关于PHP面向对象的一个问题

发布网友发布时间：2022-04-06 02:43

共2个回答

热心网友时间：2022-04-06 04:12

调试一下其实就可以很快得到答案的！

zend只类的一步步调试

第一步应该先echo sql 执行看看

第二步echo 数据集合看看有没有得到array

//以下测试成功
<?php
define('DBHOST','192.168.1.1');
define('DBUSER','asktang');
define('DBPW','asktang123');
define('DBNAME','wxt_115');

function connect($dbhost = DBHOST, $dbuser = DBUSER, $dbpw = DBPW, $dbname = DBNAME )
{
$conn = @mysql_connect($dbhost,$dbuser,$dbpw) or die("无法连接到数据库!");

mysql_query("SET character_set_connection='gbk', character_set_results='gbk', character_set_client=binary");
mysql_query("SET sql_mode=''");

mysql_select_db($dbname,$conn) or die("无法选择数据库!");
}

function close()
{
return mysql_close();
}

connect();

?>
<?

class UserInfo
{
private $userName; //属性,用户名

private $userInfo; //存储数据库返回信息的数组变量.

public function __construct($uid)
{

$sql="select count(*) from supe_spaceitems where uid=$uid";
$result = mysql_query($sql);

$this->userInfo = mysql_fetch_array($result); //返回查询结果到数组

close();

$this->getInfo(); //调用传递信息的方法.
}
// 获取信息传递给属性的方法
private function getInfo()
{
$this->userName = $this->userInfo[0]; //这边你用0看看
}

热心网友时间：2022-04-06 05:30

private function getInfo(){
$this->userName = $this->userInfo["username"];
$this->userPSW = $this->userInfo["userpsw"];
$this->userAge = $this->userInfo["userage"];
$this->userGrade = $this->userInfo["usergrade"];
}

改成

private function getInfo(){
$this->userName = $this->userInfo["name"];
$this->userPSW = $this->userInfo["password"];
$this->userAge = $this->userInfo["age"];
$this->userGrade = $this->userInfo["grade"];
}