[英]PHP-MySQL Tagging
I have a comics website which currently allows users to choose which comics they view by category: Charts, Office, Life, Misc, etc. 我有一个漫画网站,目前允许用户选择按类别查看的漫画:图表,办公室,生活,杂项等。
I'd like to implement a tagging system, similar to what we have here on StackOverflow, which will describe more of the content of each comic rather than its category. 我想实现一个标记系统,类似于我们在StackOverflow上的内容,它将描述每个漫画的更多内容而不是其类别。 Ex: In Charts category, I have several business related...
例如:在图表类别中,我有几个与业务相关的...
My simple solution would be to handle it just how I've handled my categorization- 我的简单解决方案就是如何处理我的分类 -
This approach, however, seems to limit me to one tag per category. 然而,这种方法似乎限制我每个类别一个标签。 What if I have a comic that is business and relationships related... so It'd need both of those tags.
如果我有与业务和关系有关的漫画该怎么办...所以它需要两个标签。
How would I be able to attach multiple tags per comic? 如何为每个漫画附加多个标签?
EDIT: 编辑:
A few more questions: 还有一些问题:
1) What do I insert into my new relational table... anything? 1)我在新的关系表中插入什么内容......什么?
2) Also, for while ($row = $tag->fetch_assoc()) {
, how can I loop through a table if there is a join? 2)另外,对于
while ($row = $tag->fetch_assoc()) {
,如果有连接,我如何在表中循环? Isn't that an associative array? 这不是一个关联数组吗?
3) The issue is that I am echoing out the tag choices as such, so once a user clicks on a link, how would you be able to allow them to then click on another link to assign a 2nd tag? 3)问题是我正在回应标签选择,所以一旦用户点击链接,你怎么能让他们再点击另一个链接来分配第二个标签?
function getTags() {
include 'dbconnect.php';
global $site;
$tag = $mysqli->query("SELECT tagid, tagname FROM tags");
//$tag = $mysqli->query("SELECT * FROM comics c INNER JOIN comictags ct ON (c.comicID = ct.comicID) WHERE ct.tag_id IN (1, 2, 3) GROUP BY c.comic_id");
mysqli_close($mysqli);
while ($row = $tag->fetch_assoc()) {
echo "<a href='?action=homepage&site=" . $site . "&tag=" . $row['tagid'] . "&tagname=" . $row['tagname'] . "'/>" . $row['tagname'] . "</a><br />";
}
}
Just add another table. 只需添加另一个表。 Then you have three: One for Tags, one for Comics, and one for the relationship.
然后有三个:一个用于标签,一个用于漫画,以及一个用于关系。 You have to have this indirection table to properly store a many-to-many relationship.
您必须具有此间接表才能正确存储多对多关系。 This allows each comic to have zero or more tags (and vice versa).
这允许每个漫画具有零个或多个标签(反之亦然)。
You can accomplish this with a many-to-many relationship. 您可以通过多对多关系来完成此任务。 A many-to-many relationship uses a relational join table that would look like this:
多对多关系使用的关系联接表如下所示:
+---------------+---------------+
| comic_id | tag_id |
+---------------+---------------+
| 1 | 2 |
+---------------+---------------+
| 1 | 3 |
+---------------+---------------+
| 1 | 4 |
+---------------+---------------+
Now, in your query: 现在,在您的查询中:
SELECT * FROM comics c INNER JOIN comic_tags ct ON (c.comic_id = ct.comic_id) WHERE ct.tag_id IN (1, 2, 3) GROUP BY c.comic_id
Where 1, 2, 3 are the tags the user selected that they would like to see. 其中1、2、3是用户希望查看的标签。
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