我試圖通過PHP獲取兩個輸入並將其添加到數據庫中。我得到的誤差是無法將數據添加到PHP中的表中
警告:mysqli_query()預計至少2個參數,1 C中給出:\ XAMPP \ htdocs中\形成具有db.php中上線24
警告:mysqli_query() :用C空查詢:上線25
這裏\ XAMPP \ htdocs中\形成db.php中是
<?php
if(isset($_GET['name']) || isset($_GET['id']))
{
$username = $_GET['name'];
$userid = $_GET['id'];
}
$host_name = 'localhost';
$user = 'root';
$password = '';
$database = 'test';
$connection = mysqli_connect($host_name,$user,$password,$database);
if ($connection) {
echo "Connected to the DB";
# code...
}
else
{
echo "Not Connected";
}
$sql = mysqli_query("INSERT INTO `employee` (employee_id , Name) VALUES ($userid , $username)");
if (mysqli_query($connection,$sql)) {
# code...
echo "Values added to the table";
}
else
{
echo "values not added".mysqli_error($connection);
}
?>
<form action="Form with DB.php" method="GET">
Enter Employee Name : <input type = "text" name = "name"><br>
Enter Employee ID : <input type = "text" name = "id"><br>
<input type = "submit" name="submit" value="Submit">
你的代碼以及其它諸多問題,尤其是[SQL注入攻擊(http://bobby-tables.com)漏洞,帶引號的值,具有完全不同的參數caling查詢兩次,胡說等等等等。基本上你需要在這些函數上使用RTFM,並清理所有與此代碼有關的事情。 –