Question

i m 试图利用国际笔会来换取相当于方略清单之一的结果

例如

strings = [ string1 , string2 , string3 ] c.execute( select count(*) from table where foo in ? , strings)

i 知道这一点是不正确的,而且有工作要做,但希望能强调什么是想做的。

Answer 1

You can t do that. There are three problems:

You can t have a table called table unless you use backticks around the table name.
An IN clause must have parentheses.
You want three parameters, not one.

相反:

sql =  SELECT COUNT(*) FROM yourtable WHERE foo IN (?, ?, ?)

如果体数变化,则使用:

params =  , .join( ?  for x in strings)
sql =  SELECT COUNT(*) FROM yourtable WHERE foo IN (  + params +  )

Answer 2

您可以做如下文稿:, .join(strings): @Mark Byers建议,这几段文是行文最多的。但是,如果体内数为,则会由于Kingk的询问时间有限而失败。

Another way of doing it is creating a temp table, inserting there all the strings and doing a join to perform the intersection, something like

c.execute( CREATE TEMP TABLE strings (s STRING) ) c.executemany( INSERT INTO strings (s) VALUES (?) , ((s,) for s in strings)) c.execute( SELECT COUNT(*) FROM table JOIN strings ON table.foo == strings.s )

友情链接