我正在尝试从表单插入值并将其插入到MySQL数据库中。 但它不起作用。 我是初学者,请帮忙。
这是我的代码:
<?php
function register(){
$name = $_POST['name'];
$address = $_POST['address'];
$email = $_POST['email'];
$phone = $_POST['phone'];
//echo $name;
$con = mysqli_connect("localhost", "root", "", "LinuxCreditSociety");
$rs = $con->query("insert into cust_mst (customer_id,customer_name,customer_address,customer_mobile,email_id)values(1003,'$name','$address','$phone','$email')");
$rs->free();
$con->close();
}
?>
<html>
<head>
<title>Registration</title>
<link rel="stylesheet" type="text/css" href="phpcss.css"></link>
</head>
<body>
<div style="position:absolute;left:300px;top:5px">
<h1 align="center"><font face="Purisa" size="20" color="purple">Linux Credit Society</font></h1>
</div>
<div style="position:absolute;right:160px;top:5px"><img src="linux.jpg" height="150" /></div>
<div style="position:absolute;left:140px;top:5px"><img src="linux.jpg" height="150" /></div>
<form method="post">
<div class="st1">
Name:</br></br>
Address:</br></br>
Email-id:</br></br>
Phone#:</br></br>
</div>
<div class="st2">
<div style="position:absolute;top:5px">
<input type="text" name="name">
</div>
<div style="position:absolute;top:78px">
<input type="text" name="address">
</div>
<div style="position:absolute;top:148px">
<input type="text" name="email">
</div>
<div style="position:absolute;top:218px">
<input type="text" name="phone">
</div>
<div style="position:absolute;top:290px;">
<input type="submit" value="Register">
</div>
</div>
</form>
<div style="position:absolute;top:320px;">
<?php
if(isset($_POST['submit']) && $_POST['submit'] == "Register")
register();
?>
</div>
</body>
</html>
这就是我创建数据库的方式:
create database LinuxCreditSociety;
use LinuxCreditSociety;
create table cust_mst(
customer_id int,
customer_name varchar(50),
customer_address varchar(70),
customer_mobile double,
email_id varchar(50));
insert into cust_mst values(1001, 'Jack Mathew', 'Bandra', '9998887770', 'jackm@yahoo.com');
insert into cust_mst values(1002, 'Jill Roberts', 'Dadar', '999665550', 'jillr@rediff.com');
编辑:
伙计我刚刚做了一个改变而且它起作用了。现在我也会在注射工作!!
以下是我的所作所为:
我刚刚从 - &gt;
更改了来电 <?php
if(isset($_POST['submit']) && $_POST['submit'] == "Register")
register();
?>
到此 - &gt;
<?php
if(isset($_POST['name']) && $_POST['name'] != "")
register();
?>
答案 0 :(得分:5)
当然你反过来想要这个:
$_POST[name] = $name;
$_POST[address] = $address;
$_POST[email] = $email;
$_POST[phone] = $phone;
例如:
$name = $_POST['name'];
$address = $_POST['address'];
$email = $_POST['email'];
$phone = $_POST['phone'];
答案 1 :(得分:5)
我假设您的数据来自$_POST
。变化
$_POST[name] = $name;
$_POST[address] = $address;
$_POST[email] = $email;
$_POST[phone] = $phone;
向
$name = $con->real_escape_string($_POST['name']);
$address = $con->real_escape_string($_POST['address']);
$email = $con->real_escape_string($_POST['email']);
$phone = $con->real_escape_string($_POST['phone']);
请注意使用mysqli_real_escape_string()
来阻止SQL injections。此外,由于您已使用mysqli
扩展程序,请考虑构建prepared statements,而不是将变量插入查询字符串。
最后一点注意:为了防止错误通知,例如“在/ your / script中使用未定义的常量名称 - 假定'名称',请使用字符串访问数组键(即$_POST['name']
而不是$_POST[name]
)
答案 2 :(得分:0)
试试这个,但不要忘记清理你的变量!!!
function register(){
$name = $_POST['name'];
$address = $_POST['address'];
$phone = $_POST['email'];
$email = $_POST['phone'];
$con = new mysqli("localhost", "root", "", "LinuxCreditSociety");
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
if (!$con->query("insert into cust_mst (customer_id,customer_name,customer_address,customer_mobile,email_id)values(NULL,'$name','$address','$phone','$email')")) {
printf("Errormessage: %s\n", $con->error);
}
$con->close();
}
答案 3 :(得分:-1)
function register(){
$name = $_POST['name'];
$address = $_POST['address'];
$phone = $_POST['email'];
$email = $_POST['phone'];
$con = mysqli_connect("localhost", "root", "", "LinuxCreditSociety");
$rs = $con->query("insert into cust_mst (customer_id,customer_name,customer_address,customer_mobile,email_id)values(NULL,'{$name}','{$address}','{$phone}','{$email}')");
$rs->free();
$con->close();
}