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PHP 表单未在 MySQL 中插入数据
[英] PHP Forms not inserting data in MySQL
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问题描述
我无法将数据保存到表中，我在名为 Register.php 的文件中有以下代码，我启用了 error_reporting 以查看任何错误，但没有显示任何错误.我的数据库称为注册，表称为用户，托管在 GoDaddy 上
<?phpif(isset($_Post['注册'])) {session_start();$FName = $_POST['First_Name'];$LName = $_POST['Last_Name'];$用户名 = $_POST['用户名'];$Email = $_POST['Email'];$PW = $_POST['密码'];$sql = $con->query("INSERT INTO users (Fname, Lname, Username, Email, Password)Values('{$Fname}', '{$LName}', '{$Username}', '{$Email}', '{$PW}')");}?><!doctype html><头><meta http-equiv='Content-Type' content='text/html;字符集=utf-8'/><link href="Style/Master.css" rel="stylesheet" type="text/css"/><title>注册</title><身体><form action="" method="post" name="RegisterForm" id="RegisterForm"><div class="FormElement"><input name="First_Name" type="text" required="required" class="TField" id="First_Name" placeholder="First Name">
<div class="FormElement"><input name="Last_Name" type="text" required="required" class="TField" id="Last_Name" placeholder="Last Name">
<div class="FormElement"><input name="Username" type="text" required="required" class="TField" id="Username" placeholder="Username">
<div class="FormElement"><input name="Email" type="text" required="required" class="TField" id="Email" placeholder="Email">
<div class="FormElement"><input name="Password" type="text" required="required" class="TField" id="Password" placeholder="Password">
<div class="FormElement"><input name="Register" type="submit" class="button" id="Register" placeholder="Register">
</html>
这是Connections.php中的代码，只是没有我的真实用户名和密码
任何帮助将不胜感激！
解决方案
您需要编辑 Connection.php
mysql_connect("localhost", "USERNAME", "PASSWORD");mysql_select_db("注册");//你的数据库名称
而且你还必须编辑
mysql_query("INSERT INTO users (Fname, Lname, Username, Email, Password)Values('$Fname', '$LName', '$Username', '$Email', '$PW')") 或死(mysql_error());
I'm having trouble with data not saving to a table, I've got the following code in a file called Register.php, I have enabled error_reporting to see any errors, but none are showing. My database is called Registration and the table is called users and is being hosted on GoDaddy
<?php
require 'Connections/Connections.php';
?>
<?php
if(isset($_Post['Register'])) {
session_start();
$FName = $_POST['First_Name'];
$LName = $_POST['Last_Name'];
$Username = $_POST['Username'];
$Email = $_POST['Email'];
$PW = $_POST['Password'];
$sql = $con->query("INSERT INTO users (Fname, Lname, Username, Email, Password)Values('{$Fname}', '{$LName}', '{$Username}', '{$Email}', '{$PW}')");
}
?>
<!doctype html>
<html>
<head>
<meta http-equiv='Content-Type' content='text/html; charset=utf-8'/>
<link href="Style/Master.css" rel="stylesheet" type="text/css" />
<title>Register</title>
</head>
<body>
<form action="" method="post" name="RegisterForm" id="RegisterForm">
<div class="FormElement">
<input name="First_Name" type="text" required="required" class="TField" id="First_Name" placeholder="First Name">
</div>
<div class="FormElement">
<input name="Last_Name" type="text" required="required" class="TField" id="Last_Name" placeholder="Last Name">
</div>
<div class="FormElement">
<input name="Username" type="text" required="required" class="TField" id="Username" placeholder="Username">
</div>
<div class="FormElement">
<input name="Email" type="text" required="required" class="TField" id="Email" placeholder="Email">
</div>
<div class="FormElement">
<input name="Password" type="text" required="required" class="TField" id="Password" placeholder="Password">
</div>
<div class="FormElement">
<input name="Register" type="submit" class="button" id="Register" placeholder="Register">
</div>
</body>
</html>
This is the code in the Connections.php just without my real username and password
<?php
$con = mysql_connect("localhost", "USERNAME", "PASSWORD", "Registration");
?>
Any help would be very much appreciated!
解决方案
You need to edit the Connection.php
mysql_connect("localhost", "USERNAME", "PASSWORD");
mysql_select_db("Registration"); //your data base name
And also you have to edit
mysql_query("INSERT INTO users (Fname, Lname, Username, Email, Password)Values('$Fname', '$LName', '$Username', '$Email', '$PW')") or die(mysql_error());
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