psycopg2 PSQL SELECT 语句返回 0L 而不是 0答案

【问题标题】：psycopg2 PSQL SELECT statement returning 0L instead of 0psycopg2 PSQL SELECT 语句返回 0L 而不是 0
【发布时间】：2016-05-16 07:19:15
【问题描述】：

我的 SELECT 语句出现意外结果。

我有以下代码：

db_cursor.execute("""SELECT player.player_id, player.player_name, 
                       COUNT(match.winner_id) as wins, COUNT(match.loser_id + match.winner_id) as total_matches
                       FROM player LEFT JOIN match ON player.player_id = match.winner_id
                       GROUP BY player.player_id
                       ORDER BY wins ASC""")
results = db_cursor.fetchall()
print results

我在打印时得到以下输出。

[(317, 'Bruno Walton', 0L, 0L), (318, "Boots O'Neal", 0L, 0L), (319, 'Cathy Burton', 0L, 0L), (320, 'Diane Grant', 0L, 0L)]
[(318, "Boots O'Neal", 0L, 0L), (320, 'Diane Grant', 0L, 0L), (317, 'Bruno Walton', 1L, 1L), (319, 'Cathy Burton', 1L, 1L)]

什么是0L和1L？我认为“L”仅代表非常大的数字，而不是单个数字。

抱歉，我是初学者，所以可能很明显，但我们将不胜感激。如果有帮助，这是表的 psql：

CREATE TABLE IF NOT EXISTS player
(player_id SERIAL PRIMARY KEY, player_name TEXT);

CREATE TABLE IF NOT EXISTS match
(match_id SERIAL PRIMARY KEY, winner_id INTEGER references player(player_id),
 loser_id INTEGER references player(player_id));

【问题讨论】：

标签： python-2.7 psycopg2 psql

【解决方案1】：

您可以在查询中添加转换运算符，即 COUNT(match.winner_id)::int 和 COUNT(match.loser_id + match.winner_id)::int。否则，您将不得不在结果集中转换列。

db_cursor.execute("""SELECT player.player_id, player.player_name, 
                       COUNT(match.winner_id)::int as wins, COUNT(match.loser_id + match.winner_id)::int as total_matches
                       FROM player LEFT JOIN match ON player.player_id = match.winner_id
                       GROUP BY player.player_id
                       ORDER BY wins ASC""")

【讨论】：