基于PHP角色的auth明智的解决方案我需要

时间:2017-02-15 17:40:48

标签: php mysql roles user-permissions

我需要明智地解决基于我的项目的角色。我有桌子;

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

首先,我可以将一个表单添加到department表中。并且,有一个带有选择框的表单来获取权限0或1.在此表单中,还有部门权限部分。我将获得department.ID的部门的许可

$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

我将$ departmentID_PUTDATABASE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"'; }
    }   
}

这就是它的全部。这很难管理。例如,我需要经理角色,角色具有不同部门的权限。

有没有可能让它变得简单,请帮助我。提前谢谢。

0 个答案:

没有答案