[英]How to return parameters from different table rows
我有三張桌子(桌子、桌子 2、桌子 3)。 從 table1 我想返回三行:Title、Desc、time、'products' AS type
從 table2 我想返回三行:group_title、group_desc、created、'groups' AS 類型
從 table3 我想返回三行:姓名、職業、出生、'用戶'AS 類型
$sql = "SELECT DISTINCT *
FROM
(SELECT table1.title, table1.desc, table1.time,'products' AS type
FROM table1 ORDER BY rand() LIMIT 5) AS T
UNION ALL
(SELECT table2.group_name,table2.group_desc,table2.created,'groups' AS type
FROM tabl2
JOIN table1 ON table2.id = table1.id
ORDER BY rand() LIMIT 5)
UNION ALL
(SELECT table3.name,table3.occupation,table3.birth,'users' AS type
FROM table3
JOIN table1 ON table3.id = table1.id
ORDER BY rand() LIMIT 5)";
我正在使用 Ionic 將數組的這三個參數推送到下一頁,因此如果我使用 AS 並放置相同的名稱。 我設置了一個我需要的參數 AS 類型,以便我可以使用不同的類型。
例如
(SELECT table1.title AS title, table1.desc AS desc, table1.time AS time
FROM table1 ORDER BY rand() LIMIT 5) AS T
UNION ALL
(SELECT table2.group_name AS title,table2.group_desc AS desc,table2.created AS time.....
如果我這樣做,它會顯示對象參數,例如 {{item.title}}
但我不能將參數作為標題推送或導航,它必須是 group_name。 我希望我說清楚了。 它必須是 table1 的“產品”AS 類型,table2 的“組”AS 類型,table3 的用戶類型我想將對象與每個表分開
請試試這個查詢。
$sql = "SELECT table1.title, table1.desc, table1.time , table2.group_name,table2.group_desc,table2.created table3.name,table3.occupation,table3.birth FROM table1 LEFT JOIN table2 ON table2.id = table1 .id LEFT JOIN table3 ON table3.id = table1.id"
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