HttpURLConnection GET调用参数不起作用

My *_*God 6 java url httpurlconnection

我有一个非常简单的代码,无法正常工作.

HttoCon类:

package test;

import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.URL;
import java.net.URLConnection;
import java.net.URLEncoder;

public class HttpCon {


public static void main(String[] args) {
    try {
        sendGet();
    } catch (Exception e) {
        e.printStackTrace();
    }
}
    // HTTP GET request
        private static void sendGet() throws Exception {

            String url = "http://myweb.com/public/resource/data/stream";

            URL obj = new URL(url);
            HttpURLConnection con = (HttpURLConnection) obj.openConnection();

            // optional default is GET
            con.setRequestMethod("GET");


            //add request header
            con.setRequestProperty("begin", "1430295300000");
            con.setRequestProperty("end", "1430297279988");
            con.setRequestProperty("id", "140621");
            con.setRequestProperty("time", "FIFTEEN_MINUTE");

            int responseCode = con.getResponseCode();
            System.out.println("\nSending 'GET' request to URL : " + url);
            System.out.println("Response Code : " + responseCode);

            BufferedReader in = new BufferedReader(
                    new InputStreamReader(con.getInputStream()));
            String inputLine;
            StringBuffer response = new StringBuffer();

            while ((inputLine = in.readLine()) != null) {
                response.append(inputLine);
            }
            in.close();

            //print result
            System.out.println(response.toString());

        }
}
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错误跟踪:

{"fault":{"exceptionType":"MissingServletRequestParameterException","host":"12.205.101.123","cause":"Required Long parameter 'id' is not present","status":"400","query":"/public/resource/data/stream","stackTrace":"org.springframework.web.bind.MissingServletRequestParameterException: Required Long parameter 'id' is not present\n\tat org.springframework.web.servlet.mvc.annotation.AnnotationMethodHandlerAdapter$ServletHandlerMethodInvoker.raiseMissingParameterException(AnnotationMethodHandlerAdapter.java:774)\n\tat org.springframework.web.bind.annotation.support.HandlerMethodInvoker.resolveRequestParam(HandlerMethodInvoker.java:509)\n\tat 
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注意:

如果我直接在浏览器中点击网址,它可以正常工作.

http://myweb.com/public/resource/data/stream?begin=1430295300000&end=1430297279988&id=140621&time=FIFTEEN_MINUTE

更新:

使用curl:

curl -X GET http://myweb.com/public/resource/data/stream?begin=1430295300000&end=1430297279988&id=140621&time=FIFTEEN_MINUTE
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以上卷曲不起作用,例外是相同的 -

curl -X GET 'http://myweb.com/public/resource/data/stream?begin=1430295300000&end=1430297279988&id=140621&time=FIFTEEN_MINUTE'
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这个卷曲很好.

小智 5

您创建一个包含参数的String,然后将其附加到URL上.从中创建一个URL Obj.例如 :-

String this.url = "http://myweb.com/public/resource/data/stream";
this.url += "?begin=1430295300000&end=1430297279988&id=140621
    &time=FIFTEEN_MINUTE"; 
this.urlObj = new URL(this.url);
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然后像在原始示例中一样创建连接.