gpt4 book ai didi

php - 从选择查询中设置变量

转载 作者:行者123 更新时间:2023-11-29 02:27:57 24 4
gpt4 key购买 nike

不是最有 PHP 经验的人,但是由于需要和要求,我已经慢慢开始为内部/外部项目构建一个“登录”系统。

但是,用户已登录,然后可以基于三数据字段搜索扫描的 TIF 图像;由于表中有多个客户,我想将查询限制为登录用户“客户代码”

$sql = "SELECT * FROM $tbl_name WHERE username = '$username' and password = '$md5'";

$result = mysql_query($sql);

$count = mysql_num_rows($result);

if($count == 1) {

$_SESSION['username'] = strtoupper($username);
$_SESSION['loggedin'] = TRUE;
$_SESSION['customercode'] = $result['customercode'];

header('location:home.php');

} else {

echo '<h1>Wrong Username or Password</h1>';
}

简单的登录机制按预期工作并应用适当的 $_SESSION['username'] = strtoupper($username); 但是似乎无法从 customercode $result 是否正确显示?

是否可以将来自 mysql 数据库的 SELECT 查询用作 session 变量?

谢谢-

最佳答案

您需要使用 mysql_fetch_* 获取客户代码:

$result = mysql_query($sql);

$count = mysql_num_rows($result);

if($count == 1) {

$data = mysql_fetch_array($result); //add this line

$_SESSION['username'] = strtoupper($username);
$_SESSION['loggedin'] = TRUE;
$_SESSION['customercode'] = $data['customercode'];

}

Note: As on most of the question all are recommend, I also recommend to you to use the mysqli_* or PDO instead of mysql_*

关于php - 从选择查询中设置变量,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/18058953/

24 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com