[英]MySQL query which has multiple subqueries, each with different join types
首先,让我给出一个我的mysql查询中使用的表之间的关系图:
(来源: r717.net )
我有一个看起来像这样的查询:
SELECT *
FROM `permissions`
WHERE `id` IN (
SELECT pr.perm_id
FROM `user_roles` as ur
LEFT JOIN `permissions_role` as pr
ON ur.role_id = pr.role_id
WHERE ur.user_id = '$userid'
)
OR `id` IN (
SELECT `perm_id`
FROM `permissions_user`
WHERE `user_id` = '$userid'
)
$userid
是当前用户的用户表中的id。 我将结果中的权限名称存储到一个数组中,该数组表示根据用户的角色和他/她的ID分配给用户的所有权限:
<?php
$user_perms = array();
if(mysql_num_rows($query) > 0):
while($result = mysql_fetch_array($query):
$user_perms[] = $result('name');
endwhile;
endif;
?>
print_r($user_perms);
生成一个看起来像这样的输出:
Array (
[0] => ACCESS_TELEPHONELIST_PAGE
[1] => ACCESS_VACATIONSCHED_PAGE
[2] => ACCESS_TOURSCHED_PAGE
[3] => ACCESS_WORKSCHED_PAGE
[4] => ACCESS_RESOURCES_LINKS
[5] => ACCESS_PMTOOL_PAGE
[6] => ACCESS_TOOLSTOOL_PAGE
[7] => ACCESS_SHOPTOOLLIST_PAGE
[8] => ACCESS_TOOLINVENTORY_PAGE
[9] => ACCESS_MANAGETOOLLIST_PAGE
[10] => ACCESS_TOOLREPORTS_PAGE
[11] => ACCESS_JOBSLIST_LINKS
[12] => MAIN_TAB_TOOLSTOOL
[13] => ADMIN_TAB_PODMANAGEMENT
[14] => TOOL_TAB_SHOPTOOLLIST
)
我想要做的是也将所有用户的角色存储到另一个数组中而不进行第二次查询。 我以为我需要为子查询使用别名,所以我尝试了这个查询:
SELECT permissions.*, usersroles.role_id
FROM `permissions`
INNER JOIN (
SELECT ur.user_id, pr.perm_id, ur.role_id
FROM `user_roles` as ur
LEFT JOIN `permissions_role` as pr
ON ur.role_id = pr.role_id
WHERE ur.user_id = '$userid'
) AS usersroles ON usersroles.perm_id = permissions.id
INNER JOIN (
SELECT `perm_id`, `user_id`
FROM `permissions_user`
WHERE `user_id` = '$userid'
) AS userperms ON userperms.user_id = usersroles.user_id
AND userperms.perm_id = permissions.id
并使用类似于上面代码的代码...
<?php
$user_perms = array();
$user_roles = array();
if(mysql_num_rows($query) > 0):
while($result = mysql_fetch_array($query):
$user_perms = $result('name');
$user_roles = $result('role_id');
endwhile;
endif;
?>
......我收到这个警告:
Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given
但是,我想print_r($user_roles);
并生成如下所示的输出:
Array (
[0] => administrator
[1] => humanresources
[2] => podmanager
)
任何人都可以告诉我我做错了什么,或者建议一种更好的方法将我需要的数据从一个查询转换为2个数组?
编辑:仔细考虑后,我改变了我的代码,使用ImreL建议的2个查询。 生成的代码可以很好地运行并快速执行。 我编辑了我的答案,以显示我使用的最终代码,并添加了支持代码来演示我如何使用2个查询。 非常感谢ImreL!
该查询将需要在用户加载的每个页面上运行,并且我们拥有超过30,000个权限和3,000个角色。 我只是非常努力地将我的查询数量保持在最低限度。 我们还在办公室的服务器上托管我们的7个站点,服务器似乎无法处理我们生成的流量(遗憾的是我无法控制)
我看到你的意图是好的,但让我告诉你:
“查询数量”不是衡量网站效果的正确方法。
很多时候,2个简单查询比1个复杂使用更少的资源。
还有其他方法可以加快您的网站:
最后,我试图提出查询以满足所要求的内容:
select * from (
select ur.role_id, p.*
from user_roles ur
left join permissions_role pr on ur.role_id = pr.role_id
left join permissions p on p.id = pr.perm_id
where ur.user_id = '$userid'
union all
select null as role_id, p.*
from permissions_user pu
join permissions p on p.id = pu.perm_id
where pu.user_id = '$userid'
) sub
group by ifnull(name,role_id) -- group by to eliminate duplicates
但这对性能不利。 使用2个查询会更好:第1个获取用户的所有权限
select p.* from permissions p
join permissions_role pr on pr.perm_id = p.id
join user_roles ur on ur.role_id = pr.role_id and ur.user_id = '$userid'
union
select p.* from permissions p
join permissions_user pu on pu.perm_id = p.id and pu.user_id = '$userid';
和第二个获得所有角色。
编辑:这是最后的工作查询。 此外,我认为查看查询的使用方式可能对其他人有用,因此包含了完整的解释。
//build the array(s)
$userperms = array();
$userroles = array();
if(isset($_SESSION['userid'])):
//get the userid
$thisuserid = $_SESSION['userid'];
//get the permissions for each of the user's roles
$Cpermissions_role = mysql_query("
SELECT r.type, ur.user_id, pr.perm_id, ur.role_id, p.name
FROM user_roles ur
LEFT JOIN permissions_role pr ON ur.role_id = pr.role_id
LEFT JOIN roles r ON ur.role_id = r.id
LEFT JOIN permissions p ON p.id = pr.perm_id
WHERE ur.user_id = '$userid'") or die(mysql_error());
//get the extra permissions for the user
$Cpermissions_user = mysql_query("
SELECT pu.user_id, pu.perm_id, p.name
FROM permissions_user pu
LEFT JOIN permissions p ON p.id = pu.perm_id
WHERE pu.user_id = '$userid'") or die(mysql_error());
//build an array of the user's roles & an array of the user's permissions
if(mysql_num_rows($Cpermissions_role) > 0):
while($Rpermissions_role = mysql_fetch_array($Cpermissions_role)):
if(empty($userperms)):
$userperms[] = $Rpermissions_role['name'];
elseif(!in_array($Rpermissions_role['name'],$userperms)):
$userperms[] = $Rpermissions_role['name'];
endif;
if(empty($userroles)):
$userroles[] = $Rpermissions_role['type'];
elseif(!in_array($Rpermissions_role['type'],$userroles)):
$userroles[] = $Rpermissions_role['type'];
endif;
endwhile;
endif;
if(mysql_num_rows($Cpermissions_user) > 0):
while($Rpermissions_user = mysql_fetch_array($Cpermissions_user)):
if(empty($userperms)):
$userperms[] = $Rpermissions_user['name'];
elseif(!in_array($Rpermissions_user['name'],$userperms)):
$userperms[] = $Rpermissions_user['name'];
endif;
endwhile;
endif;
endif;
/**
* Determines if the user has permission for the page or parts of page
* @param string $perm the permission constant
* @return boolean true if user has access, false if not
*/
function hasPermission($perm){
global $userperms;
if(empty($userperms)):
return false;
else:
if(is_array($userperms)):
if(in_array($perm,$userperms)):
return true;
else:
return false;
endif;
else:
if($perm == $userperms):
return true;
else:
return false;
endif;
endif;
endif;
}
然后,我使用以下决策声明:
if(hasPermission("ACCESS_HOME_PAGE")):
//perform circus tricks here
endif;
那么,现在print_r($userperms);
输出:
Array (
[0] => ACCESS_TELEPHONELIST_PAGE
[1] => ACCESS_VACATIONSCHED_PAGE
[2] => ACCESS_TOURSCHED_PAGE
[3] => ACCESS_WORKSCHED_PAGE
[4] => ACCESS_RESOURCES_LINKS
[5] => ACCESS_PMTOOL_PAGE
[6] => ACCESS_TOOLSTOOL_PAGE
[7] => ACCESS_SHOPTOOLLIST_PAGE
[8] => ACCESS_TOOLINVENTORY_PAGE
[9] => ACCESS_MANAGETOOLLIST_PAGE
[10] => ACCESS_TOOLREPORTS_PAGE
[11] => ACCESS_JOBSLIST_LINKS
[12] => MAIN_TAB_TOOLSTOOL
[13] => ADMIN_TAB_PODMANAGEMENT
[14] => TOOL_TAB_SHOPTOOLLIST
[15] => ACCESS_HOME_PAGE
)
并且, print_r($userroles);
输出:
Array (
[0] => administrator
[1] => humanresourcees
[2] => podmanager
)
而且,为了彻底,这是我用来访问php数组的JavaScript:
var js_userperms = new Array();
js_userperms = ["<?php echo join("\", \"", $userperms); ?>"];
并且,jQuery决策声明:
if(jQuery.inArray("ACCESS_HOME_PAGE", js_userperms) != -1){
//perform circus tricks here
}
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