開発環境
- OS: OS X Lion - Apple
- データベース言語: SQL
- リレーショナルデータベース: MySQL
『初めてのSQL』(Alan Beaulieu 著、株式会社クイープ 翻訳、オライリー・ジャパン、2006年、ISBN4-8733-181-8) の8章(グループ化と集約化), 8.5(練習問題)8-3を解いてみる。
8-3.
SQL文(TextWrangler)
SELECT cust_id, count(*) FROM account GROUP BY cust_id HAVING COUNT(*) >= 2;
入出力結果(Terminal)
$ mysql -u lrngsql -p bank Enter password: Reading table information for completion of table and column names You can turn off this feature to get a quicker startup with -A Welcome to the MySQL monitor. Commands end with ; or \g. Your MySQL connection id is 1 Server version: 5.5.24 MySQL Community Server (GPL) Copyright (c) 2000, 2011, Oracle and/or its affiliates. All rights reserved. Oracle is a registered trademark of Oracle Corporation and/or its affiliates. Other names may be trademarks of their respective owners. Type 'help;' or '\h' for help. Type '\c' to clear the current input statement. mysql> SELECT cust_id, count(*) -> FROM account -> GROUP BY cust_id -> WHERE COUNT(*) >= 2; ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE COUNT(*) >= 2' at line 4 mysql> SELECT cust_id, count(*) -> FROM account -> GROUP BY cust_id -> HAVING COUNT(*) >= 2; +---------+----------+ | cust_id | count(*) | +---------+----------+ | 1 | 3 | | 2 | 2 | | 3 | 2 | | 4 | 3 | | 6 | 2 | | 8 | 2 | | 9 | 3 | | 10 | 2 | +---------+----------+ 8 rows in set (0.05 sec) mysql> quit Bye $
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