Question

我的手机APP在文字模式下完美下载内容。下面是一个代码来做到这一点。我调用Communicator类和exectueHttpGet：

URL_Data = new Communicator().executeHttpGet("Some URL");

public class Communicator {
public String executeHttpGet(String URL) throws Exception 
{
    BufferedReader in = null;
    try 
    {
        HttpClient client = new DefaultHttpClient();
        client.getParams().setParameter(CoreProtocolPNames.USER_AGENT, "android");
        HttpGet request = new HttpGet();
        request.setHeader("Content-Type", "text/plain; charset=utf-8");
        request.setURI(new URI(URL));
        HttpResponse response = client.execute(request);
        in = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));

        StringBuffer sb = new StringBuffer("");
        String line = "";

        String NL = System.getProperty("line.separator");
        while ((line = in.readLine()) != null) 
        {
            sb.append(line + NL);
        }
        in.close();
        String page = sb.toString();
        //System.out.println(page);
        return page;
    } 
    finally 
    {
        if (in != null) 
        {
            try 
            {
                in.close();
            } 
            catch (IOException e)    
            {
                Log.d("BBB", e.toString());
            }
        }
    }
}
}

我所接受的是这个（URL的源代码）：

[{"id_country":"3","country":"AAA"},{"id_country":"8","country":"BBB"},
{"id_country":"66","country":"CCC"},{"id_country":"14","country":"DDD"},
{"id_country":"16","country":"EEE"},{"id_country":"19","country":"FFF"},
{"id_country":"24","country":"GGG"},{"id_country":"33","country":"HHH"},
{"id_country":"39","country":"III"},{"id_country":"44","country":"JJJ"},
{"id_country":"45","country":"KKK"},{"id_country":"51","country":"LLL"},
{"id_country":"54","country":"MMM"},{"id_country":"55","country":"NNN"},
{"id_country":"57","country":"OOO"},{"id_country":"58","country":"PPP"},
{"id_country":"63","country":"RRR"},{"id_country":"65","country":"SSS"}]

此响应是一个字符串。在服务器上，它作为JSON对象输出（使用PHP），现在在我的Android PHP中，我希望将此字符串转换为JSON。这可能吗？

Answer 1

您收到的是InputStream中附加到StringBuffer并在结尾处转换为String的一系列字符 - 因此String的结果为好的:)）

您想要的是通过org.json.*类（如

）对此字符串进行后处理

String page = new Communicator().executeHttpGet("Some URL");
JSONObject jsonObject = new JSONObject(page);

然后继续jsonObject。由于您收到的数据是一个数组，您实际上可以说

String page = new Communicator().executeHttpGet("Some URL");
JSONArray jsonArray = new JSONArray(page);
for (int i = 0 ; i < jsonArray.length(); i++ ) {
  JSONObject entry = jsonArray.get(i);
  // now get the data from each entry
}

Answer 2

修改

为了进一步解决我之前链接到的问题，请使用此示例。把它放在一个返回JSONArray的函数中（这样你就可以遍历数组并使用array.getString）。这适用于大多数数据。它会将正确的压缩标头发送到Web服务器并检测gzip压缩结果。试试吧：

URL url = new URL('insert your uri here'); HttpURLConnection urlConn = (HttpURLConnection) url.openConnection(); urlConn.setRequestProperty("Accept-Encoding", "gzip"); HttpURLConnection httpConn = (HttpURLConnection) urlConn; httpConn.setAllowUserInteraction(false); httpConn.connect(); if (httpConn.getContentEncoding() != null) { String contentEncoding = httpConn.getContentEncoding().toString(); if (contentEncoding.contains("gzip")) { in = new GZIPInputStream(httpConn.getInputStream()); } // else it is encoded and we do not want to use it... } else { in = httpConn.getInputStream(); } BufferedInputStream bis = new BufferedInputStream(in); ByteArrayBuffer baf = new ByteArrayBuffer(1000); int read = 0; int bufSize = 1024; byte[] buffer = new byte[bufSize]; while (true) { read = bis.read(buffer); if (read == -1) { break; } baf.append(buffer, 0, read); } queryResult = new String(baf.toByteArray()); return new JSONArray(queryResult);

/结束编辑

尝试阅读我在这个问题上发布的解决方案：

Create list in android app

H个，

斯图

Answer 3

使用org.json.JSONObject，如

JSONObject json = new JSONObject(oage);

需要注意的是响应只是“真实”或“错误”。可能想创建一个检测这些情况的util函数，否则只需加载JSONObject。

好的，在这种情况下你会使用JSONArray

JSONArray jsonArray = new JSONArray(page); 
for (int i = 0; i < jsonArray.length(); ++i) {
  JSONObject element = jsonArray.getJSONObject(i);
  ..... 
}

Answer 4

您是否尝试将内容类型设置为application/json？

Answer 5

假设我们有一个名为发布

的POJO类

<select ng-model="ubrandname">
    <option ng-repeat="brandname in brandfunction" value="{{brandname}}">{{brandname}}</option>
</select>

Answer 6

只是尝试

... ///

JSONArray jsonArray = new JSONArray（responseString）;

for（JSONObject jsonObject：jsonArray） { .........
}

//// ..

在Android上从URL检索JSON

6 个答案:

只是尝试