5

这是较早问题的更高级的续集。

这是一个 SQLFiddle 的链接，使事情更清楚

我需要返回类似的东西（对于每个位置）：

一楼
---------------
车库
-
暖气厨房
- 炊具
- 冰箱

会有比这更多的关卡，但如果我能开始，我希望能够根据需要走得更远。

谢谢

4

3 回答 3

2

由于 SQL 无法返回您在上面格式化的嵌套记录，因此您只需要返回一个查询，该查询按楼层然后按 sub_location 对您的行进行排序。你会得到每一行的location和sub_location重复的，但是在你的应用程序代码循环中打印结果时，你可以像上面那样格式化它们。

SELECT
  location.location_name,
  sublocation.sub_location_name,
  asset.asset_name
FROM
  location
  LEFT JOIN sub_location ON location.location_key = sublocation.location_key
  LEFT JOIN assets ON sub_location.sub_location_key = assets.sub_location_key
ORDER BY
  location.location_name,
  sub_location.sub_location_name

在应用程序中循环遍历行集时，您只需在新位置或 sub_location 更改时打印它。格式化是在代码中完成的。

假设您的行都在数组中$rowset：

// Store location, sub_location for each loop
$current_loc = "";
$current_subloca = "";
foreach ($rowset as $row) {
  // If the location changed, print it
  if ($row['location'] != $current_loc) {
    echo $row['location_name'] . "\n";
    // Store the new one
    $current_loc = $row['location_name'];
  }
  // If the sub_location changed, print it
  if ($row['sub_location'] != $current_subloc) {
    echo $row['sub_location_name'] . "\n";
    $current_subloc = $row['sub_location_name'];
  }
  echo $row['asset_name'] . "\n";      
}
于 2012-07-23T11:16:40.810 回答
2

您可以使用此解决方案显示已知深度的层次结构：

SELECT a.name
FROM
(
    SELECT 
        CONCAT('- - - - - - - - - -> ', a.asset_name) AS name,
        CONCAT(c.location_key, c.location_name, b.sub_location_name, a.asset_name) AS orderfactor
    FROM asset a
    INNER JOIN sub_location b ON a.sub_location_key = b.sub_location_key
    INNER JOIN location c ON b.location_key = c.location_key

    UNION ALL

    SELECT 
        CONCAT('- - - - -> ', a.sub_location_name),
        CONCAT(b.location_key, b.location_name, a.sub_location_name)
    FROM sub_location a
    INNER JOIN location b ON a.location_key = b.location_key

    UNION ALL

    SELECT
        location_name,
        CONCAT(location_key, location_name)
    FROM location
) a
ORDER BY a.orderfactor

对于更多级别，您可以在子选择中添加更多联合。

SQLFiddle 演示

示例结果集：

Ground Floor
-----> Garage
----------> Radiator
-----> Kitchen
----------> Cooker
----------> Fridge
First Floor
-----> Bedroom
----------> Radiator
----------> Taps
Second Floor
-----> Bathroom
----------> Shower
----------> Taps
----------> Toilet
Third Floor
于 2012-07-23T11:33:50.467 回答
1

如果是在 MySQL5

SELECT L.location_name, GROUP_CONCAT(DISTINCT S.sub_location_name), GROUP_CONCAT(A.asset_name) 
FROM location L
INNER JOIN sub_location S ON L.location_key = S.location_key
INNER JOIN asset A ON A.sub_location_key = S.sub_location_key
GROUP BY S.sub_location_name ORDER BY L.location_key

http://sqlfiddle.com/#!2/db932/9

于 2012-07-23T11:32:32.827 回答