Question

我正在尝试使用两个列名'id'和'easy_high_score'将信息解析为SELECT语句，以便可以在程序中操纵两个列的值，但是尝试获取列的值时“ easy_high_score”，应为46或20这样的整数，它返回的字符串为（'easy_high_score'，）。

即使表中没有提及[['easy_high_score'，）]，它仍然会打印出来。在表中，id 1具有我尝试获取的正确值和信息，但无济于事。我对SQLite3相当陌生。

if mode == "Easy":
    mode = 'easy_high_score'

if mode == "Normal":
    mode = "normal_high_score"

if mode == 'Hard':
    mode == "hard_high_score"


incrementor = 1 ##This is used in a for loop but not necessary for this post
c.execute("SELECT ? FROM players WHERE id=?", (mode, incrementor))
allPlayers = c.fetchall()
print(allPlayers)  #This is printing [('easy_high_score',)], when it should be printing an integer.

预期结果：20（或代表简单模式得分最高的整数）

实际结果：[（'easy_high_score'，）]

Answer 1

不能使用参数指定列名，它应在查询中逐字出现。修改执行查询的行，如下所示：

c.execute("SELECT %s FROM players WHERE id=?" % mode, (incrementor,))

SELECT语句返回列名而不是VALUE（对于所述列）

1 个答案: