檢查是否在另一個表中存在一個id

Question

我現在所擁有的一個相當大的SQL查詢：

SELECT a . * , ap . * , i . * , appstatus . * ,(SELECT SUM(rating)/count(case when rating > 0 then 1 end) AS total_rating FROM review AS r WHERE r.applyid=a.applyid GROUP BY applyid) AS total_rating, (SELECT count(rating) FROM review AS r WHERE r.applyid=a.applyid GROUP BY applyid) as count 
       FROM apply AS a, applicant AS ap, interest AS i, application_status AS appstatus, apply_tags as at 
       WHERE a.applicantid = ap.applicantid 
       AND a.applicantid = i.applicantid 
       AND a.application_status_id = appstatus.application_status_id 
       AND a.archive = 'false' 
       AND a.jobid =27 

我需要修改這個讓另一臺也檢查，看看是否在它存在a.applyid或不。

根據我現有的查詢，我想對其進行修改，因此它只會輸出apply_tags表中沒有數據的行。

我不知道最好的方法將其整合到我的查詢中。我應該使用連接嗎？

我基本上希望增加一個與聲明是這樣的：

AND a.applyid NOT EXIST at.applyid（at是apply_tags表）

我知道我這樣做不對，但就是那種東西的例子我正在努力。

Answer 1

對不起，我看你已經選擇一個正確答案。但是你的查詢與完美無關。因此，這裏只爲你一個想法，您的查詢可能是怎麼樣的，如果你想優化它第一天：

SELECT a . * , 
ap . * , 
i . * , 
appstatus . * , 
r.total_rating, 
r.count 
FROM apply AS a 
LEFT JOIN (
    SELECT applyid, 
    SUM(rating)/SUM(IF(rating > 0,1,0)) AS total_rating, 
    SUM(IF(rating > 0,1,0)) as count 
    FROM review 
    GROUP BY applyid) AS r 
ON r.applyid=a.applyid 
LEFT JOIN applicant AS ap 
ON a.applicantid = ap.applicantid 
LEFT JOIN interest AS i 
ON a.applicantid = i.applicantid 
LEFT JOIN application_status AS appstatus 
ON a.application_status_id = appstatus.application_status_id 
LEFT JOIN apply_tags as at 
ON a.applyid = at.applyid 
WHERE a.archive = 'false' 
    AND a.jobid =27 
    AND at.applicantid IS NULL 

小說明在原來的查詢，你做這樣的事情：

SELECT a . * , 
..., 
(SELECT SUM(rating)/count(case when rating > 0 then 1 end) AS total_rating FROM review AS r WHERE r.applyid=a.applyid GROUP BY applyid) AS total_rating, 
(SELECT count(rating) FROM review AS r WHERE r.applyid=a.applyid GROUP BY applyid) as count 
.... 

你問這個技術有什麼問題？所以問題在於MySQL會嘗試從a.table接收到的每條記錄上執行這2 SELECT。與我的LEFT JOIN相反，這將只需要一次所有記錄。如果你在a有100條記錄，MySQL會調用你的子查詢100x2 = 200倍，並且只有1個我的子查詢。

與WHERE a.applicantid = i.applicantid和LEFT JOIN i ON a.applicantid = i.applicantid幾乎相同的原因。最後一個由MySQL絕對優化。第一個是非常緩慢和沉重。如果你忽略了MySQL的特性，那麼它就是爲它而發明的。你可以從所有5個表中將SELECT *轉換爲php數組，並執行php端的所有邏輯。

你可以只是谷歌一點。例如：https://stackoverflow.com/a/2242015/4421474

Answer 2

您的嘗試非常接近。只需添加：

and not exists (select 1 from apply_tags where applyid = a.applyid) 

這是在使用了文件EXISTS/NOT EXISTS：

https://dev.mysql.com/doc/refman/5.0/en/exists-and-not-exists-subqueries.html