Question

这是python代码

import urllib.request as urllib2
import json 

data =  {
    "Inputs": {
        "input1": {
            "ColumnNames": ["id", "regex"],
            "Values": [ [ "0", "the regex value" ],]
        },
    },
    "GlobalParameters": {
        "Database query": "select * from expone",
    }
}

body = str.encode(json.dumps(data))

url = 'https://ussouthcentral.services.azureml.net/workspaces/4729545551a741e1a2e606d37' \ 
      'ae61ce0/services/ac7c34ad134d43ca9fdc65e292ce35d3/execute?api-version=2.0&details=true'
api_key = '8ku5P6fR3F8ykgMHK5Y8+PL8dn+Zi2Ajmwyjk9ENsomzzkDfuT8CtgKS7dF4yjaJfYxARe+1iLjh' \ 
          'Tv1R0qOTvw=='
headers = {
    'Content-Type': 'application/json',
    'Authorization': ('Bearer '+ api_key)
}

req = urllib2.Request(url, body, headers) 

try:
    response = urllib2.urlopen(req)
    result = response.read()
    print(result) 
except Exception as e:
    print("The request failed with status code: ", e)

这是我在Java中的尝试

public static void main(String[] args) {   
    System.out.println("MachineLearning main");
    try{                         
        //connections settings
        URL url = new URL("https://ussouthcentral.services.azureml.net/workspaces/4729545551a741e1a2e606d37ae61ce0/services/ac7c34ad134d43ca9fdc65e292ce35d3/execute?api-version=2.0&details=true");
        HttpURLConnection con = (HttpURLConnection)url.openConnection();
        con.setDoInput(true);
        con.setDoOutput(true);             
        String requestMethod = "GET";            
        con.setRequestMethod(requestMethod);
        con.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");

        String data=URLEncoder.encode("input1", "UTF-8") + "="
            + URLEncoder.encode("\"ColumnNames\": [\"id\", \"regex\"]", "UTF-8") + "&" 
            + URLEncoder.encode("GlobalParameters", "UTF-8") 
            + URLEncoder.encode("Database query\": \"select * from expone\"", "UTF-8");

        //make the request
        OutputStreamWriter writer = new OutputStreamWriter(con.getOutputStream());                
        writer.write(data);                
        writer.flush(); 

        //read the request
        BufferedReader reader=new BufferedReader(new InputStreamReader(con.getInputStream()));
        String response;
        while ((response=reader.readLine())!=null) 
            System.out.println(response);        
    }
    catch(Exception e) {
        System.out.println("Exception in MachineLearning.main " + e);
    }
}

java中请求的代码不成功并返回异常：服务器返回HTTP响应代码：URL为

问题是我不知道如何将python中的数据变量转换为java中的数据变量，如何传递apiKey以及如何将其放入标题中？

Answer 1

这个python代码看起来不错，但是我试图将其放在程序代码中并得到消息： Response code in Catch block

是不是因为我需要使用正确的用户名和密钥或实施WSSE auth？

如何将Python urllib.request代码翻译为Java代码

1 个答案: