我制作了一个网络应用程序，需要让用户更改密码，所以我用临时代码制作了一列。问题是，我的查询确实更新了我的临时代码，但不会将密码更新为新密码。请帮忙，提前谢谢！）
<?php
if(isset($_POST['submit'])) {
$code = $_SESSION['code'];
$email = $_SESSION['emailneger'];
$password = $_POST['password2'];
$password1 = $_POST['password3'];
$reset = $_POST['reset'];
$tempcode = "";
if($reset == $reset) {
if($password == $password1) {
$query = "UPDATE consultant SET password = ?
AND tempcode = ?
WHERE email = ? ";
$qry=$db->prepare($query);
$qry->execute(array($password, $code, $email));
echo "<p>Uw wachtwoord is hersteld!</p>";
$query = 'UPDATE consultant SET tempcode = ? WHERE email = ?';
$qry=$db->prepare($query);
$qry->execute(array($tempcode, $email));
} else {
echo "<p>Wachtwoorden komen niet overeen.</p>";
}
} else {
echo "<p>Je link is ongeldig, probeer nogmaals alstublieft!</p>";
}
}
?>
您的更新不正确。语法为
UPDATE yourtable SET field1=value1, field2=value, ...
^---
注意字段/值对之间的,。您使用的是AND，它将被解析/执行为
UPDATE consultant SET password = (? AND tempcode = ?) WHERE email = ?
UPDATE consultant SET password = ('foo' AND tempcode=123) WHERE email = 'foo@example.com')
UPDATE consultant SET password = ('foo' AND true/false) WHERE ...
UPDATE consultant SET password = true/false WHERE ...