity); 
    } catch (UnsupportedEncodingException e1) { 
     // TODO Auto-generated catch block 
     e1.printStackTrace(); 
    } 
    HttpResponse response; 
    try {   
     response = client.execute(httpPost); 
     int statusCode = response.getStatusLine() 
        .getStatusCode(); 
     System.out.println(statusCode); 
     String responseString = new BasicResponseHandler().handleResponse(response); 
     System.out.println(responseString); 
    } catch (ClientProtocolException e) { 
     // TODO Auto-generated catch block 
     e.printStackTrace(); 
    } catch (IOException e) { 
     // TODO Auto-generated catch block 
     e.printStackTrace(); 
    }catch (Exception e){ 
     e.printStackTrace(); 
    } 

试图以此作为回应，我得到

org.apache.http.client.HttpResponseException: Unauthorized 
at org.apache.http.impl.client.AbstractResponseHandler.handleResponse(AbstractResponseHandler.java:70) 
at org.apache.http.impl.client.BasicResponseHandler.handleResponse(BasicResponseHandler.java:66) 
at fastbillAPI.Fastbill.createCustomer(Fastbill.java:93) 
at main.Mar.main(Mar.java:38) 

现在，我不知道我做错了什么。

Answer 1

我找到了解决

此方法执行请求。 Fastbill接受JSON或XML请求，所以我做了一个弯路与JSONString建设者

private String performFastbillRequest(String InputJsonRequest) { 
    String responseString = null; 

    CloseableHttpClient client = HttpClients.createDefault(); 
    try { 
     URI uri = URI.create(URL_SECURED_BY_BASIC_AUTHENTICATION); 
     HttpPost post = new HttpPost(uri); 
     String auth = getAuthentificationString(); 

     post.addHeader("Content-Type", "application/json"); 
     post.addHeader("Authorization", auth); 

     StringEntity stringEntity = new StringEntity(InputJsonRequest); 
     post.setEntity(stringEntity); 

     CloseableHttpResponse response = client.execute(post); 

     HttpEntity entity = response.getEntity(); 
     responseString = EntityUtils.toString(entity, "UTF-8"); 
    } catch (Exception e) { 
     e.printStackTrace(); 
    } 
    return responseString; 
} 

此方法创建了请求authentificationstring，Fastbill接受base64编码字符串。对于这个项目，我需要储存的电子邮件和API在config.xml文件

private String getAuthentificationString() { 
    String authentificationString = "Basic "; 
    String fastbillUsernameAndApiKey = null; 
    XmlParser getFromXml = new XmlParser(); 
    fastbillUsernameAndApiKey = getFromXml.getApiEmailFromConfigFile() + ":" + getFromXml.getApiKeyFromConfigFile(); 
    //if email and apikey are not stored in config.xml you need following string containing the emailadress and ApiKey seperated with ':' 
    //f.ex. fastbillUsernameAndApiKey = "*****@****.***:*********"; 
    authentificationString = authentificationString + base64Encoder(fastbillUsernameAndApiKey); 
    return authentificationString; 
} 

private String base64Encoder(String input) { 
    String result = null; 
    byte[] encodedBytes = Base64.encodeBase64(input.getBytes()); 
    result = new String(encodedBytes); 
    return result; 
} 

此方法创建从我的请求JSON文件JSON字符串

private String createCustomerJson(String firstname, 
    String lastname) { 
     String createCustomerJsonStringBuilder = "{\"SERVICE\":\"customer.create\",\"DATA\":{\"CUSTOMER_TYPE\":\""; 
    createCustomerJsonStringBuilder += "consumer"; 
    createCustomerJsonStringBuilder += /* "\", */"\"LAST_NAME\":\""; 
    createCustomerJsonStringBuilder += lastname; 
    createCustomerJsonStringBuilder += "\",\"FIRST_NAME\":\""; 
    createCustomerJsonStringBuilder += firstname; 
    createCustomerJsonStringBuilder += "\"}}"; 

    return createCustomerJsonStringBuilder; 
} 

Answer 2

我遇到过类似的问题，我设法通过使用Matt S.示例解决方案解决了Apache HTTP BasicScheme.authenticate deprecated?。

正在验证通过：

UsernamePasswordCredentials creds = new UsernamePasswordCredentials("admin", "admin"); 
URI uriLogin = URI.create("http://localhost:8161/hawtio/auth/login/"); 
HttpPost hpPost = new HttpPost(uriLogin); 
Header header = new BasicScheme(StandardCharsets.UTF_8).authenticate(creds , hpPost, null); 
hpPost.addHeader(header);