我正在尝试将用户IP地址存储到VARCHAR(39)下的mysql表中,但它只存储为"::1"
我正在使用此代码:
<?php
session_start();
$db_hostname = 'localhost';
$db_database = 'hewden1';
$db_username = 'root';
$db_password = '';
$db_server = mysql_connect($db_hostname, $db_username, $db_password)
or die("Unable to connect to MySQL: " . mysql_error());
mysql_select_db($db_database)
or die("Unable to select database: " . mysql_error());
$cname = $_POST['cname'];
$creg = $_POST['creg'];
$address = $_POST['address'];
$post = $_POST['post'];
$contactn = $_POST['contactn'];
$contactt = $_POST['contactt'];
$email = $_POST['email'];
$vat = $_POST['vat'];
$ipaddress = $_SERVER["REMOTE_ADDR"];
$sql="INSERT INTO supplier_registration (company_name, company_reg_number, company_address, company_postcode, contact_name, contact_number, contact_email, company_vat_number, date_time, user_ip)
VALUES ('$cname', '$creg', '$address', '$post', '$contactn', '$contactt', '$email', '$vat', NOW(), '$ipaddress')";$result = mysql_query($sql);
if($result){
echo "jobs a gooden";
}else {
echo "ERROR";
}
?>
有人可以告诉我我哪里出错了,谢谢
答案 0 :(得分:2)
简单的解决方案,将您的ip
列更改为varbinary (16)
数据类型,然后使用以下内容存储ips:
$ip = bin2hex(inet_pton($_SERVER['REMOTE_ADDR']));
注意:请确保您的php版本是最新的,因为inet_pton()
是一个新功能。