我试过这个PHP代码,出现很多错误,我无法识别 哪里是我的初学者错误,我真的很希望你的帮助尽快: (未定义索引+没有选择数据库
错误:未定义指数::名称在C:\ XAMPP \ htdocs中\ phpfile.php上线23
说明:未定义指数: /*成功
通知电子邮件连接用C :\ xampp \ htdocs \ phpfile.php on line 24
注意:未定义指数:电话在C:\ XAMPP \ htdocs中\ phpfile.php上线25
说明:未定义指数:口令在C:\ XAMPP \ htdocs中\ phpfile.php上线26 错误:INSERT INTO用户(姓名,电子邮件,电话,密码)VALUES( '', '', '', '') 没有数据库中选择
*/ // HTML表单:
<html>
<head>
<title> event reg </title>
<link rel="stylesheet" href="style.css">
</head>
<body>
<br /><br /><br /><br /><br /><br /><br /><br /><br /><br />
<form enctype="from-data" action="phpfile.php" method="post">
<label for ="name"> User Name :</label>
<input type="text" name="name" id="name" maxlength="30" Size="30"/>
<br />
<br />
<label for ="email"> E-mail : </label>
<input type="email" name="email" id="email" maxlength="255" Size="30"/>
<br />
<br />
<label for ="phone"> Phone :</label>
<input type="text" name="phone" id="phone" maxlength="10" Size="30"/>
<br />
<br />
<label for ="password"> password : </label>
<input type="password" name="password" id="password" maxlength="12" Size="30"/>
<br />
<br />
<label for ="event"> Event name: </label>
<select name="event" id="event">
<option value="Inceptionopening"> Inception </option>
</select>
<br />
<br />
<br />
<br />
<input type="submit" name="submit" id="submit" value="submit" />
<input type="reset" name="reset" id="reset" value="Cnacel" />
</form>
</body>
</html>
//连接到它的PHP代码:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "userdb";
// Create connection
$conn = mysqli_connect($servername, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
echo "Connected Successfully <br>";
$name=$_POST['name'];
$email=$_POST['email'];
$phone=$_POST['phone'];
$password=$_POST['password'];
$sql="INSERT INTO user(name,email,phone,password)VALUES('$name','$email','$phone','$password')";
if (mysqli_query($conn, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);
?>
你应该检查'$ _POST'中有你试图使用它之前预计的数据。另外,看起来好像你正在连接到MySQL正确 - 试试'$ conn = mysqli_connect($ servername,$ username,$ password,$ dbname);' –
我连接到一个html格式的PHP我试过填写表格并显示此错误。我也写了$ conn = mysqli_connect($ servername,$ username,$ password,$ dbname);它也会显示相同的错误:\ –
'var_dump($ _ POST)' - 查看发送到您的表单的内容。此外,由于您尚未共享您的表单的代码,我认为假设您有'method =“get”'或未指定,而不是使用'method =“post”'可能是安全的? –