我的网站上有一个表单标签,该表单标签指向带电子邮件和/或/没有描述的PHP页面。基于该代码生成查询,该查询需要更新这些凭证。该部分代码有效并且已经过测试。问题在于数据库未更新电子邮件凭证,但是如果我将其更新为描述,它将这样做。该代码有3个检查项,如果用户仅放置他的电子邮件,是否仅放置他的描述或两者都放置。基于此,代码的工作方式如下:<?phpsession_start();include_once 'connection.php';$id = $_SESSION['user_id'];if(isset($_POST['emailChange']) || isset($_POST['descChange'])){ $desc = $_POST['descChange']; $email = $_POST['emailChange']; if(empty($email)){ $query = "UPDATE users SET description = :descr WHERE user_id= :id ;"; $stmt = $conn->prepare($query); $stmt->bindParam(":descr", $desc); } else if(empty($desc)){ $query = "UPDATE users SET user_email= :email WHERE user_id= :id ;"; $stmt = $conn->prepare($query); $stmt->bindParam(":email", $email); } else{ $query = "UPDATE users SET description = :descr AND user_email = :email WHERE user_id= :id;"; $stmt = $conn->prepare($query); $stmt->bindParam(":email", $email); $stmt->bindParam(":descr", $desc); } if(!filter_var($email, FILTER_VALIDATE_EMAIL)){ header("Location: ../profile.php?error=invalidEmail"); exit(); } $stmt->bindParam(":id", $id); $stmt->execute(); }表单本身看起来像这样: <form action="assets/upload.php" method="POST"> <input type="text" name="emailChange" class="inputs" id="changeEmail" placeholder = "Enter your new E-mail"> <input type="text" name="descChange" class="inputs" id="changeDesc" placeholder="Enter your description"> <button type="submit" id="btnconfirmCreds" name="changeCreds">Confirm Changes</button> </form>数据库中的名称如下所示:[user_id][user_username][user_email][user_password][role_id][user_image][description][num_of_posts
2 回答
慕工程0101907
TA贡献1887条经验 获得超5个赞
有关逻辑的一些注意事项:
if(isset($_POST['emailChange']) || isset($_POST['descChange']))
{
$desc = $_POST['descChange'];
$email = $_POST['emailChange'];
...
首先,检查是否至少存在一个参数,然后再访问两个参数。您可能会争辩说,该表单总是同时发送,但永远不要相信用户输入:操作数据是如此简单!
要么将您更改if(...)为:
if( isset($_POST['emailChange']) && isset($_POST['descChange']) )
以下是具有相同语义的较短形式:
if( isset( $_POST['emailChange'], $_POST['descChange'] ) )
其他方法是更改其他2行,例如:
$desc = isset($_POST['descChange']) ? $_POST['descChange'] : '';
$email = isset($_POST['emailChange']) ? $_POST['emailChange'] : '';
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