Question

作为标题，我想知道如何从MySQL数据库中检索单个变量并分配给变量。

   try{
        HttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost("http://192.168.1.25/userdatabase/include/GetSource.php");
        HttpResponse response = httpClient.execute(httpPost);
        BasicResponseHandler handler = new BasicResponseHandler();
        String data = handler.handleResponse(response);
        JSONObject jObj = new JSONObject(data);
        String path= jObj.getString("sources");

          }catch (Exception e) {
        Log.e("Log_tag", "Error converting result" + e.toString());
    }
            mVideoView.setVideoPath(path);
            mVideoView.setMediaController(new MediaController(getActivity()));
            mVideoView.setVideoChroma(MediaPlayer.VIDEOCHROMA_RGB565);
            mVideoView.requestFocus();

            progressBar.setVisibility(View.VISIBLE);
            mVideoView.setOnPreparedListener(new MediaPlayer.OnPreparedListener() {
                @Override
                public void onPrepared(MediaPlayer mediaPlayer) {
                    // optional need Vitamio 4.0
                    progressBar.setVisibility(View.GONE);
                    mediaPlayer.setPlaybackSpeed(1.0f);

                }
            });

这就是我试图从MySQL中检索路径的方法。但无法检索。请帮忙

<?php
header('Content-type: application/json; charset=UTF-8');

mysql_connect("127.0.0.1","root","");
mysql_select_db("streaming");

$sql=mysql_query("select * from iptv where id = 1");

while($row=mysql_fetch_assoc($sql))
$output[]=$row;
json_encode($output);
print(json_encode($output, JSON_UNESCAPED_SLASHES));
mysql_close();
?>

[{"sources":"http://210.200.79.25:1935/live/ttv.sdp/chunklist.m3u8"}]

Answer 1

对不起，请不要回答一个真正的答案，但我还不准发表评论。

您没有＆＃34;数据＆＃34; JSON响应中的字段如果我没有弄错，那么在Android代码中调用它将会失败。所以改变这个

String data = handler.handleResponse(response);
JSONObject jObj = new JSONObject(data);
JSONObject j_user = jObj.getJSONObject("data");  // why do you do this?
String path = j_user.getString("sources");

到

String data = handler.handleResponse(response);
JSONObject jObj = new JSONObject(data);
String path = jObj.getString("sources");

我觉得你的JSON响应以[]括号开头和结尾似乎很奇怪。也许你必须首先将它转换为JSON数组并从中获取第一个JSON对象？所有这些都应该导致Android代码出错（应该调用catch子句），但是你说你没有收到错误，所以我不确定我的想法是否可以帮到你。

修改：

你必须在try和catch之外放置路径声明或者在其中包含所有代码。还尝试首先将您的响应解析为JSON数组。试试这个：

try{ HttpClient httpClient = new DefaultHttpClient(); HttpPost httpPost = new HttpPost("http://192.168.1.25/userdatabase/include/GetSource.php"); HttpResponse response = httpClient.execute(httpPost); BasicResponseHandler handler = new BasicResponseHandler(); String data = handler.handleResponse(response); JSONArray array = new JSONArray(data); JSONObject jObj = array.getJSONObject(0); String path= jObj.getString("sources"); mVideoView.setVideoPath(path); mVideoView.setMediaController(new MediaController(getActivity())); mVideoView.setVideoChroma(MediaPlayer.VIDEOCHROMA_RGB565); mVideoView.requestFocus(); progressBar.setVisibility(View.VISIBLE); mVideoView.setOnPreparedListener(new MediaPlayer.OnPreparedListener() { @Override public void onPrepared(MediaPlayer mediaPlayer) { // optional need Vitamio 4.0 progressBar.setVisibility(View.GONE); mediaPlayer.setPlaybackSpeed(1.0f); } }); } catch (Exception e) { Log.e("Log_tag", "Error converting result" + e.toString()); }

Answer 2

尝试以下代码。这应该是您确切的try... catch块。您现在可以在path块之外访问try变量。

String path = null;
try{
        HttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost("http://192.168.1.25/userdatabase/include/GetSource.php");
        HttpResponse response = httpClient.execute(httpPost);
        BasicResponseHandler handler = new BasicResponseHandler();
        String data = handler.handleResponse(response).replace('"','\'');  //edited this line
        JSONArray ja = new JSONArray(data); //added this line
        JSONObject jObj = ja.getJSONObject(0); //edited this line
        path= jObj.getString("sources");

} catch (Exception e) {
    Log.e("Log_tag", "Error converting result" + e.toString());
}

它让我回到了你期待的道路。

Android：如何从MySQL检索数据并将其分配给变量

2 个答案: