blocks|key|2102545|text|只需做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2102546|SELECT+
++++date,+sum
FROM
++++t_fb_data+
UNION+SELECT+
++++date,+sum
FROM
++++t_google_data|code-block|syntax|javascript|2102547|或者将其放入子查询中：|2102548|SELECT+
++++*
FROM
++++(SELECT+
++++++++date,+sum
++++FROM
++++++++t_fb_data+UNION+SELECT+
++++++++date,+sum
++++FROM
++++++++t_google_data)+t;|2102549|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Simply do:

<pre><code>SELECT 
 date, sum
FROM
 t_fb_data 
UNION SELECT 
 date, sum
FROM
 t_google_data
</code></pre>

or put it in a subquery:

<pre><code>SELECT 
 *
FROM
 (SELECT 
 date, sum
 FROM
 t_fb_data UNION SELECT 
 date, sum
 FROM
 t_google_data) t;
</code></pre>

blocks|key|505331|text|在MySQL中，如果您想从不同的数据集中总结相同的值，则需要使用关键字ALL，因为只有联合才能在唯一的数据集中总结数据集并在其上应用不同的数据集。您的查询应该如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|505332|SELECT+date,sum+FROM+t_fb_data+Union+ALL+SELECT+date,sum+FROM+t_google_data|code-block|syntax|javascript|505333|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|$]]

In MySQL, if you want to sum up identical values from different data sets, you need to use the keyword ALL, because the union alone will sum up the data sets in an unique data sets and apply a DISTINCT on it.
Your query should look like this:

<pre><code>SELECT date,sum FROM t_fb_data Union ALL SELECT date,sum FROM t_google_data
</code></pre>

blocks|key|2102625|text|您可以肯定地使用UNION+ALL，但是如果您只想坚持使用UNION，那么请尝试这样做，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2102626|SELECT+*+
++FROM+
+++++(+SELECT+date
++++++++++++,+sum,+'fb'+identifier+
+++++++++FROM+t_fb_data
+++++)+T1+
+Union+
+++++(+SELECT+date
++++++++++++,+sum,+'google'+identifier+
+++++++++FROM+t_google_data
+++++)+T2+|code-block|syntax|javascript|2102627|这是因为，来自两个查询的数据将提供相同的记录，这些记录将合并到单个记录中，因此，如果要写入标识符，两个表都会有不同的记录。|2102628|我希望这能帮上忙。|2102629|编辑|offset|length|style|BOLD|2102630|+(+SELECT+date
++++++++++++,+sum,+'fb'+identifier+
+++++++++FROM+t_fb_data
+++++)++
+Union+
+++++(+SELECT+date
++++++++++++,+sum,+'google'+identifier+
+++++++++FROM+t_google_data
+++++)+|2102631|entityMap^0|0|0|0|0|0|2|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|Y|8|@$M|Z|N|10|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

You can use UNION ALL for sure,
but if you want to stick to UNION only then try this,

<pre><code>SELECT * 
 FROM 
 ( SELECT date
 , sum, 'fb' identifier 
 FROM t_fb_data
 ) T1 
 Union 
 ( SELECT date
 , sum, 'google' identifier 
 FROM t_google_data
 ) T2 
</code></pre>

This is because, your data from two queries will give identical records which will be merged into single record, so if you will write identifer both tables will have differential records.

I hope this will help.

EDIT

<pre><code> ( SELECT date
 , sum, 'fb' identifier 
 FROM t_fb_data
 ) 
 Union 
 ( SELECT date
 , sum, 'google' identifier 
 FROM t_google_data
 ) 
</code></pre>

blocks|key|505370|text|你的问题是错的。您尝试合并两个具有SELECT语句的子查询，在顶部引用第一个子查询。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|505371|您正在寻找的代码：|505372|SELECT+date,+sum+FROM+t_fb_data
UNION
SELECT+date,+sum+FROM+t_google_data|code-block|syntax|javascript|505373|这将从两个表返回非重复的值。|505374|如果您想要将它用作子查询，可以简单地用另一个SELECT语句包围它，如下例所示：|505375|SELECT+sub.*
FROM+(
SELECT+date,+sum+FROM+t_fb_data
UNION
SELECT+date,+sum+FROM+t_google_data
)+sub|505376|在这种情况下，如果需要，可以在顶级SELECT语句中进行附加计算。|505377|entityMap^0|H|6|0|0|0|0|M|6|0|0|H|6|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|J|7|15|8|@]|D|@]|E|$K|L]]|$1|S|3|T|5|6|7|16|8|@$9|17|A|18|B|C]]|D|@]|E|$]]|$1|U|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|V|$]]

Your query is wrong. You try to union two subqueries having <code>SELECT</code> statement at top refering to first subquery.

Code that You are looking for:

<pre><code>SELECT date, sum FROM t_fb_data
UNION
SELECT date, sum FROM t_google_data
</code></pre>

That will return non duplicated values from both tables.

If you want to use this as a subquery, you can simply surround that with another <code>SELECT</code> statement as on example below:

<pre><code>SELECT sub.*
FROM (
SELECT date, sum FROM t_fb_data
UNION
SELECT date, sum FROM t_google_data
) sub
</code></pre>

In this case, You can do additonal calculations in your top <code>SELECT</code> statement, if needed.

I dont get it. I try to sum up two identical tables with union but it doesnt works.

<pre><code>t_fb_data:

id | date | sum
---------------------
1 | 2017.01.01| 2
2 | 2017.01.02| 1

t_google_data:

id | date | sum
---------------------
1 | 2017.01.01| 7
2 | 2017.01.02| 1
3 | 2017.01.03| 2

result could looks like:

| date | sum
------------------
| 2017.01.01| 2
| 2017.01.02| 1
| 2017.01.01| 7
| 2017.01.02| 1
| 2017.01.03| 2
</code></pre>

This is my Query:

<pre><code>SELECT * 
 FROM 
 ( SELECT date
 , sum 
 FROM t_fb_data
 ) T1 
 Union 
 ( SELECT date
 , sum 
 FROM t_google_data
 ) T2 
</code></pre>

T2 should be wrong because my workbench want an parenthesis in front of T2. But if i do this its wrong two. Do i have to setup something on my workbench?

MySQL union missing parenthesis

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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我不明白。我试着把两张相同的表和union放在一起，但是它不起作用。t_fb_data:id | date      | sum---------------------1  | 2017.01.01| 22  | 2017.01.02| 1t_google_data:id | date      | sum-----...

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