下面是我的表格结构:
菜单表
id title position
--------------------
1 Test home
2 Test2 home
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类别
cid name parent parent_menu
--------------------------------
1 ABC 0 1
2 DEF 0 2
3 GHI 1 0
4 JKL 2 0
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类别说明
id cat_id catdesc slug
-------------------------------
1 1 ABC_DESC abc
2 2 DEF_DESC def
3 3 GHI_DESC ghi
4 4 JKL_DESC jkl
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Menu table 处理菜单标题的位置.Category table处理类别名称和其他参数.(如果parent = 0则表示这是主要类别,依此类推......)Category Description table 处理描述,slug和其他参数.现在我想显示如下数据
Name Description Edit Delete /*table headings*/
-------------------------------------------------------
Menu Title: (Test) Main Category Name: (ABC)
------------------------------------------------------
GHI GHI_DESC edit_icon delete_icon
______________________________________________________
Menu Title: (Test2) Main Category Name: (DEF)
------------------------------------------------------
JKL JKL_DESC edit_icon delete_icon
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我尝试使用JOINS并在PHP中操作数据,但没有运气.
SELECT * FROM `category` t1 LEFT JOIN `category_description` t2 ON t1.cid = t2.cat_id WHERE 1
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然后在PHP中我尝试如下
<?php $i = 1; foreach($subcat as $sub) { ?>
<?php if($sub->parent == 0) { ?>
<tr><td><?php echo $sub->name ?></td></tr>
<?php } ?>
<?php if($sub->parent != 0) { ?>
<tr><td><?php echo $sub->name ?></td><td><?php echo $sub->catdesc ?></td>
<td>Edit</td><td>Delete</td></tr>
<?php } ?>
<?php } ?>
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以上打印表格如下:
Main Category Name: ABC
Main Category Name: DEF
------
GHI GHI_DESC
JKl JKL_DESC
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请根据需要建议如何打印.
假设parent_menu来自菜单表中的id,请尝试以下操作:
SELECT t1.title, t2.name, t2.parent, t2.parent_menu, t3.catdesc
FROM menu t1
LEFT JOIN category t2 ON t1.id=t2.parent_menu
LEFT JOIN description t3 ON t2.cid=t3.cat_id
GROUP BY t2.name
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