Dim*_*13i 8 php mysqli prepared-statement sql-update
这是我的情况:
$sql = 'UPDATE user SET password = ? WHERE username = ? AND password = ?';
if($stmt->prepare($sql)) {
$stmt->bind_param('sss', $newPass, $_SESSION['username'], $oldPass);
$stmt->execute();
}
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现在,如何查看UPDATE查询是否成功执行?更确切地说,如何查看旧密码和用户名是否正确以便我可以存储新密码?我试过这样做:
$res = $stmt->execute();
echo 'Result: '.$res;
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但我总是得到:
Result: 1
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即使旧密码不正确.