回答編集履歴
2
回答修正
answer
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@@ -1,23 +1,34 @@
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自分のサーバー(MySQL)でテスト環境を作り検証しましたところ、
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いくつか間違いがあったので修正しました。
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実機動作の確認が出来たので、下記SQLを修正しました。
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（MySQLでは[]が使用できないので``に置換が必要です。）
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---
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1
7
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こんな感じでしょうか？
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未検証ですがたぶん動くと思います。
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```SQL
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SELECT
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-
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[parent].[col_group]
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-
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, [parent].[path] AS root_path
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-
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, [parent].[root_size]
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-
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, [children].[size_sum]
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FROM
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(SELECT group
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(SELECT [T_table].[group] AS col_group
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, [path]
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, [size] AS root_size
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FROM
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FROM [t_table]
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WHERE path_hierarchy = 0) parent
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WHERE [path_hierarchy] = 0) parent
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LEFT JOIN
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(SELECT [t_table].[group] AS col_group
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-
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, SUM([size]) AS size_sum
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FROM
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FROM [t_table]
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WHERE path_hierarchy <> 0
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WHERE [path_hierarchy] <> 0
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GROUP BY group) children
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GROUP BY [t_table].[group]) children
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ON [parent].[
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ON [parent].[col_group] = [children].[col_group]
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;
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```
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34
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以上、
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誤記訂正
answer
CHANGED
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@@ -12,7 +12,7 @@
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FROM T_table
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WHERE path_hierarchy = 0) parent
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LEFT JOIN
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(SELECT group, SUM(size) AS size_sum
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(SELECT group, SUM(size) AS size_sum
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FROM T_table
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WHERE path_hierarchy <> 0
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GROUP BY group) children
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