掙扎寫作我的第一個連接查詢

Question

我有3個表，我試圖在下面佈局。我無法堅持。我沒有做過很多的查詢，之前從來沒有加入過，我真的很困惑這是怎麼回事。示例查詢我本來會用簡單的英語，

得到所有portfolio_item_ids從skill_area其中web_design =「1」， 然後讓所有從與portfolio_item_ids 則每行的ID值portfolio_items行，得到標籤的行，其中portfolio_item_id等於portfolio_item_id和插畫家之一的值，Photoshop或CSS是1

QUERY我設法拿出

---

SELECT portfolio_item_id

FROM skill_area小號

WHERE web_design = 1

---

SELECT *

FROM portfo lio_items p

WHERE p.id = s.portfolio_item_id

---

SELECT *

FROM標籤噸

WHERE s.portfolio_item_id =噸。portfolio_item_id

---

所以我認爲這是正確的，我只需要加入它現在一起

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投資組合項目

ID

ITEM_NAME

描述

技能領域

portfolio_item_id

網頁設計

品牌

打印

標籤

portfolio_item_id

插畫

的Photoshop

CSS

Answer 1

打破它的步驟，就像你在你的 '純英文' 語句有：

1. 得到web_design skill_area的

   select * from skill_area where web_design = '1' 
   

2. 與portfolio_items加入

   select * 
   from skill_area s, portfolio_items p 
   where web_design = '1' 
       and p.id = s.portfolio_item_id 
   

3. 加入與標籤，選擇行其中value = 1，並返回剛剛從標籤表中的列）

   select t.* 
   from skill_area s, portfolio_items p, tags t 
   where web_design = '1' 
       and p.id = s.portfolio_item_id 
       and p.id = t.portfolio_item_id 
       and t.value = '1' 
   

Answer 2

試試這個：

SELECT sa.portfolio_item_id, pi.*, subtags.* 
FROM skill_area sa 
INNER JOIN portfolio_items pi ON sa.portfolio_item_id = pi.id 
INNER JOIN (SELECT CASE WHEN illustrator = '1' THEN 'illustrator' 
         WHEN photoshop = '1' THEN 'photoshop' 
         ELSE 'css' END as tag, portfolio_item_id 
      FROM tags) subtags ON sa.portfolio_item_id = subtags.portfolio_item_id 
WHERE sa.web_design = '1' 

你的評論，你想有一個替代的，你只得到標籤的所有列。

SELECT sa.portfolio_item_id, pi.*, tags.* 
FROM skill_area sa 
INNER JOIN portfolio_items pi ON sa.portfolio_item_id = pi.id 
INNER JOIN tags ON sa.portfolio_item_id = tags.portfolio_item_id 
WHERE sa.web_design = '1' 

Answer 3

SELECT  P.* 
FROM  SkillArea S 
INNER JOIN Portfolio P ON S.portfolio_item_id = P.id 
INNER JOIN Tags T ON T.portfolio_item_id = P.id 
WHERE  S.web_design = 1 
AND   P.id = 1 

Answer 4

這裏是這樣的查詢可能是什麼樣子。我試圖逐字逐句翻譯您的英語變體，以幫助您理解，並將註釋用於描述表情。

select s.portfolio_item_id, p.*, t.* 
from Skill_Area s         -- get all portfolio_item_ids from skill_area 
inner join Portfolio_Items p on     -- then get all the rows from portfolio_items 
    p.portfolio_item_ids = s.portfolio_item_ids -- with the id values of portfolio_item_ids 
inner join Tags t on        -- then for each row, get the row from tags 
    t.portfolio_item_id = p.portfolio_item_id  -- where portfolio_item_id is equal to portfolio_item_id 
    and t.portfolio_item_id = 1     -- and the value of it is 1 
where s.web_design = "1"       -- where web_design = "1"