Question

php连接数据库的类，但是选择数据库的时候总是出错，怎么回事呢？

<?php

class register{
private $host; //The host address
private $user; //The user
private $pass; //The password
private $database; //The database

//Connect with the database
function __construct(){
$host = $this -> host;
$user = $this -> user;
$pass = $this -> pass;
// $database = $this -> database;
// echo $db;

$conn = mysql_connect($host,$user,$pass);
$db = mysql_select_db($this -> database,$conn);
// if($conn){
// echo "连接成功";
// }else{
// echo "连接失败";
// }
if($db){
echo "数据库成功";
}else{
echo "数据库失败";
}
}

//Add a user
function addUser(){
$dataObj = $_POST["dataObj"];
$userInfo = json_decode($dataObj,true); //change the data from the type of json to array.
$username = $userInfo[0];
$password = $userInfo[1];
$repassword = $userInfo[2];

//The sql to add the new user
@$sql_addUser = <<<EOF
insert into users values("","$username","$password","$repassword");
EOF;
// var_dump($sql_addUser);break;
$result = mysql_query($sql_addUser);
// var_dump($result);break;
if($result){
echo 1;
}else{
echo 0;
}
}
}
// echo addUser();
$register = new register("localhost","root","","hotel");
$register -> addUser();
// break;

?>

Answer 1

构造函数错咯

        function __construct($host,$user,$pass,$database){
            $this -> host=$host;
            $this -> user=$user;
            $this -> pass=$pass;
            $this -> database=$database;
            // echo $db;
            
            $conn = mysql_connect($host,$user,$pass);
            $db = mysql_select_db($this -> database,$conn);
            if($db){
                echo "数据库成功";
            }else{
                echo "数据库失败";
            }
        }
//给你改咯哈这个类
class register{
 private $host;        //The host address
 private $user;        //The user
 private $pass;    //The password
 private $database;          //The database
 private $conn;
 
 //Connect with the database
 function __construct($host,$user,$pass,$database){
  $this -> host=$host;
  $this -> user=$user;
  $this -> pass=$pass;
  $this -> database=$database;          
  $conn = mysql_connect($host,$user,$pass)or die("连接失败!");
  mysql_select_db($this -> database,$conn) or die("选择数据库失败!");
  $this->conn=$conn;
 }
//Add a user
 function addUser($data){
  $dataObj    =     $data;//类里边不要把变量写得太死
  $userInfo   =     json_decode($dataObj,true);  //change the data from the type of json to array.
  $username   =     $userInfo[0];
  $password   =     $userInfo[1];
  $repassword =     $userInfo[2];
  
  //The sql to add the new user
  @$sql_addUser = <<<EOF
   insert into users values("","$username","$password","$repassword");
EOF;
  // var_dump($sql_addUser);break;
  $result = mysql_query($sql_addUser,$this->conn);//指定连接，少些麻烦
  // var_dump($result);break;
  if($result){
   echo  1;
  }else{
   echo 0;
  }
 }
}

Answer 2

$register = new register("localhost","root","","hotel"); //对构造函数传入了4个参数

//定义构造函数时未定义接收传入参数的变量。(括号中是空的)
function __construct(){
$host = $this -> host;
$user = $this -> user;
$pass = $this -> pass;
// $database = $this -> database;
// echo $db;
...
}