我正在整理一份服务提供商列表。我希望将它们列出如下:
水管工: 乔的水管 迈克的管道
电工: 斯科特电气 约翰电气
等等...
如何从我的查询中提取并分组这些信息。另外,当我尝试调用 pt.name 时,它说它是 NULL,有什么想法吗?
谢谢
$qry = "SELECT pt.name, p.name, p.phone
FROM provider_type AS pt
LEFT JOIN providers_provider_type AS ppt ON pt.id = ppt.provider_type_id
LEFT JOIN providers AS p ON ppt.providers_id = p.id
WHERE installation_id = $installation_id
GROUP BY ppt.provider_type_id";
//$qry = "SELECT * FROM providers WHERE installation_id = $installation_id";
$res = mysqli_query($mysqli, $qry) or die('-1'.mysqli_error($mysqli));
$categories = array();
while ($row_rsCategories = mysqli_fetch_assoc($res)) {
///////////////////////////////////////////////
$categories[] = array(
'id' => $row_rsCategories['id'],
'provider_type' => $row_rsCategories['pt.name'],
'provider_name' => $row_rsCategories['name']
);
}
?>
<div class="container">
<div class="content">
<?php if ($msgBox) { echo $msgBox; } ?>
<div class="row">
<?php var_dump($categories); ?>
<?php if (empty($categories)) { ?>
<div class = "big-icon">
<i class="fa fa-meh-o fa-5x"></i>
</div>
<p><center>Sorry! It looks like we don't have a Scout in this area. <br />
Please check back in as we continue to add new Scouts to the network all the time.</center></p>
<?php } else {
?>
<table class="table table-striped">
<tr>
<th>Type</th>
<th>Name</th>
<th>Phone</th>
<th>More Info</th>
</tr>
<?php foreach ($categories as $category) { ?>
<tr>
<td><?php echo $category['provider_type']; ?> </td>
<td><?php echo $category['provider_name']; ?> </td>
<td> </td>
<td> </td>
</tr>
<?php }
}
?>
</table>
最佳答案
您的查询对于每种提供程序类型仅返回一行,而您想要列出所有提供程序(每种类型多个提供程序)。
一种选择是使用 GROUP_CONCAT,它将每种类型的所有提供程序连接成一个逗号分隔的字符串:
SELECT pt.name AS type, GROUP_CONCAT(p.name) AS providers
FROM provider_type AS pt
LEFT JOIN providers_provider_type AS ppt ON pt.id = ppt.provider_type_id
LEFT JOIN providers AS p ON ppt.providers_id = p.id
WHERE installation_id = $installation_id
GROUP BY ppt.provider_type_id;
这将为您提供一个如下所示的结果集:
type | providers |
------------+---------------------------------+
plumber | Joe's Plumbing,Mike's Plumbing |
electrician | Scott electric,John electric |
就你的情况而言,我认为有一个更好的选择 - 因为无论如何你都将所有结果检索到 php 中,所以你可能想要一一获取所有提供者,然后在代码中进行分组,并显示它任何你喜欢的方式。
查询将与您的查询相同,但没有分组:
SELECT pt.name AS type, p.name, p.phone
FROM provider_type AS pt
LEFT JOIN providers_provider_type AS ppt ON pt.id = ppt.provider_type_id
LEFT JOIN providers AS p ON ppt.providers_id = p.id
WHERE installation_id = $installation_id;
结果集只是一个提供商列表,每个提供商都有他的姓名和电话号码。
type | name | phone |
------------+-----------------+---------------+
plumber | Joe's Plumbing | 123... |
plumber | Mike's Plumbing | 123... |
electrician | Scott electric | 123... |
electrician | John electric | 123... |
至于pt.name中的NULL -
在 MySQL 的结果集中,您不会得到 pt.name
和 p.name
作为列名称。您将得到 name
和 name1
(两列具有相同的名称,无论它们最初来自哪个表,因此数据库只需添加 1 即可获得唯一的名称)。
您应该为查询中的列指定唯一的名称,例如pt.name AS type
然后使用这些名称来读取数据,例如'provider_type' => $row_rsCategories['type']
关于PHP 结果与 MYSQL 多对多表和 GROUP By,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/28506790/