java - 错误 : Connecting MySQL database in Android app using PHP

Question

我一直在尝试使用 android 应用程序中的 php 连接 MySQL 数据库。我不知道下面的代码有什么问题。谁能告诉我我需要做什么。

这是我的代码:

1. sql查询

   CREATE TABLE user_detail
   (
   name varchar(30),
   age int(2),
   email varchar(30)
   );
   

2.getdata.php类

<?php

$con=mysql_connect('localhost','root','');
mysql_select_db('hellodb',$con); 

$name = $_POST['name'];
$age = $_POST['age'];
$email = $_POST['email'];

$mysql_query("insert into user_detail(name,age,email) values('{$name}','{$age}','{$email}')");

?>

3.MainActivity.java类

public class MainActivity extends ActionBarActivity {

    EditText eName,eAge,eEmail;
    Button inButton;
    InputStream is=null;
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        StrictMode.ThreadPolicy tp = new StrictMode.ThreadPolicy.Builder().permitAll().build();
        StrictMode.setThreadPolicy(tp);
        setContentView(R.layout.main);

        eName= (EditText) findViewById(R.id.etname);
        eAge= (EditText) findViewById(R.id.etage);
        eEmail= (EditText) findViewById(R.id.etemail);
        inButton=(Button)findViewById(R.id.ibutton);

        inButton.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
               String name=""+eName.getText();
               String age=""+eAge.getText();
               String email=""+eEmail.getText();

                List<NameValuePair> nameValuePairs=new ArrayList<NameValuePair>(1);
                nameValuePairs.add(new BasicNameValuePair("name","name"));
                nameValuePairs.add(new BasicNameValuePair("age","age"));
                nameValuePairs.add(new BasicNameValuePair("email","email"));

                try {
                    HttpClient httpClient=new DefaultHttpClient();
                    HttpPost httpPost=new HttpPost("http://10.0.2.2:8888/demo/getdata.php");
                    httpPost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
                    HttpResponse response=httpClient.execute(httpPost);

                    HttpEntity entity=response.getEntity();

                    is=entity.getContent();

                } catch (ClientProtocolException e) {
                    Log.e("ClientProtocol","LogTag");
                    e.printStackTrace();
                }catch (IOException e) {
                    Log.e("LogTag","IOException");
                    e.printStackTrace();
                }
            }
        });
    }

}

1. AndroidManifest.xml

2. ma​​in.xml类

   <EditText
       android:id="@+id/etname"
       android:hint="Enter Your Name"
       android:textAlignment="center"
       android:layout_width="fill_parent"
       android:layout_height="wrap_content" />
   
   <EditText
       android:id="@+id/etage"
       android:hint="Enter Your Age"
       android:layout_width="fill_parent"
       android:layout_height="wrap_content" />
   <EditText
       android:id="@+id/etemail"
       android:hint="Enter Your Email"
       android:layout_width="fill_parent"
       android:layout_height="wrap_content" />
   
   <Button
       android:layout_width="169dp"
       android:layout_height="wrap_content"
       android:text="Insert"
       android:id="@+id/ibutton"
       android:layout_gravity="center_horizontal" />
   

Answer 1

查找有关 PDO 的更多信息 Here

$dbtype     = "mysql";
$dbhost     = "localhost";
$dbname     = "hellodb";
$dbuser     = "root";
$dbpass     = "";

$name  = $_POST['name'];
$age   = $_POST['age'];
$email = $_POST['email'];

// database connection
$conn = new PDO("mysql:host=$dbhost;dbname=$dbname",$dbuser,$dbpass);

$sql = "INSERT INTO user_detail(name,age,email) VALUES (:name,:age,:email)";

$q = $conn->prepare($sql);

$q->execute(array(':name'=>$name,
                  ':age'=>$age,
                  ':email'=>$email));