在oop php中,我創建了構造函數mysql連接(我知道它將被棄用,並且你sugest使用PDO等),但我遇到了問題。連接完成後,一切都會好起來的。但插入無法完成不知道爲什麼,代碼運行到結束。看起來對象不接受連接,但它不可能。我使用PHP 5.4.3。代碼如下:在啓動對象後PHP OOP連接mysql不起作用
Table (Coach):
Coach_id INT (AutoIncrement)
Coach_name char(30)
Coach_nationality char(30)
class League
{
public $con;
public function MySQLCon()
{
$this->con = mysql_connect("localhost","root","") or mysql_error($this->con);
mysql_select_db("basket",$this->con) or mysql_error($this->con);
return $this->con;
}
public $coach,$coachNationality;
public function NewCoach($coach,$coachNationality)
{
$this->coach = $coach;
$this->coachNationality = $coachNationality;
$Query = "insert into Coach_name (Coach_name,Coach_nationality) VALUES ('".$this->coach."','".$this->coachNationality."')";
//this query doesn't do anything but prints yes
mysql_query($Query,$this->con) or mysql_error($this->con);
echo "yes";
}
}
//no data about mike Brown in database, database engine InnoDB
$LG = new League;
$LG->MySQLCon();
$LG->NewCoach("Mike Brown","USA");
你不能從構造函數中使用返回值......你也使用php4和php5風格的屬性聲明。此外您的查詢是錯誤的。乍一看,你的代碼可能有更多的錯誤。 – PeeHaa 2013-02-27 14:12:37
好吧,沒有正確理解你的迴應,好吧,我編輯我的代碼,因爲它應該是。仍然沒有插入。正如我理解你的意思是關於var $ con和public $ coach的php4和php5樣式屬性聲明。我在這裏用錯了是嗎? – ArnasGo 2013-02-27 14:21:44
顯示我們更新的代碼 – Adder 2013-02-27 14:22:34