我有以下两个表：
表1
id name
---------
A3 B2
A3 B400
A5 B100
A7 B200
A8 B6
A8 B2
A8 B3
和表2
id name company
-------------------
A1 company1
A2 company2
A3 B1 company3
A4 company4
A5 B2 company5
A6 company6
A7 B3 company7
A8 B4 company8
A9 company9
A10 B6 company10
我正在启动三个查询以获得我想要的内容，但有三个查询。我的问题是如何在一个查询中以更有效的方式完成所有这些工作？
查询1：
SELECT DISTINCT t1.id as ID,
t2.company as Company,
'FOUND' AS status
FROM table1 t1
JOIN table2 t2
ON t1.id = t2.id
group by ID
查询1结果：
ID Company Status
-----------------------
A3 company3 FOUND
A5 company5 FOUND
A7 company7 FOUND
A8 company8 FOUND
查询2：
SELECT DISTINCT t2.id as ID,
t2.company as Company,
'FOUND' AS status
FROM table1 t1
JOIN table2 t2
ON t1.name = t2.name
group by ID
查询2结果：
ID Company Status
-----------------------
A5 company5 FOUND
A10 company10 FOUND
A7 company7 FOUND
查询3：
SELECT t1.name as ID,
t1.name as Company,
'NOT FOUND' AS status
FROM table1 t1
WHERE t1.name NOT IN (SELECT t2.name
FROM table2 t2)
GROUP BY ID
查询3结果：
ID Company Status
-----------------------
B400 B400 NOT FOUND
B100 B100 NOT FOUND
B200 B200 NOT FOUND
和最终结果输出如下：
ID Company Status
---------------------------
A3 company3 FOUND
A5 company5 FOUND
A7 company7 FOUND
A8 company8 FOUND
A10 company10 FOUND
B100 B100 NOT FOUND
B200 B200 NOT FOUND
B400 B400 NOT FOUND
注意：A5和A7在第一个查询结果中弹出，也在第二个查询结果中弹出！所以我们只需要保留一个。
如果需要，可以进行一些解释：
我们从 table1 获取唯一id的列表，从 table2 获取相应的company。我们对 table1 的第二列执行了类似的查询：我们在第二列中查找 table1 中第二列的值，即name table2 ，如果我们找到它，那么我们会从 table2 ，获得相应的id和company，但是如果{{1}已经存在于我们之前的查询中然后我们放弃它，不需要重复它。第三，如果我们在 table2 的id中找不到 table1 的name值，那么我们没有相应的name和id，因此我们将company值归为name和id。一般情况下，如果我们在 table2 中找到 table1 中的company和id，那么我们会为其提供状态name如果没有，FOUND。
提前致谢
顺便说一句，我已经尝试了两次使用NOT FOUND，但查询需要很长时间而且效率不高。
答案 0 :(得分：3)
这些要求令人困惑，可能值得重新评估您的数据模型。我认为UNION解决方案是您最好的选择，可以修改为使用UNION ALL来提高效率。
我确实整理了一个基于互斥锁的hack，它可能会出现与此页面上任何其他查询一样多的微妙问题。
select
coalesce(t2.id, t1.name) AS ID,
coalesce(t2.company, t1.name) AS Company,
if(isnull(t2.id), 'NOT FOUND', 'FOUND') as Status
from (select 0 as mutex union select 1) as m
left join table1 as t1 on 1 = 1
left join table2 as t2 on t1.name = t2.name or (t1.id = t2.id and mutex)
group by coalesce(t2.id, t1.name)
也就是说，请仔细测试这些查询并查看您的数据和结果。根据您的输入数据，存在很大的错误空间。
答案 1 :(得分：2)
尝试使用UNION DISTINCT：
SELECT DISTINCT t1.id as ID,
t2.company as Company,
'FOUND' AS status
FROM table1 t1
JOIN table2 t2
ON t1.id = t2.id
group by ID
union distinct
SELECT DISTINCT t2.id as ID,
t2.company as Company,
'FOUND' AS status
FROM table1 t1
JOIN table2 t2
ON t1.name = t2.name
group by ID
union distinct
SELECT t1.name as ID,
t1.name as Company,
'NOT FOUND' AS status
FROM table1 t1
WHERE t1.name NOT IN (SELECT t2.name
FROM table2 t2)
GROUP BY ID
答案 2 :(得分：2)
您的查询有些不清楚，因为它们不应该像您显示的那样执行（由于sess.run(y_, feed_dict={x: [np.flatten(image_you_loaded_in)]})
包含不在select中的非聚合）。但根据你对你要做的事情的解释......
您可以使用外连接，然后使用案例逻辑和/或合并来确定在每种情况下使用哪个值。
group by请注意，我使用了SELECT DISTINCT
coalesce(t2_id.id, t2_name.id, t1.name) as ID
, coalesce(t2_id.company, t2_name.company, t1.name) as Conpany
, case when t2_id.id is not null or t2_name.name is not null
then 'FOUND'
else 'NOT FOUND'
end status
FROM table1 t1
LEFT JOIN table2 t2_id
ON t1.id = t2_id.id
LEFT JOIN table2 t2_name
ON t1.name = t2_name.name
来确保完全相同的行不会多次出现;但这可能会为ID（具有不同的公司值）返回多行，具体取决于数据。我无法确切地说出是什么意思，因为在问题中发布的三个查询中DISTINCT和DISTINCT的使用似乎并没有为我增加。
答案 3 :(得分：1)
我认为您可以使用如下查询：
SELECT DISTINCT IF(name2 IS NULL, name, ID) AS ID,
IF(name2 IS NULL, name, Company) AS Company,
IF(name2 IS NULL, 'NOT FOUND', 'FOUND') AS Status
FROM (
SELECT DISTINCT
CASE
WHEN t1.id = t2.id THEN t1.id
WHEN t1.name = t2.name THEN t2.id
ELSE t1.id
END AS ID,
CASE
WHEN t1.id = t2.id THEN t2.company
WHEN t1.name = t2.name THEN t2.company
ELSE t1.name
END AS Company,
t1.name,
(SELECT Table2.name
FROM Table2
WHERE Table2.name = t1.name LIMIT 1) AS name2
FROM Table1 AS t1
LEFT JOIN Table2 AS t2 ON (t1.id = t2.id) OR (t1.name = t2.name)) AS t
ORDER BY ID;
查询使用单个LEFT JOIN操作加上相关子查询。