无法使用Android将数据插入MySQL数据库

Question

EditText tv_username;
    EditText tv_firstname;
    EditText tv_age;
    Button reg;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        tv_username = (EditText) findViewById(R.id.username);
        tv_firstname = (EditText) findViewById(R.id.firstname);
        tv_age = (EditText) findViewById(R.id.age);
        reg = (Button) findViewById(R.id.register);



        reg.setOnClickListener(new View.OnClickListener() {

            @Override
            public void onClick(View arg0) {
                // TODO Auto-generated method stub

                HttpClient httpclient = new DefaultHttpClient();
                HttpPost httppost = new HttpPost("http://10.0.2.2/regandroid.php");
                if(httppost != null)
                {
                    Context context = getApplicationContext();
                    CharSequence text = "Connected";
                    int duration = Toast.LENGTH_SHORT;

                    Toast toast = Toast.makeText(context, text, duration);
                    toast.show();

                }
                try
                {
                    List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
                    nameValuePairs.add(new BasicNameValuePair("username", tv_username.getText().toString()));
                    nameValuePairs.add(new BasicNameValuePair("firstname", tv_firstname.getText().toString()));
                    nameValuePairs.add(new BasicNameValuePair("age", tv_age.getText().toString()));

                    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
                    HttpResponse response = httpclient.execute(httppost);

                }
                catch(Exception e)
                {
                    e.printStackTrace();
                }

This is my phpcode

            $username = $_POST['username'];
             $firstname = $_POST['firstname'];
            $age = $_POST['age'];

           $query = mysql_query($connect, "insert into users
               (username, firstname,age) values             ('$username' 
                ,'$firstname','$age') ");


                ?>

Maybe my php code have a problem I couldn't find what the problem is.This is my Php and java code I am unable to insert data although it's connected to the database. I can't figure it out what mistake I have done.

Answer 1

Your parameters for mysql_query are in the wrong order.

mysql_query ( string $query [, resource $link_identifier = NULL ] )

The link identifier / connection should be added via the second parameter, whereas the query should be added via the first parameter.

Change:

$query = mysql_query($connect, "insert into users(username, firstname,age) values ('$username', '$firstname', '$age')");

to

$query = mysql_query("insert into users(username, firstname,age) values ('$username', '$firstname', '$age')", $connect);

Also, your code is open to SQL injection as you're inserting external data directly into your query. Use prepared statements with PDO or at least use mysql_real_escape_string if that is not an option.

Answer 2

The database link isn't really required. If passed however, it is only passed at the end of the mysql_query statement. ie:

$query = mysql_query("insert into users (username, firstname, age) values ('$username', '$firstname', '$age')", $connect);

Hope this helps!