Question

我有一个JSONParser类，可以让我创建HTTPRequests。所以这是班级

package com.thesis.menubook;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.UnsupportedEncodingException;
import java.util.List;

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.client.utils.URLEncodedUtils;
import org.apache.http.impl.client.DefaultHttpClient;
import org.json.JSONException;
import org.json.JSONObject;

import android.util.Log;

public class JSONParser {

    static InputStream is = null;
    static JSONObject jObj = null;
    static String json = "";

    // constructor
    public JSONParser() {

    }

    // function get json from url
    // by making HTTP POST or GET mehtod
    public JSONObject makeHttpRequest(String url, String method,
            List<NameValuePair> params) {

        // Making HTTP request
        try {

            // check for request method
            if(method == "POST"){
                // request method is POST
                // defaultHttpClient
                DefaultHttpClient httpClient = new DefaultHttpClient();
                HttpPost httpPost = new HttpPost(url);
                httpPost.setEntity(new UrlEncodedFormEntity(params));

                HttpResponse httpResponse = httpClient.execute(httpPost);
                HttpEntity httpEntity = httpResponse.getEntity();
                is = httpEntity.getContent();

            }else if(method == "GET"){
                // request method is GET
                DefaultHttpClient httpClient = new DefaultHttpClient();
                String paramString = URLEncodedUtils.format(params, "utf-8");
                url += "?" + paramString;
                HttpGet httpGet = new HttpGet(url);
                Log.d("URL",url);
                HttpResponse httpResponse = httpClient.execute(httpGet);
                HttpEntity httpEntity = httpResponse.getEntity();
                is = httpEntity.getContent();
            }          

        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        } catch (ClientProtocolException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

        try {
            BufferedReader reader = new BufferedReader(new InputStreamReader(
                    is, "iso-8859-1"), 8);
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null) {
                sb.append(line+ "n" );
            }
            is.close();
            json = sb.toString();
        } catch (Exception e) {
            Log.e("Buffer Error", "Error converting result " + e.toString());
        }


        // try parse the string to a JSON object
        try {
            jObj = new JSONObject(json);
        } catch (JSONException e) {
            Log.e("JSON Parser", "Error parsing data " + e.toString());
            Log.e("JSON Parser", "json string :" +json);
        }

        // return JSON String
        return jObj;

    }
}

我这样访问它。

params.add(new BasicNameValuePair("category", "MAIN DISH"));
            JSONObject json = jsonParser.makeHttpRequest("http://"+ipaddress+"/MenuBook/selectMenu.php", "GET", params);

我想保持参数的格式，MAIN DISH。但是当我查看我的LogCat时，它会返回一个这样形成的URL。

http://192.168.10.149:80/MenuBook/selectMenu.php?category=MAIN+DISH

然后导致我的应用程序失败并强制关闭，因为我没有像MAIN + DISH

这样的类别

我希望我的网址形成如下。

http://192.168.10.149:80/MenuBook/selectMenu.php?category='MAIN DISH'

然后会返回正确的结果。我在网上搜索过，只发现了使空格+和%20无法返回正确结果的解决方案。

您可以提出任何解决方案吗？

Answer 1

你的问题体现了矛盾。已经定义了URL编码（实际上是表单编码），并且已经定义了用'+'替换空格，而不是引用相关的值元素。服务器端软件需要了解并遵循相应的行为。 Java提供的所有服务器端软件，例如HttpServletRequest，已经这样做了。如果您的代码不符合RFC，请将其修复。

Answer 2

我通过用'MAIN+DISH'

对其进行硬编码来解决它

如何在java中编码url字符串以不在空格中放置+符号？

2 个答案: