Question

我目前正在使用http://imdbapi.org的imdb api来获取有关电影的一些信息。当我使用API并尝试在java中打开此url时，它会给出403错误。 url应该以JSON格式返回数据。这是我到目前为止的代码（Java 7）：

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.net.MalformedURLException;
import java.net.URL;

public class Test {
    public static void main(String[] args) {
        URL url =null;
        try {
            url = new URL("http://imdbapi.org/?q=batman");
        } catch (MalformedURLException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        InputStream is =null;
        try {
            is = url.openConnection().getInputStream();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        BufferedReader reader = new BufferedReader( new InputStreamReader( is )  );
        String line = null;
        try {
            while( ( line = reader.readLine() ) != null )  {
               System.out.println(line);
            }
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        try {
            reader.close();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        System.out.println(line);
    }
}

Answer 1

您应该设置User-Agent：

System.setProperty("http.agent", "Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/28.0.1500.29 Safari/537.36");

或

URLConnection connection = url.openConnection();
connection.setRequestProperty("User-Agent", "Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/28.0.1500.29 Safari/537.36");
is = connection.getInputStream();

我尝试打开URL时为什么会收到403错误

1 个答案: