Question

我有一个用户登录它的系统。每次特定用户登录时，我都会使用TIMESTAMP：

将登录日期时间存储在MySQL数据库中

db中的每条记录都有一个user_id和login_date，它存储在我的att_log表中。但是，我想确保用户每天只能根据日期登录一次。到目前为止我的代码

        <?php
        include('conn.php');
        if (isset($_POST['submit'])) {
        $UserName=$_POST['id_emp']; 
        $Password=$_POST['password'];
        $result=mysql_query("select * from employee where id_emp='$UserName' and password='$Password'" )or die (mysql_error()); 

        $count=mysql_num_rows($result);
        $row=mysql_fetch_array($result);

        if ($count > 0){
        $_SESSION['emp_id']=$row['emp_id'];
        mysql_query("insert into att_log (username,login_date,id)values('$UserName',NOW(),".$row['emp_id'].")")or die(mysql_error());
?>

如果user_id在相同日期登录，如何在将其插入att_log表之前拒绝，如何编写正确的语句来检查案例。使用当前代码，它将在同一天不断地连续插入同一个用户，只是时间不同。

我想在特定日期过滤掉特定用户，因为我确实计算了他们在月末的频率，例如

id_attlog  username   logindate           id   
1           2010    2015-06-20 19:08:19   33
2           2011    2015-06-20 19:08:19   34
3           2011    2015-06-21 19:08:19   34
4           2011    2015-06-22 19:08:19   34


User ID   Day of Attendance
33          3
34          1

用户ID 34已登录3天，用户33仅登录1天。这就是为什么我希望用户每天只能登录一次。我仍在寻找正确的方法来实现对用户的限制，让他们每天登录一次。

谢谢！

Answer 1

// Compare username/password and check last login date is less than today
$result = mysql_query('select a.*, max(b.login_date) as login_date from employee AS a LEFT JOIN att_log AS b ON a.emp_id = b.id where a.emp_id = "' . mysql_real_escape_string($UserName) . '" and a.password = "' . mysql_real_escape_string($Password) . '" and login_date < DATE(NOW()) GROUP BY a.emp_id' ) or die (mysql_error());

if (mysql_num_rows($result) > 0) {
    $row = mysql_fetch_array($result);
    mysql_query('insert into att_log (username, login_date, id) values("' . mysql_real_escape_string($UserName) . '", NOW(), "' . mysql_real_escape_string($row['emp_id']) . '"') or die(mysql_error());
} else {
    echo 'Access denied';
}

或：

// Compare username/password
$result = mysql_query('select a.*, max(b.login_date) as login_date from employee AS a INNER JOIN att_log AS b ON a.emp_id = b.id where a.emp_id = "' . mysql_real_escape_string($UserName) . '" and a.password = "' . mysql_real_escape_string($Password) . '" GROUP BY a.emp_id' ) or die (mysql_error());

$row = mysql_fetch_array($result);

// If last login date is greater or equal than the start of today, show access denied message
if (strtotime($row['login_date']) >= strtotime(date('Y-m-d 00:00:00')) {
    echo 'Sorry you have already logged in today';
} else {
    mysql_query('insert into att_log (username, login_date, id) values("' . mysql_real_escape_string($UserName) . '", NOW(), "' . mysql_real_escape_string($row['emp_id']) . '"') or die(mysql_error());
}

在PHP中不允许多次登录

1 个答案: