Question

我在SQLite中有两个表：

Table1:
-------
id
name

Table2:
-------
id
temp_name

我的问题是，如何编写一个返回Table2中不在Table1中的名称的SQL查询？

例如：

Table1:
-------
1, 'john'
2, 'boda',
3, 'cydo',
4, 'linus'

Table2:
-------
1123, 'boda'
2992, 'andy',
9331, 'sille',
2,    'cydo'

在此示例中，SQL查询应返回andy中的元素sille和Table2，因为它们不在Table1中。

Answer 1

这是如何在“明显的”标准SQL中实现的：

select *
from table2
where temp_name not in (select name from table1)

还有其他方法，例如在left outer join子句中使用exists，where和except操作。

Answer 2

select name
from table2
except
select temp_name
from table1

Answer 3

我很好奇哪个选项对我的用例最有效，并认为如果我分享我的结果可能对其他人有帮助：

简而言之，选择中最快的地方（对我而言！）。值得注意的是，使用所有选项都会返回重复项，除了除了。

Option 0: SELECT {c2} FROM {t2} WHERE {c2} not in (SELECT {c1} FROM {t1});
    Entries returned: 1098, Unique entries returned: 357
    Average: 1.8680 seconds
    Entries returned: 0, Unique entries returned: 0
    Average: 0.6664 seconds
Option 1: SELECT {c2} FROM {t2} EXCEPT SELECT {c1} FROM {t1};
    Entries returned: 357, Unique entries returned: 357
    Average: 3.9455 seconds
    Entries returned: 0, Unique entries returned: 0
    Average: 3.3074 seconds
Option 2: SELECT {t2}.{c2} FROM {t2} LEFT OUTER JOIN {t1} ON {t1}.{c1} = {t2}.{c2} WHERE {t1}.{c1} IS null;
    Entries returned: 1098, Unique entries returned: 357
    Average: 2.3330 seconds
    Entries returned: 0, Unique entries returned: 0
    Average: 1.1982 seconds
Option 3: SELECT {c2} FROM {t2} WHERE NOT EXISTS (SELECT 1 FROM {t1} WHERE {c1} = {t2}.{c2});
    Entries returned: 1098, Unique entries returned: 357
    Average: 2.6945 seconds
    Entries returned: 0, Unique entries returned: 0
    Average: 0.9737 seconds

以下是我用来运行数字的代码：

import sqlite3
import timeit

# Database path here
database = "database.db"

# Your table and column names here
t1, c1 = 'Table_1', 'name'
t2, c2 = 'Table_2', 'temp_name'

# Reverse the test
dbs = [{'t1': t1, 'c1':c1, 't2': t2, 'c2': c2},
       {'t1': t2, 'c1':c2, 't2': t1, 'c2': c1}]

commands = ["SELECT {c2} FROM {t2} WHERE {c2} not in (SELECT {c1} FROM {t1});",
"SELECT {c2} FROM {t2} EXCEPT SELECT {c1} FROM {t1};",
"SELECT {c2} FROM {t2} LEFT OUTER JOIN {t1} ON {t1}.{c1} = {t2}.{c2} WHERE {t2}.{c2} IS null;",
"SELECT {c2} FROM {t2} WHERE NOT EXISTS (SELECT 1 FROM {t1} WHERE {c1} = {t2}.{c2});",]

for i, c in enumerate(commands):
    print("Option {}: {}".format(i, c))
    for db in dbs:
        co = c.format(**db)
        foo = sqlite3.connect(database).execute(co).fetchall()
        # Sanity check that entries have been found and how many
        print("\tEntries returned: {}, Unique entries returned: {}".format(len(foo), len({a[0] for a in foo})))
        # Reconnect to the database each time - I can't remember if there's any caching
        t = timeit.repeat(lambda: sqlite3.connect(database).execute(co).fetchall(), repeat=5, number=1)
        print('\tAverage: {:.4f} seconds'.format(statistics.mean(t)))

Answer 4

左连接版本：

select t2.* from Table2 t2
left outer join Table1 t1 on t1.name = t2.temp_name
where t2.temp_name is null

Answer 5

戈登提到的EXISTS方法：

SELECT *
FROM Table2
WHERE NOT EXISTS (SELECT 1
                  FROM Table1
                  WHERE Table1.name = Table2.temp_name)

在SQLite中，如何在一个表中选择不在其他表中的元素？

5 个答案: