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我正在开发一个 android 应用程序，其想法是当您按下按钮时，该应用程序将连接到数据库并显示数据库内容。这里的问题是什么都没有发生。

这是 onclick() 方法

Button butt1=(Button)findViewById(R.id.buttonUN);
butt1.setOnClickListener( new OnClickListener(){

    @Override
    public void onClick(View v) {
        // TODO Auto-generated method stub
        Toast.makeText(getApplicationContext(),
            "chargement en cours", Toast.LENGTH_LONG).show();

        StringBuilder responseHTTP = new StringBuilder();//stocker la reponse HTTP
        HttpClient client =new DefaultHttpClient();      //envoi de requete HTTP

        HttpGet httpGet =new HttpGet                     //recuperer l'url de fichier PHP
            ("http://192.168.12.5/BaseUne/connect.php");     // via la methode HttpGet
            try{
                HttpResponse response= client.execute(httpGet); // recevoir la reponse
                StatusLine statusline=response.getStatusLine(); // StatusLine: methode pour savoir le statut de 
                int statuscode = statusline.getStatusCode();    // de la connexion
                if (statuscode==200){
                    HttpEntity entity =response.getEntity();
                    InputStream content = entity.getContent();   //InputStream :recuperer les données binaires
                    //InputStreamReader: construction des caracteres

                    BufferedReader reader= new BufferedReader(new InputStreamReader(content));
                    String line;
                    while((line=reader.readLine())!=null){
                        responseHTTP.append(line);
                    }
                    JSONObject jsonObject =new JSONObject(responseHTTP.toString());  // creation d'un objet JSON
                    int    ID       =jsonObject.getInt("ID_contenu");
                    String Descrip  =jsonObject.getString("Description");

                    Intent a =new Intent(ChoixDuSite.this,AdminInterface.class);

                    a.putExtra("String", Descrip);
                    a.putExtra("int", ID);

                    startActivity(a);

                }
            } catch (Exception e){
                Toast.makeText(getApplicationContext(),
                    e.getMessage(), Toast.LENGTH_LONG).show();
                e.printStackTrace();
            }
     }
});

这是我的php文件： 

<?php 
mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("DBname") or die(mysql_error());
$sql=mysql_query("SELECT * FROM TABLEname");
while($row=mysql_fetch_assoc($sql))
    print(json_encode($row));
mysql_close();
?>
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0 回答 0