我正在尝试使用php和表单将项目添加到我的数据库(sql),但数据没有被添加,似乎没有任何事情发生,我只是留在create.php页面。
php code
<?php
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "PolyTest";
// Create connection
$conn = mysql_connect($servername, $username, $password);
mysql_select_db($dbname)
$doorName = $_POST['doorName'];
$doorDes = $_POST['doorDes'];
$doorPrice = $_POST('doorPrice');
$doorColour = $_POST('doorColour');
$doorImage = $_POST['doorImage'];
if(!$_POST['submit']){
echo "please fill in the boxs";
header('Location: dooradd.php');
} else {
mysql_query("INSERT INTO Doors ('ID', 'name', 'description', 'price', 'colour', 'image') VALUES(NULL, '$doorName', '$doorDes', '$doorPrice', '$doorColour', '$doorImage')") or die(mysql_error());
echo "Door been added!";
header('Location: doorlist.php');
}
?>
HTML表格
<form class="add" action="doorCreate.php" method="post">
<input type="text" name="doorName" value="doorName">
<input type="text" name="doorDes" value="doorDes">
<input type="text" name="doorPrice" value="doorPrice">
<input type="text" name="doorColour" value="doorColour">
<input type="text" name="doorImage" value="doorImage">
<input type="submit" name="submit">
</form>
答案 0 :(得分:3)
使用mysql_select_db($dbname)
mysql_select_db($dbname);
并改变;
$doorPrice = $_POST('doorPrice');
$doorColour = $_POST('doorColour');
与
$doorPrice = $_POST['doorPrice'];
$doorColour = $_POST['doorColour'];