在MySQL數據庫從CSV場
CHILDREN
|NAME |LIKES (LONGTEXT)
|Sam |75
|John |58,64,75
FRUITS
|ID |LABEL
|58 |Apple
|64 |Banana
|75 |Cherry
爲了獲得水果名兒童,我有以下IN選擇我做的:
> select * from fruits where id in
(select LIKES from CHILDREN where name="Sam");
|ID |LABEL
|75 |Cherry
OK,但
> select * from fruits where id in
(select LIKES from CHILDREN where name="John");
|ID |LABEL
|58 |Apple
我也試過以下結果相同:
> select * from fruits where id in ("58,64,75")
|ID |LABEL
|58 |Apple
[Main Instruction]
Your query produced 1 warnings.
Warning: Truncated incorrect DOUBLE value: '58,64,75'
[OK]
>select * from fruits where id in (58,64,75)
|ID |LABEL
|58 |Apple
|64 |Banana
|75 |Cherry
怎麼可能來解決這個問題不改變表定義,事情是這樣的...
現在找什麼哇你喜歡混搭你不。 – Drew
你需要什麼輸出結構? –
如果你被困在桌子上,那是一回事。如果你設計了它們,你應該學會如何正確設計SQL表格。 –