OOP php问题，没有任何错误，但没有打印数据
OOP php问题，没有任何错误，但没有打印数据
提问于 2018-12-19 04:25:46
我编写了一个类，用于从数据库获取有关我的客户的一些信息(MySQL)。如下所示：
<?php
class Player {
public $username;
public $inf;
public function __construct ($username, $inf){
$this->username = $username;
$this->inf = $inf;
}
public function getInfuser (){
include('connectdatabase.php');
$sql = "SELECT * FROM member WHERE email = '.$this->username.'";
$result = $conn->query($sql);
$row = $result->fetch_assoc();
switch ($this->inf){
case ('email'):
$result = $row["email"];
break;
case ('playerid'):
$result = $row["playerid"];
break;
case ('inviteid'):
$result = $row["inviteid"];
break;
case ('hash'):
$result = $row["hash"];
break;
default:
$result = "Error";
break;
}
return $result;
}
}
?>在connectdatabase.php页面中，我编写了一个连接以连接到我的数据库，我正确地测试了它的工作。
我想要从我的类中打印信息的页面是：
<?php
include ('classtest.php');
$username = 'chance';
$inf = 'inviteid';
$keyvan = new Player($username, $inf);
echo $keyvan->getInfuser ();
?>我修复了一些存在的语法错误。
[18-Dec-2018 20:14:52 UTC] PHP Parse error: syntax error, unexpected 'function__construct' (T_STRING), expecting variable (T_VARIABLE) in /home/jokerpoker021/public_html/classtest.php on line 8
[18-Dec-2018 20:14:52 UTC] PHP Parse error: syntax error, unexpected 'function__construct' (T_STRING), expecting variable (T_VARIABLE) in /home/jokerpoker021/public_html/classtest.php on line 8
[18-Dec-2018 20:14:53 UTC] PHP Parse error: syntax error, unexpected 'function__construct' (T_STRING), expecting variable (T_VARIABLE) in /home/jokerpoker021/public_html/classtest.php on line 8
[18-Dec-2018 20:16:34 UTC] PHP Fatal error: Call to a member function query() on null in /home/jokerpoker021/public_html/classtest.php on line 17
[18-Dec-2018 20:16:35 UTC] PHP Fatal error: Call to a member function query() on null in /home/jokerpoker021/public_html/classtest.php on line 17现在没有任何错误，但我的信息不打印，只显示一个白页，我不知道为什么。我的代码有什么问题？
更新:我的connecttodatabase.php文件：
<?php
$servenm = "localhost";
$usnme = "username";
$passnm = "********";
$dbname = "jokerpok";
// Create connection
$conn = new mysqli($servenm, $usnme, $passnm, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
?>回答 1
Stack Overflow用户
发布于 2018-12-19 06:17:17
我认为你应该这样写：
$sql = "SELECT * FROM member WHERE email = '".$this->username."'";或者：
$sql = "SELECT * FROM member WHERE email = '{$this->username}'";页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/53840533
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