SELECT c.*
,cc.*
,uc.*
FROM customers_companies cc
JOIN customers c
ON cc.customer_id = c.id
JOIN users_companies uc
ON cc.company_id = uc.id
WHERE uc.id = cc.company_id
这样的东西?
题
我尝试了一些东西,但似乎没有任何作用,我会解释它必须做什么,
首先,我有3个MySQL表,
TableName:客户
客户的所有信息(名称,zipcode等)
tableName:customers_companies
仅来自餐桌客户的客户_id和餐桌公司的company_id
tableName:users_companies
仅来自表用户的user_id和company_id来自表公司
首先,用户将公司添加到数据库中,它将user_id和company_id添加到users_companies
当您添加客户时,它将customer_id and company_id添加到cunituder_companies
它必须与PHP中的客户_companies的所有客户相呼应,但只有当您参与公司的一部分时,首先是MySQL检查customer_id并从cunituter_companies获取company_id
使用该Company_ID搜索所有允许在users_companies中查看客户的用户
当找到所有允许的用户时,必须从表客户中获取有关客户的所有信息,并在HTML/PHP的表中回声
有人能帮我吗?如果您需要更多信息,请询问,我会回答
谢谢
编辑:
<tbody>
<? $result = mysql_query("SELECT customers.* FROM `customers` LEFT JOIN `customers_companies` ON (`customer`.`id` = `customers_companies`.`company_id`) INNER JOIN `user_companies` (`customers_companies`.`company_id` = `customers_companies`.`company_id`) WHERE `customers_companies`.`bedrijf_id`");?>
<? $i = 1;?>
<? while($row = mysql_fetch_array($result)): ?>
<tr>
<td class="align-center"><? echo $i; ?></td>
<td> <?= $row['naam']; ?></td>
<td> <?= $row['adres']; ?></td>
<td> <?= $row['postcode']; ?></td>
<td> <?= $row['plaats']; ?></td>
<td> <?= $row['land']; ?></td>
<td> <?= 0,$row['telefoon']; ?></td>
<td>
<a href="#" class="table-icon edit" title="Edit"></a>
<a href="#" class="table-icon archive" title="Archive"></a>
<a onclick="return confirm('Weet je zeker dat je dit bedrijf wilt verwijderen? alle gekoppelde items worden ook verwijdert zoals facturen, klanten, enz')" href="user.php?p=bedrijven&del=<?= $row['id'];?>" class="table-icon delete" title="Delete"></a>
</td>
</tr>
<? $i += 1; ?>
<? endwhile; ?>
</tbody>
解决方案
SELECT c.*
,cc.*
,uc.*
FROM customers_companies cc
JOIN customers c
ON cc.customer_id = c.id
JOIN users_companies uc
ON cc.company_id = uc.id
WHERE uc.id = cc.company_id
这样的东西?