Share about PHP Write app interface and return JSON data instance

The following small series for everyone to bring a PHP write app interface and return the JSON data instance (share). Small series feel very good, now share to everyone, also for everyone to make a reference. Let's take a look at it with a little knitting.

The first step: Conn. PHP file to connect to the database and define the interface format, the code is as follows:

<?php  Header ("Charset=utf-8");  $servername = "localhost";  $username = "root";  $password = "root";  $dbname = "Test";  $conn = mysql_connect ($servername, $username, $password);  if (! $conn) {   echo "database connection failed! ";  }  mysql_select_db ($dbname);  Class response{public   static function json ($code, $message = "", $data =array ()) {    $result =array (     ' code ' = = $code,     ' message ' = $message,     ' data ' and $data    );    Output JSON    echo json_encode ($result);    Exit;   }  }  ? >

Step two: text.php, which converts the data in the database into a JSON string and outputs:

<?php  require_once (' conn.php ');  /   * * Select data table * */  $sqla = "SELECT * from user";  $result = mysql_query ($sqla, $conn);  $dataarr = Array ();  while ($row = Mysql_fetch_array ($result)) {   $dataarr []= $row;  }  $id =$_get[' id '];  if ($id ==1) {   Response::json (1, ' data returned successfully ', $dataarr);  } else if ($id ==2) {   Message::json (0, ' failed ');  }   ? >

Step three: Text.html,ajax loads the JSON data and displays:

<! DOCTYPE html> 
This makes it possible to write JSON interfaces using PHP.