尝试在html中运行PHP插入文件时出现错误404
我对MVC编码相当陌生,我正在尝试编写一个简单的web应用程序来存储我的成员信息。我有一个html表单,我想调用另一个名为“insertmember.php”的文件,“insertmember.php”应该接受从表单传入的参数并插入到我的mysql数据库中。问题是,我不确定是否将“insertmember.php”放在了错误的位置,或者是否有其他设置丢失,以便将页面拉起 所有东西(Apache服务器和MySQL)都在运行XAMPP 我的“insertmember.php”文件当前位于 E:\xampp\htdocs\Firstwebapp\application\views 我的PHP文件尝试在html中运行PHP插入文件时出现错误404,php,codeigniter,Php,Codeigniter,我对MVC编码相当陌生,我正在尝试编写一个简单的web应用程序来存储我的成员信息。我有一个html表单,我想调用另一个名为“insertmember.php”的文件,“insertmember.php”应该接受从表单传入的参数并插入到我的mysql数据库中。问题是,我不确定是否将“insertmember.php”放在了错误的位置,或者是否有其他设置丢失,以便将页面拉起 所有东西(Apache服务器和MySQL)都在运行XAMPP 我的“insertmember.php”文件当前位于 E:\xa
<?php
defined('BASEPATH') OR exit('No direct script access allowed');
class Welcome extends CI_Controller {
function __construct()
{
parent::__construct();
}
function index()
{
$servername = "localhost";
$username = "";
$password = "";
$dbname = "test";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error)
{
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT member_ID, member_name, city_level, attack_power FROM member";
//$result = $conn->query($sql);
$data['result'] = $conn->query($sql);
$data['title'] = "GreenForest Member";
$data['heading'] = "List of GreenForest members:";
$this->load->view('home', $data);
$conn->close();
}
}
?>
我的HTML文件
<html>
<head>
<title>
<?=$title?>
</title>
</head>
<body>
<h3>Add New member</h3>
<form action = "insertmember.php" method ="post">
member_name : <input type = "text" name = "member_name">
<br/>
city_level : <input type = "text" name = "city_level">
<br/>
attack_power : <input type = "text" name = "attack_power">
<br/>
<input type = "submit" value = "INSERT">
</form>
<h3><?=$heading?></h3>
<table style="width:100%">
<tr>
<th>Member_ID</th>
<th>Member_name</th>
<th>City_level</th>
<th>Attack_power</th>
</tr>
<?php while($row = mysqli_fetch_assoc($result)) { ?>
<tr>
<th><?php echo$row['member_ID'] ?></th>
<th><?php echo$row['member_name'] ?></th>
<th><?php echo$row['city_level'] ?></th>
<th><?php echo$row['attack_power'] ?></th>
</tr>
<?php } ?>
</table>
</body>
</html>
添加新成员
成员姓名:
城市层面:
攻击力:
会员身份证
成员姓名
城市级
攻击力
我的insertmember.php文件
<html>
<body>
<? php
$servername = "localhost";
$username = "";
$password = "";
$dbname = "test";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error)
{
die("Connection failed: " . $conn->connect_error);
}
$member_name = $_POST['member_name'];
$city_level = $_POST['city_level'];
$attack_power = $_POST['attack_power'];
$sql = "INSERT INTO member (member_name, city_level, attack_power) VALUES ( '$member_name', '$city_level', '$attack_power')";
if(!mysqli_query($con, $sql))
{
echo 'not inserted';
}
else
{
echo 'inserted';
}
mysql_close($con)
header("refresh:2; url=index.html");
?>
</body>
</html>
这已经解决了。。我得换衣服
发件人:
致:
向Swati竖起大拇指,指向我检查目录路径的方向也许这不是问题所在,因为您正在Windows中运行,但我建议您在文件名的大小写上保持一致。您在表单请求中使用“insertmember.php”,在文件名中使用“insertmember.php”。您是否在这里->echo$row['member_ID']
echo
和$row
之间必须有很少的空间,而且您没有添加代码>最后,它应该是这样的->echo$row['member_ID'代码>,对所有其他语句执行相同的操作。谢谢Marcelino,Sawti,大写和语法已修复,当我尝试访问此页“”时仍然看到相同的问题,是否还有其他需要执行/添加的操作?
form action = "insertmember.php" method ="post"
form action = "index.php/welcome/insertmember" method ="post"