blocks|key|1371726|text|添加一个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1371727|AS+some_alias|code-block|syntax|javascript|1371728|像这样：|1371729|SELECT+name,+email,+COUNT(*)+
FROM+(SELECT+DISTINCT+name,+email+
FROM+people
WHERE+degree+=+'Gradiate')+as+my_alias|1371730|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|$]]

Add an

<pre><code>AS some_alias
</code></pre>

Like:

<pre><code>SELECT name, email, COUNT(*) 
FROM (SELECT DISTINCT name, email 
FROM people
WHERE degree = 'Gradiate') as my_alias
</code></pre>

blocks|key|2770745|text|您根本不需要子查询(如前所述，这将需要一个表别名)。这对你需要的东西是没有意义的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2770746|像这样使用GROUP+BY：|offset|length|style|CODE|2770747|SELECT+name,+email,+count(*)+AS+ct
FROM+++people
WHERE++degree+=+'Graduate'
GROUP++BY+1,+2;|code-block|syntax|javascript|2770748|在本例中，GROUP++BY+1,+2是GROUP++BY+name,+email的缩写。Example+with+more+details.|2770749|当然，这里必须匹配name和email，这与您的示例中只显示name和count不同(这与您的查询相矛盾)。|2770750|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/20021602/find-movies-with-highest-number-of-awards-in-certain-year-code-duplication/20090089#20090089^0|0|5|8|0|0|5|E|K|L|19|Q|0|0|9|4|E|5|U|4|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@$D|10|E|11|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|12|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|13|8|@$D|14|E|15|F|G]|$D|16|E|17|F|G]]|9|@$D|18|E|19|1|1A]]|A|$]]|$1|O|3|P|5|6|7|1B|8|@$D|1C|E|1D|F|G]|$D|1E|E|1F|F|G]|$D|1G|E|1H|F|G]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

You do not need the subquery at all (which would require a table alias as has been pointed out). It wouldn't make sense for what you need.

Use <code>GROUP BY</code> like this:

<pre><code>SELECT name, email, count(*) AS ct
FROM people
WHERE degree = 'Graduate'
GROUP BY 1, 2;
</code></pre>

<code>GROUP BY 1, 2</code> is shorthand for <code>GROUP BY name, email</code> in this case. <a href="https://stackoverflow.com/questions/20021602/find-movies-with-highest-number-of-awards-in-certain-year-code-duplication/20090089#20090089">Example with more details.</a>

Of course, <code>name</code> and <code>email</code> have to match here, unlike your example where you only display <code>name</code> and count (which contradicts your query).

blocks|key|1376059|text|SELECT+name,+count(*)+AS+count
FROM+++people
WHERE++degree+=+'Gradiate'
GROUP++BY+name;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1376060|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>SELECT name, count(*) AS count
FROM people
WHERE degree = 'Gradiate'
GROUP BY name;
</code></pre>

I'm trying write a Postgres query to display a distinct number of people but count the actual (non-distinct) number of people. 

So if I have

<pre><code>1 Ray ray@gmail.com
2 Ray ray@gmail.com
3 Kate kate@gmail.com
</code></pre>

I'd want to show:

<pre><code>Ray 2
Kate 1
</code></pre>

==

<pre><code>SELECT name, email, COUNT(*) 
FROM (SELECT DISTINCT name, email 
FROM people
WHERE degree = 'Gradiate') 
</code></pre>

I get:

<pre><code>ERROR: subquery in FROM must have an alias
LINE 3: FROM (SELECT DISTINCT name, email
</code></pre>

How to fix this?

select name and email and count using subquery in postgresql

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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我正在尝试编写Postgres查询，以显示不同的人数，但计算实际的(非不同的)人数。所以如果我有1  Ray ray@gmail.com2  Ray ray@gmail.com3  Kate kate@gmail.com我想展示的是：Ray 2Kate 1==SELECT name, email, COUNT(*) F...

问使用子查询在postgresql中选择姓名、电子邮件和计数
EN

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问使用子查询在postgresql中选择姓名、电子邮件和计数EN