質問編集履歴
1
テーブルを2つ見なければならない場合の条件を追加
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@@ -85,3 +85,169 @@
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+---------+----------+
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```
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↑の条件は解決できましたが、テーブル2つを参照する場合などはどうすればよいでしょうか。
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テーブルが2つあり、
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JOINするか副問い合わせするかして、
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username in ('A', 'D', 'E', 'X', 'Y') という条件に絞った上で、
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loginname を ('AA', 'DD', 'EE', 'notfound', 'notfound') と表示させたいのです。
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```MySQL
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/* login TABLE */
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+--------+---------+-----------+
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| id | user_id | loginname |
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+--------+---------+-----------+
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| 21 | 1 | AA |
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| 22 | 2 | BB | /* ★ */
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| 23 | 3 | CC |
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| 24 | 4 | DD | /* ★ */
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| 25 | 5 | EE | /* ★ */
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| 26 | 6 | FF |
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+--------+---------+--------+
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/* user TABLE */
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+---------+----------+
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| id | username |
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+---------+----------+
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| 1 | A |
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| 2 | B | /* ★ */
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| 3 | C |
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| 4 | D | /* ★ */
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| 5 | E | /* ★ */
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| 6 | F |
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+---------+----------+
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/* usernameが'X', 'Y'のレコードが該当しない */
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SELECT
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u.loginname
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FROM
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login AS l
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LEFT JOIN
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user AS u ON u.id = l.user_id
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WHERE
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username in ('A', 'D', 'E', 'X', 'Y') ;
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+-----------+
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| loginname |
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+-----------+
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| AA |
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| DD |
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| EE |
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+-----------+
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/* 下記のようなクエリで、'B','C','F'を出さず、'X','Y'も表示したい */
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/* 案1 */
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SELECT
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CASE
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WHEN u.loginname IS NULL THEN 'notfound'
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ELSE u.loginname
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END
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FROM
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login AS l
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LEFT JOIN
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user AS u ON u.id = l.user_id
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WHERE
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username in ('A', 'D', 'E', 'X', 'Y') ;
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+-----------+
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| loginname |
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+-----------+
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| AA |
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| DD |
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| EE |
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| notfound |
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| notfound |
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+-----------+
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/* 案2 */
|
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234
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+
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SELECT
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236
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+
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+
CASE
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238
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+
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WHEN loginname IS NULL THEN 'notfound'
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+
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ELSE loginname
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+
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END
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+
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+
FROM
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+
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247
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+
login
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248
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+
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249
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+
WHERE
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250
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+
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251
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+
user_id in (select id from user where username in ('A', 'D', 'E', 'X', 'Y')) ;
|
|
252
|
+
|
|
253
|
+
```
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