如何在MYSQL到select中使用CASE語句從基於條件的結果中獲取一行？

Question

我是 sql 的新手。我有一個 SELECT 查詢，它返回 2 行，其中有一個 rid（兩行均為 1）和一個路徑列，其二進制值為 0 或 1 我想使用 CASE 語句，這樣它將 SELECT 一行從基於條件s1.pid < s2.pid THEN (SELECT S2.rid WHERE r.path = 0) ELSE (SELECT S2.rid WHERE r.path=1 )的第一個查詢的結果中，下面的代碼返回兩行 rid 1 和路徑 0 和 1 我如何編寫案例陳述？ 我不知道正確的語法。

架構：數據庫架構

• stopno rid busno pid 總線鍵路徑
  4 1 1111 4 鍵 1 0
  4 1 2222 4 鍵 2 1

這是我從運行第一個查詢得到的結果，現在我想要 select 基於條件的行之一 where s1.pid< s2.pid 然后 select 行 where path = 0 否則我該怎么做？

$sql="SELECT s2.stopno, s2.rid, busno, s2.pid, buskey, `path`
FROM `stop` s2
INNER JOIN `route` r ON s2.rid = r.rid
INNER JOIN bus b ON r.bid = b.bid
INNER JOIN place p ON s2.pid = p.pid
WHERE p.place = 'place2'
AND s2.rid IN (SELECT  s1.rid
                 FROM `stop` s1
                 INNER JOIN `route` r ON s1.rid = r.rid
                 INNER JOIN bus b ON r.bid = b.bid
                 INNER JOIN place p ON s1.pid = p.pid
                 AND p.place = 'place1')
";

我已經試過了，但它給了我錯誤 Unknown column 's2.pid' in 'field list'。 似乎我無法在外部查詢中訪問內部子查詢別名，或者我做錯了嗎？ s2的scope是什么？ 有什么解決方法嗎？

$sql=" SELECT (CASE 
WHEN s2.pid > s1.pid
THEN (SELECT s3.rid 
     FROM `stop` s3
    INNER JOIN `route` r ON s3.rid = r.rid
    INNER JOIN bus b ON r.bid = b.bid
    INNER JOIN place p ON s3.pid = p.pid
    WHERE s3.rid IN 
          (SELECT s2.rid
          FROM `stop` s2
          INNER JOIN `route` r ON s2.rid = r.rid
          INNER JOIN bus b ON r.bid = b.bid
          INNER JOIN place p ON s2.pid = p.pid
          WHERE p.place = 'place2'
          AND s2.rid IN 
                (SELECT  s1.rid
                FROM `stop` s1
                INNER JOIN `route` r ON s1.rid = r.rid
                INNER JOIN bus b ON r.bid = b.bid
                INNER JOIN place p ON s1.pid = p.pid
                AND p.place = 'place1'))
                AND r.path = '0')
ELSE (SELECT s3.rid 
      FROM `stop` s3
      INNER JOIN `route` r ON s3.rid = r.rid
      INNER JOIN bus b ON r.bid = b.bid
      INNER JOIN place p ON s3.pid = p.pid
      WHERE  s3.rid IN
              (SELECT  s2.rid
              FROM `stop` s2
              INNER JOIN `route` r ON s2.rid = r.rid
              INNER JOIN bus b ON r.bid = b.bid
              INNER JOIN place p ON s2.pid = p.pid
              WHERE p.place = 'place2'
              AND s2.rid IN 
                    (SELECT  s1.rid
                    FROM `stop` s1
                    INNER JOIN `route` r ON s1.rid = r.rid
                    INNER JOIN bus b ON r.bid = b.bid
                    INNER JOIN place p ON s1.pid = p.pid
                    AND p.place = 'place1'))
                    AND r.path = '1')
                    END)AS pathfinder
FROM `stop` s
INNER JOIN `route` r ON s.rid = r.rid
INNER JOIN bus b ON r.bid = b.bid
INNER JOIN place p ON s.pid = p.pid
";

編輯 2：我試過下面的代碼，它返回 p1.pid 和 p2.pid 但在 IF 語句中給出語法錯誤，任何人都可以解決它嗎？

$sql="SELECT s2.stopno, s2.rid, b.busno, p1.pid as id1, p2.pid as id2, b.buskey, r.path, p1.place as place1, p2.place as place2,
   IF(p1.pid < p2.pid, IF(r.path=0, s2.*, null), IF(r.path=1, s2.*,null))
       
FROM `stop` s2

JOIN `route` r
ON s2.rid = r.rid
JOIN bus b 
ON r.bid = b.bid
JOIN place p1
ON s2.rid = p1.pid
JOIN place p2
ON s2.pid = p2.pid
WHERE 
p1.place = 'place1'
AND p2.place = 'place 2'

";

Answer 1

你的描述真的很混亂。 無論如何，我試圖重寫查詢。 而不是使用

AND s2.rid IN (
                    SELECT  s1.rid

我使用了具有相同stop表的自連接並將其別名為s1並將連接條件用作s2.rid = s1.rid and s1.place = 'place1'

$sql="SELECT s2.stopno, s2.rid, b.busno, s2.pid, b.buskey, r.path,
        IF(s1.pid < s2.pid, IF(r.path=0, s2.rid, null), IF(r.path=1, s2.rid,null))
    FROM `stop` s2, bus b, `route` r, place p, `stop` s1
    WHERE s2.rid = r.rid
    AND r.bid = b.bid
    AND s2.pid = p.pid
    AND s2.rid = s1.rid and s1.place = 'place1'
    AND p.place = 'place2'"

您還可以進一步簡化它，而不是使用多個 IF 或 CASE 語句。

s1.pid < s2.pid THEN (SELECT S2.rid WHERE r.path = 0) ELSE (SELECT S2.rid WHERE r.path=1 )

您可以使用r.path in (0,1)來定義它

IF(s1.pid < s2.pid), IF(r.path in (0,1),s2.rid, NULL), NULL)