我需要基于PHP角色的身份验证解决方案

我需要关于我的项目基于角色的明智解决方案。我有桌子

- users [ID, ..., roleID,..]
- role [roleID, role_name]
- permission_department[ID, roleID, departmentID]
- department [ID, department_name]

首先，有一个表单可以将新部门添加到部门表中。此外，还有一个带有选择框的表单，用于获取权限0或1。在此表单中，还有“部门权限”部分。我会得到属于部门的部门的许可。ID是

$conn_departmentlist = mysql_query("SELECT * FROM department");
$num_rows_departmentlist = mysql_num_rows($conn_departmentlist);
$y = $num_rows_departmentlist + 1;

for($x = 1; $x < $y; $x++) {
    $departmentIDs[$x] = @$_POST['menu_Department'][$x];
    if(isset($departmentIDs[$x])) { $departmentIDs[$x] = $x; }
}
$departmentID_permission = array();

for($x = 1; $x < $y; $x++) {
    if($departmentIDs[$x]) { array_push($departmentID_permission, $departmentIDs[$x]); }
}

$departmentID_PUTDATABASE = implode(",", $departmentID_permission); // example: 1,3,5

$conn\u departmentlist=mysql\u查询（“选择*来自部门”）；
$num_rows_departmentlist=mysql_num_rows（$conn_departmentlist）；
$y=$num_rows_departmentlist+1；
对于（$x=1；$x<$y；$x++）{
$departmentIDs[$x]=@$\u POST['menu\u Department'][$x]；
if（isset（$departmentIDs[$x]）{$departmentIDs[$x]=$x；}
}
$departmentID_permission=array（）；
对于（$x=1；$x<$y；$x++）{
if（$departmentIDs[$x]）{array_push（$departmentID_permission，$departmentIDs[$x]）}
}
$departmentID_PUTDATABASE=内爆（“，”，$departmentID_权限）；//示例：1,3,5

我将$departmentID_数据库值记录到permission_department.deparmentID及其roleID。当我需要显示部门的权限时：

**HTML:**
<?php
$conn_departmentlist = mysql_query("SELECT * FROM department");
while($get_departmentlist = mysql_fetch_array($conn_departmentlist)) {
    $departmentID = $get_departmentlist['ID'];
    $department_name = $get_departmentlist['department_name'];
?>

<input type="checkbox" class="flat" id="menu_Department[<?php echo $departmentID; ?>]" name="menu_Department[<?php echo $departmentID; ?>]"  <?php get_department_permission_sql($departmentID, $Selected_RoleID); ?> />

<?php } ?>

**PHP:**

function get_department_permission_sql($departmentID, $Selected_RoleID) {
    $conn_permission_department = mysql_query("SELECT departmentID FROM permission_department WHERE roleID = '$Selected_RoleID'");
    $get_permission_department = mysql_fetch_array($conn_permission_department);

    $departmentID_permission = array($get_permission_department['departmentID']);
    $departmentID_permission = explode(',', $departmentID_permission[0]);

    $y = count($departmentID_permission);

    for($x=0; $x < $y; $x++) {
        if($departmentID == $departmentID_permission[$x]) { echo 'checked = "checked"'; }
    }   
}

**HTML：**
为什么要使用已有十年历史的数据库API？它已被弃用、未维护且不安全，尤其是您使用它的方式。切换到PDO。我知道，但是六个月前开始的。我将很快用CakePhp更改它。我知道了。