我的表类有这些列:存储数据库值到变量
idcategory categorySubject users_idusers
我有一个简单的单选按钮和文本框的形式。 我有类别全选语句,并需要获得idcategory存储到一个变量($ getCatId),所以我可以用这个语句:
$sql="INSERT INTO topic(subject, topicDate, users_idusers, category_idcategory, category_users_idusers) VALUES('($_POST[topic])', '$date', '$_SESSION[userid]', '$getCatId', '$_SESSION[userid]');";
是什么,以获得最佳的方式和存储类别ID?
if($_SERVER['REQUEST_METHOD'] != 'POST') //show form if not posted
{
$sql = "SELECT * FROM category;";
$result = mysqli_query($conn,$sql);
?>
<form method="post" action="createTopic.php">
Choose a category:
</br>
</br>
<?php
while ($row = mysqli_fetch_assoc($result)) {
echo "<div class= 'choice'><input type='radio' name='category' value='". $row['idcategory'] . "'>" . $row['categorySubject'] ."</div></br>";
}
echo 'Topic: <input type="text" name="topic" minlength="3" required>
</br></br>
<input type="submit" value="Add Topic" required>
</form>';
}
if ($_POST){
if(!isset($_SESSION['signedIn']) && $_SESSION['signedIn'] == false)
{
echo 'You must be signed in to contribute';
}
else{
$sql="INSERT INTO topic(subject, topicDate, users_idusers, category_idcategory, category_users_idusers) VALUES('($_POST[topic])', '$date', '$_SESSION[userid]', '$getCatId', '$_SESSION[userid]');";
$result = mysqli_query($conn,$sql);
echo "Added!";
** WARNING **:当使用'mysqli'你应该使用[参数化查询(http://php.net/manual/en/mysqli.quickstart.prepared-statements.php)和['bind_param'](http://php.net/manual/en/mysqli-stmt.bind-param.php)将用户数据添加到您的查询中。 **不要**使用字符串插值或连接来完成此操作,因为您创建了严重的[SQL注入漏洞](http://bobby-tables.com/)。 **绝不**将'$ _POST'或'$ _GET'数据直接放入查询中,如果有人试图利用您的错误,这会非常有害。 – tadman