blocks|key|4005705|text|如果您知道当前和以前的日期，那么您可以：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4005706|SELECT+SUM(visited_date+=+CURDATE())+as+today,
+++++++SUM(visited_date+=+CURDATE()+-+interval+1+day)+as+yesterday
FROM+visitors
WHERE+visited_date+>=+CURDATE()+-+interval+1+day;|code-block|syntax|javascript|4005707|如果您不知道这两天，那么您可以做一些类似的事情，获取数据中的最新日期：|4005708|SELECT+SUM(v.visited_date+=+m.max_vd)+as+today,
+++++++SUM(v.visited_date+<+m.max_vd)+as+yesterday
FROM+visitors+v+CROSS+JOIN
+++++(SELECT+MAX(v2.visited_date)+as+max_vd+FROM+visitors+v2)+as+m
WHERE+v.visited_date+>=+m.max_vd+-+interval+1+day|4005709|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|$]]

If you know the current and previous date, then you can do:

<pre><code>SELECT SUM(visited_date = CURDATE()) as today,
 SUM(visited_date = CURDATE() - interval 1 day) as yesterday
FROM visitors
WHERE visited_date &gt;= CURDATE() - interval 1 day;
</code></pre>

If you don't know the two days, then you can do something similar, getting the latest date in the data:

<pre><code>SELECT SUM(v.visited_date = m.max_vd) as today,
 SUM(v.visited_date &lt; m.max_vd) as yesterday
FROM visitors v CROSS JOIN
 (SELECT MAX(v2.visited_date) as max_vd FROM visitors v2) as m
WHERE v.visited_date &gt;= m.max_vd - interval 1 day
</code></pre>

blocks|key|4994006|text|这应该能起作用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4994007|SELECT+COUNT(CASE+
++++++++++++++++WHEN+visited_date+=+CURDATE()+THEN+1+
+++++++++++++END)+AS+todayCount+,
+++++++COUNT(CASE+
++++++++++++++++WHEN+visited_date+=+CURDATE()+-+INTERVAL+1+DAY+THEN+1+
+++++++++++++END)+AS+yesterdayCount+
FROM+visitors++
WHERE+visited_date+IN+(CURDATE(),+CURDATE()+-+INTERVAL+1+DAY)
GROUP+BY+visited_date
ORDER+by+visited_date+|code-block|syntax|javascript|4994008|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|$]]

This should do the trick:

<pre><code>SELECT COUNT(CASE 
 WHEN visited_date = CURDATE() THEN 1 
 END) AS todayCount ,
 COUNT(CASE 
 WHEN visited_date = CURDATE() - INTERVAL 1 DAY THEN 1 
 END) AS yesterdayCount 
FROM visitors 
WHERE visited_date IN (CURDATE(), CURDATE() - INTERVAL 1 DAY)
GROUP BY visited_date
ORDER by visited_date 
</code></pre>

blocks|key|4005855|text|只需尝试这个简单的查询|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4005856|select+visited_date+as+date,+COUNT(*)+as+count+from+`visitors`+
group+by+`visited_date`+order+by+`visited_date`+asc|code-block|syntax|javascript|4005857|它将产生输出|4005858|​|4005859|📷|atomic|offset|length|4005860|4005861|它会对你有用的。|4005862|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://developer.qcloudimg.com/http-save/yehe-900000/6122cbf5965d5677f1f746e084815fae.png|imageAlt^0|0|0|0|0|0|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|13|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|14|8|@]|9|@]|A|$]]|$1|K|3|L|5|M|7|15|8|@]|9|@$N|16|O|17|1|18]]|A|$]]|$1|P|3|J|5|6|7|19|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1A|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z|10|-4]]]]

Just try this simple query

<pre><code>select visited_date as date, COUNT(*) as count from `visitors` 
group by `visited_date` order by `visited_date` asc
</code></pre>

It will produce output as

<a href="https://i.stack.imgur.com/DElsB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DElsB.png" alt="enter image description here"></a>

It will work for you.

blocks|key|4994206|text|尝试如下：$sqlToday+=“从menjava+(Date_submitted)=CURRENT_DATE()中选择计数(*)”；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4994207|$sqlYesterday+=“从菜单中选择计数(*)，其中日期(Dc_created)=+CURDATE()+-间隔1天”；|4994208|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Try this:
$sqlToday = "Select COUNT(*) FROM menjava WHERE DATE(date_submitted)=CURRENT_DATE()";

$sqlYesterday = "Select COUNT(*) FROM menjava WHERE DATE(dc_created) = CURDATE() - INTERVAL 1 DAY";

<a href="https://i.stack.imgur.com/7IET3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7IET3.png" alt="enter image description here"></a>

I want to get the today count of users and yesterday's users count for that i want to write only one query how can i do that..?

these are my queries I want only one query:

<pre><code>SELECT COUNT(*) FROM visitors group by visited_date ORDER by visited_date DESC limit 1,1 as todayCount

SELECT COUNT(*) FROM visitors group by visited_date ORDER by visited_date DESC limit 1,0 as yesterdayCount
</code></pre>

My expected results or only 2 columns

<pre><code>todayCount yesterdayCount
 2 4
</code></pre>

how can i write as one query to get count in mysql

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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​​我想得到用户的今天计数和昨天的用户计数--我只想写一个查询--我怎么能这样做呢？这些是我的查询，我只想要一个查询：SELECT COUNT(*) FROM visitors group by visited_date ORDER by visited_date DESC limit 1,1 as todayCoun...

问我如何写成一个查询来获取mysql中的计数
EN

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问我如何写成一个查询来获取mysql中的计数EN