这是我的数据库连接文件。 (DB_CONFIG enter code here
。PHP)
<?php
$con = mysqli_connect("localhost","root","","oop");
if(mysqli_connect_errno())
{
echo "failed to connect to mysql:" . mysqli_connect_error();
}
?>
这是我的user.class.php代码
<?php
include "db_config.php";
class User
{
public function registration($fname,$lname,$username,$email,$password)
{
$sql = $con->query("INSERT INTO users_registration(fname,lname,username,email,password)
VALUES('$fname','$lname','$username','$email','$password',)");
return $sql;
}
}
?>
最后我的registrtaion.php文件就在这里
<?php
include "class.user.php";
$user = new User();
if (isset($_POST['submit']))
{
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$username = $_POST['username'];
$email = $_POST['email'];
$password = $_POST['password'];
$user->registration($fname,$lname,$username,$email,$password);
echo "registration success";
}
?>
<!DOCTYPE HTML>
<html>
<head>
<title>Registration Form</title>
<link rel="stylesheet" href="css/style.css" type="text/css">
</head>
<body>
<div class="wrapper"><!--wrapper start here-->
<form action="" method="POST">
<input type="text" name="fname" placeholder="First name"/><br>
<input type="text" name="lname" placeholder="Lastst name"/><br>
<input type="text" name="username" placeholder="User name"/><br>
<input type="text" name="email" placeholder="Email id"/><br>
<input type="text" name="password" placeholder="Password"/><br>
<input type="text" name="cpassword" placeholder="Confirm Password"/><br>
<input type="button" value="submit" name="submit" class="btn"/>
</form>
</div><!--wrapper ends here-->
</body>
</html>
我无法将数据插入数据库,当我提交按钮时没有发生任何事情.Plz帮我解决了问题。我是新的php。 还想在同一个字段中添加图像。 提前谢谢。
答案 0 :(得分:2)
您的查询中有一个不受欢迎的,
。
$sql = $con->query("INSERT INTO users_registration(fname,lname,username,email,password)
VALUES('$fname','$lname','$username','$email','$password',)");
^
此外,要提交HTML表单,按钮类型应为submit
。改变
<input type="button" value="submit" name="submit" class="btn"/>
到
<input type="submit" value="submit" name="submit" class="btn"/>
答案 1 :(得分:0)
<?php
class dbins
{
function insert_data($name,$age)``
{
$runn = mysqli_connect("localhost","root","","oop");
$ins = "INSERT into ajax(name,age) values('$name','$age')";
$run = mysqli_qyery($runn,$ins);
return $run;
}
}
?>
<form name="insert" method="POST">
<table align="center">
<thead>
<h1 align="center">Simple Insert Use OOP</h1>
<tr>
<th>Name</th>
<td><input type="text" name="name" required="" placeholder="Enter Your Name"></td>
</tr>
<tr>
<th>Age</th>
<td><input type="text" name="age" required="" placeholder="Enter Your Age"></td>
</tr>
<tr align="center">
<td><input type="submit" name="insert" value="Insert" ></td>
</tr>
</thead>
</table>
</form>
<?php
$con = new dbins();
if(isset($_POST['insert']))
{
$name = $_POST['name'];
$age = $_POST['age'];
$con->insert_data($name,$age);
}
?>