我怎样才能构建一个“has-many-through”?链接两个以上模型的关系?
我有 3 个型号,例如;
TABLE `users`
`id` INT
`username` VARCHAR(32)
...
TABLE `books`
`id` INT
`title` VARCHAR(100)
`author` INT (foreign ket constraint)
TABLE `rights`
`id` INT
`name` VARCHAR(32)
现在我希望用户拥有特定的权利,例如。阅读或编辑一本书。所以权限表应该看起来像(很像 ORM 角色表):
|----|------|
| id | name |
|----|------|
| 1 | view |
| 2 | edit |
| .. | ... |
我会有第四个表链接所有三个表;
TABLE user_book_rights
|---------|---------|----------|
| user_id | book_id | right_id |
|---------|---------|----------|
| 1 | 1 | 2 |
| 1 | 2 | 1 |
| 2 | 1 | 1 |
| ... | ... | ... |
因此,如果用户想要阅读一本书,我想检查 id 为 1 的登录用户是否具有 id 1 阅读 id 2 书籍的权限。
但是我到底如何使用 ORM 来实现这一点呢?当然,我可以编写自己的查询;
SELECT COUNT(*) as `has_right` FROM `user_book_rights` WHERE user_id=1 AND book_id=2 AND right_id=1
if($result['has_right']) {
echo 'Yeah, read the book!';
} else {
echo 'Sorry mate, this book is not for dummies...';
}
但我宁愿做这样的事情:
$has_right = $user->has('book_rights', ORM::factory('user_book_right', array('book_id' => '2', 'right_id' => 1));
或者甚至更好:
$book = ORM::factory('book', 1);
$right = ORM::factory('right', array('name' => 'view'));
$has_right = $user->has('book_rights', ORM::factory('user_book_right', array($book, $right)));
我找不到我的问题的答案。想要将三个模型链接为 Many_through 关系是不是很奇怪?或者 ORM 没有能力,我应该编写自己的查询吗?
谢谢。为了您的见解!
I have 3 models eg;
TABLE `users`
`id` INT
`username` VARCHAR(32)
...
TABLE `books`
`id` INT
`title` VARCHAR(100)
`author` INT (foreign ket constraint)
TABLE `rights`
`id` INT
`name` VARCHAR(32)
Now I want a user to have particular rights to eg. read or edit a book. So the rights table should look like (much like ORM roles table):
|----|------|
| id | name |
|----|------|
| 1 | view |
| 2 | edit |
| .. | ... |
And I would have a fourth table linking all three;
TABLE user_book_rights
|---------|---------|----------|
| user_id | book_id | right_id |
|---------|---------|----------|
| 1 | 1 | 2 |
| 1 | 2 | 1 |
| 2 | 1 | 1 |
| ... | ... | ... |
So if a user wants to, say, read a book, I want to check if the logged in user with id 1, has the right with id 1 for book with id 2.
But how the heck can I achieve this with ORM? Of course I can just write my own query;
SELECT COUNT(*) as `has_right` FROM `user_book_rights` WHERE user_id=1 AND book_id=2 AND right_id=1
if($result['has_right']) {
echo 'Yeah, read the book!';
} else {
echo 'Sorry mate, this book is not for dummies...';
}
But I'd rather do something like:
$has_right = $user->has('book_rights', ORM::factory('user_book_right', array('book_id' => '2', 'right_id' => 1));
Or even better:
$book = ORM::factory('book', 1);
$right = ORM::factory('right', array('name' => 'view'));
$has_right = $user->has('book_rights', ORM::factory('user_book_right', array($book, $right)));
I could not find an answer to my question. Is it weird to want to link three models as a many_through realtionship? Or is ORM just not capable and should I write my own query?
Thanks ia. for your insights!
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也许我的回答对你帮助不大。但我建议您以位表示形式对右侧进行编码。
如果用户具有读取、编辑和删除的写入权限
操作
如果您想测试用户是否拥有特定权利,您只需执行以下 :
if ($user_write & READ) { // he can read}
也许如果您使用这种设计,并消除带有这些常量的权限表,它可能会对您有所帮助。
Maybe my answer does not help you a lot. But i suggest that you encode the right in bit representation.
than, if a user has the write to read, edit and delete
you just do
If you want to test if a user has a particular right you just do:
if ($user_write & READ) { // he can read}
Maybe if you use this design, and eliminate the rights table with those constants, it may help you.