blocks|key|1781238|text|也许是像这样的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1781239|select+coalesce(t1.id,+t2.id)+as+id,+(coalesce(t1.value,+0)+%2B+coalesce(t2.value,+0))+as+value+
from+table1+t1+full+outer+join+table2+t2+on+t1.id+=+t2.id|code-block|syntax|javascript|1781240|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

maybe something like

<pre><code>select coalesce(t1.id, t2.id) as id, (coalesce(t1.value, 0) + coalesce(t2.value, 0)) as value 
from table1 t1 full outer join table2 t2 on t1.id = t2.id
</code></pre>

blocks|key|1781253|text|使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1781254|++SELECT+x.id,
+++++++++SUM(x.value)
++++FROM+(SELECT+t.id,
+++++++++++++++++t.value
++++++++++++FROM+TABLE_1+t
++++++++++UNION+ALL
++++++++++SELECT+t2.id,
+++++++++++++++++t2.value
++++++++++++FROM+TABLE_2+t2)+x
GROUP+BY+x.id|code-block|syntax|javascript|1781255|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Use:

<pre><code> SELECT x.id,
 SUM(x.value)
 FROM (SELECT t.id,
 t.value
 FROM TABLE_1 t
 UNION ALL
 SELECT t2.id,
 t2.value
 FROM TABLE_2 t2) x
GROUP BY x.id
</code></pre>

blocks|key|373968|text|你可以这样做--但其他的答案可能会更快：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|373969|SELECT+t1.id,+t1.value+%2B+t2.value+AS+value
++FROM+t1+INNER+JOIN+t2+ON+t1.id+=+t2.id
UNION
SELECT+t1.id,+t1.value
++FROM+t1
+WHERE+t1.id+NOT+IN+(SELECT+t2.id+FROM+t2)
UNION
SELECT+t2.id,+t2.value
++FROM+t2
+WHERE+t2.id+NOT+IN+(SELECT+t1.id+FROM+t1)|code-block|syntax|javascript|373970|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You could do it like this - but the other answers are probably swifter:

<pre><code>SELECT t1.id, t1.value + t2.value AS value
 FROM t1 INNER JOIN t2 ON t1.id = t2.id
UNION
SELECT t1.id, t1.value
 FROM t1
 WHERE t1.id NOT IN (SELECT t2.id FROM t2)
UNION
SELECT t2.id, t2.value
 FROM t2
 WHERE t2.id NOT IN (SELECT t1.id FROM t1)
</code></pre>

blocks|key|373982|text|SELECT
++++COALESCE(t1.id,+t2.id)+AS+id,
++++COALESCE(t1.value,+0)+%2B+COALESCE(t2.value,+0)+AS+value
FROM
++++t1
++++FULL+OUTER+JOIN
++++t2+ON+t1.id+=+t2.id|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|373983|或|unstyled|373984|SELECT
++++foo.id,
++++COALESCE(t1.value,+0)+%2B+COALESCE(t2.value,+0)+AS+value
FROM
++++(
++++SELECT+t1.id+FROM+t1
++++UNION
++++SELECT+t2.id+FROM+t2
++++)+foo
++++LEFT+JOIN
++++t1+ON+foo.id+=+t1.id
++++LEFT+JOIN
++++t2+ON+foo.id+=+t2.id|373985|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>SELECT
 COALESCE(t1.id, t2.id) AS id,
 COALESCE(t1.value, 0) + COALESCE(t2.value, 0) AS value
FROM
 t1
 FULL OUTER JOIN
 t2 ON t1.id = t2.id
</code></pre>

OR

<pre><code>SELECT
 foo.id,
 COALESCE(t1.value, 0) + COALESCE(t2.value, 0) AS value
FROM
 (
 SELECT t1.id FROM t1
 UNION
 SELECT t2.id FROM t2
 ) foo
 LEFT JOIN
 t1 ON foo.id = t1.id
 LEFT JOIN
 t2 ON foo.id = t2.id
</code></pre>

blocks|key|453249|text|SELECT+ISNULL(T1.Col1.T2.Col1)+as+ID,+(ISNULL(T1.Col2,0)+%2B+ISNULL(T2.Col2,0))+as+SumCols
FROM+T1+OUTER+JOIN+T2+ON+T1.Col1+=+T2.Col2|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|453250|没有分组或其他任何东西。它处理以下情况|unstyled|453251|如果id在T1中，但不在T2中，那么您将获得T1中的值，反之亦然。这将处理双向包含。|453252|如果ID在两者中，您将获得总和|453253|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|F|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>SELECT ISNULL(T1.Col1.T2.Col1) as ID, (ISNULL(T1.Col2,0) + ISNULL(T2.Col2,0)) as SumCols
FROM T1 OUTER JOIN T2 ON T1.Col1 = T2.Col2
</code></pre>

No grouping or anything. It handles the following cases

if an id is in T1 but not in T2 you will get the value in T1 and vice versa. This handles bi-directional inclusion.

If an ID is in both you will get the sum

Is it possible to craft a query that adds values within two tables:

For example, say you have two tables

<pre><code>id value
-- -----
a 1
c 2
d 3
f 4
g 5
</code></pre>

and 

<pre><code>id value
-- -----
a 1
b 2
c 3
d 4
e 5
</code></pre>

Then when you 'add' the two tables you would get the result where the id's match. So, a=1+1=2, and simply the same result where they don't. So the query would return:

<pre><code>id value
-- -----
a 2
b 2
c 5
d 7
e 5
f 4
g 5
</code></pre>

Merge and add values from two tables

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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是否可以手工创建一个在两个表中添加值的查询：例如，假设您有两个表id    value--    -----a       1c       2d       3f       4g       5和id     value--     -----a        1b        2c        3d     ...

问合并和添加两个表中的值
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问合并和添加两个表中的值EN