java - 使用 Java HttpURLConnection 查询 Open MapQuest API 没有给出结果

Question

我编写了这段代码来从 Java 程序访问 Open MapQuest API(故意省略 APPKEY..):

public class OpenMapQuestAPITest {

  private final static String APPKEY="...";

  public static void main(String[] args) throws Exception {
    String address="Germany, Hannover, Am Hohen Ufer 3A";
    URL url=new URL("http://open.mapquestapi.com/nominatim/v1/search.php?key="+
        APPKEY+"&format=json&q="+address.replace(' ','+'));
    System.out.println("Query: "+url.toString());
    HttpURLConnection con=(HttpURLConnection)url.openConnection();
    int responseCode = con.getResponseCode();
    System.out.println("Response Code : " + responseCode);
    StringBuffer response = new StringBuffer();
    try (BufferedReader in=
         new BufferedReader(new InputStreamReader(con.getInputStream()))) {
      String inputLine;
      while ((inputLine = in.readLine()) != null)
        response.append(inputLine);
    }
    System.out.println("Response: "+response.toString());
  }

}

我明白了(故意省略键，添加换行符):

Query: http://open.mapquestapi.com/nominatim/v1/search.php?↵
       key=...&format=json&q=Germany,+Hannover,+Am+Hohen+Ufer+3A
Response Code : 200
Response: []

从例如发出完全相同的查询Chrome 给出了正确的响应:

[{"place_id":"1528890","licence":"Data \u00a9 OpenStreetMap contributors, ODbL 1.0. http:\/\/www.openstreetmap.org\/copyright",
  "osm_type":"node","osm_id":"322834134",
  "boundingbox":["52.3729826","52.3729826","9.7304606","9.7304606"],
  "lat":"52.3729826","lon":"9.7304606",
  "display_name":"3a, Am Hohen Ufer, Mitte, Hannover, Region Hannover, Niedersachsen, 30159, Deutschland",
  "class":"place","type":"house","importance":0.511}]

当我通过Java询问时，为什么MapQuest不返回这些数据？是否有一些我不知道的字符正在逃脱？我需要设置一些特殊的用户代理吗？我是否以错误的方式查询连接对象？毕竟，服务返回 OK(HTTP 状态代码 200)，所以看起来对我的请求感觉还不错。

为什么不回复？

Answer 1

大多数时候这些问题与 url 编码有关。

<强>1。首先尝试 url 编码。

String address="Germany, Hannover, Am Hohen Ufer 3A";
URL url=new URL("http://open.mapquestapi.com/nominatim/v1/search.php?key="+
    APPKEY+"&format=json&q="+address.replace(' ','+'));

尝试:

URLEncoder.encode(address.replace(' ','+'), "UTF-8"));

2.您可以使用org.apache.commons.httpclient.HttpClient api。它有一个类 PostMethod (org.apache.commons.httpclient.methods.PostMethod)，您可以使用它添加参数。 (而不是 HttpUrlConnection) 喜欢

HttpClient hc = new HttpClient();
PostMethod pm =  new PostMethod(url);
// add parameters to it.
pm.addParameter("format","json");
hc.executeMethod(pm);