HTTP从API获取XML请求

Question

我对用java调用API请求XML特别陌生，所以我使用的是模板。从下面可以看出。

然而，现在我从另一个web API调用API，他们要求我使用我一无所知的python，或者使用第三方HTTP web客户端。相反，我希望API的调用在我的java应用程序本身中。是否可以使用以下标题更改以下代码？

示例请求URL:http://datamall.mytransport.sg/ltadataservice.svc/TravelTimeSet

请求头：AccountKey=xxx，UniqueUserId=xxx，accept=application/atom+xml

API文件/指南

NowCast.java*NowCast.jar链接到SAX解析器

import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.URL;
public class NowCast extends SAXParser {
public static void main(String[] args) {
    String datasetName, keyRef;
    datasetName = "Nowcast";
    keyRef = "781CF461BB6606ADEA6B1B4F3228DE9DB26ABF1B1B755E93";
    NowCast od = new NowCast();
    try {
        od.callWebAPI(datasetName, keyRef);
    } catch (Exception e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
}
private void callWebAPI(String datasetName, String keyRef) throws Exception {
    // Step 1: Construct URL
    String url = "http://www.nea.gov.sg/api/WebAPI?dataset=" + datasetName
            + "&keyref=" + keyRef;
    // Step 2: Call API Url
    URL obj = new URL(url);
    HttpURLConnection con = (HttpURLConnection) obj.openConnection();
    con.setRequestMethod("GET");
    int responseCode = con.getResponseCode();
    System.out.println("nSending 'GET' request to URL : " + url);
    System.out.println("Response Code : " + responseCode);
    // Step 3: Check the response status
    if (responseCode == 200) {
        // Step 3a: If response status == 200
        // print the received xml
        super.Parserr(con.getInputStream());
        System.out.println(readStream(con.getInputStream()));
    } else {
        // Step 3b: If response status != 200
        // print the error received from server
        System.out.println("Error in accessing API - "
                + readStream(con.getErrorStream()));
    }
}
// Read the responded result
private String readStream(InputStream inputStream) throws Exception {
    BufferedReader reader = new BufferedReader(new InputStreamReader(
            inputStream));
    String inputLine;
    StringBuffer response = new StringBuffer();
    while ((inputLine = reader.readLine()) != null) {
        response.append(inputLine);
    }
    reader.close();
    return response.toString();
}

}

Answer 1

我将python代码（在API文档中）与上面的代码交叉引用，并解决了问题

在java中设置请求头

URL obj=新URL（URL）；

HttpURLConnection con=（HttpURLConnection）obj.openConnection（）；

con.setRequestMethod（"GET"）；

con.setRequestProperty（"AccountKey"，"zIIkxPOqBRxnkH97G6ZGlA=="）；

con.setRequestProperty（"UniqueUserID"，"5ccef4a0-10e2-4f75-8f4d-2be67885b10f"）；

con.setRequestProperty（"accept"，"application/atom+xml"）；