PHP MySQL查询，其中x= $variable

Question

我有这个代码(我知道电子邮件已经定义好了)

 <?php
$con=mysqli_connect($host,$user,$pass,$database);
 if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT `note` FROM `glogin_users` WHERE email = '.$email.'");

while($row = mysqli_fetch_array($result))
echo $row
?>

在我的MySQL数据库中，我有以下设置(表名为glogin_users) id电子邮件备注

我尝试从数据库中提取注释文本，然后对其进行回显，但似乎没有任何回显。

Answer 1

您现在所做的是在字符串上添加.，而不是连接。它应该是，

$result = mysqli_query($con,"SELECT `note` FROM `glogin_users` WHERE email = '".$email."'");

或者简单地说

$result = mysqli_query($con,"SELECT `note` FROM `glogin_users` WHERE email = '$email'");

Answer 2

$result = mysqli_query($con,"SELECT `note` FROM `glogin_users` WHERE email = '".$email."'");
while($row = mysqli_fetch_array($result))
echo $row['note'];

Answer 3

你必须这样做才能回应它：

echo $row['note'];

(数据将以数组的形式出现)