0
我有1個表菜單標籤在它。我需要一個查詢來獲取父標籤和UnderMenu label.when我執行此:GROUP_CONCAT &&在一個表上留下外部連接?
SELECT es.Label MenuLabel, m.Label UnderLabel
FROM books_menu es
LEFT JOIN books_menu m ON es.ID = m.MenuParentID
WHERE es.ParentID =1
AND m.Type='under'
AND es.Type='main'
LIMIT 0 , 30
一切正常,但MenuLabel顯示在每record.Something像這樣:
MenuLabel UnderLabel
Home 1
Home 2
Home 3
Contacts 4
當我執行:
SELECT es.Label MenuLabel, GROUP_CONCAT(m.Label) AS UnderLabel
FROM books_menu es
LEFT JOIN books_menu m ON es.ID = m.MenuParentID
WHERE es.ParentID =1
AND m.Type='under'
AND es.Type='main'
LIMIT 0 , 30
我得到:
MenuLabel UnderLabel
Home 1,2,3,4
我怎樣才能得到這樣的:
MenuLabel UnderLabel
Home 1,2,3
Contacts 4
謝謝:)
感謝很多:) – 2012-07-31 07:14:26
歡迎,我希望它的工作原理爲你! – Omesh 2012-07-31 07:18:11