blocks|key|686782|text|这是GROUP+BY和GROUP_CONCAT()的工作。试试这样的东西。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|686783|(https://www.db-fiddle.com/f/4sVVgGk6SyEnFE4C3emMp9/0)|686784|SELECT+user_id,
+++++++GROUP_CONCAT(+CONCAT(post_id,+'(',+point,+')')+ORDER+BY+post_id+SEPARATOR+'+%7C+'+),
+++++++SUM(point)+
++FROM+tbl
+GROUP+BY+user_id|code-block|syntax|javascript|686785|entityMap|0|LINK|mutability|MUTABLE|url|https://dev.mysql.com/doc/refman/8.0/en/group-by-functions.html#function_group-concat|1|https://www.db-fiddle.com/f/4sVVgGk6SyEnFE4C3emMp9/0^0|2|8|B|E|B|E|0|0|1|1G|1|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|6|7|14|8|@]|D|@$9|15|A|16|1|17]]|E|$]]|$1|H|3|I|5|J|7|18|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]]]

This is a job for <code>GROUP BY</code> and <a href="https://dev.mysql.com/doc/refman/8.0/en/group-by-functions.html#function_group-concat" rel="nofollow noreferrer"><code>GROUP_CONCAT()</code></a>. Try something like this.

(<a href="https://www.db-fiddle.com/f/4sVVgGk6SyEnFE4C3emMp9/0" rel="nofollow noreferrer">https://www.db-fiddle.com/f/4sVVgGk6SyEnFE4C3emMp9/0</a>)

<pre><code>SELECT user_id,
 GROUP_CONCAT( CONCAT(post_id, '(', point, ')') ORDER BY post_id SEPARATOR ' | ' ),
 SUM(point) 
 FROM tbl
 GROUP BY user_id
</code></pre>

blocks|key|3387579|text|运行一个查询，在其中GROUP+BY+user_id。使用GROUP_CONCAT()和CONCAT()生成点/帖子列。CONCAT()将字符串组合在一起形成postID+(points)格式，GROUP_CONCAT()将按该组分组的所有值(在本例中为user_id)放在一个字符串中，该字符串由分隔符%7C分隔。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3387580|SELECT+user_id,+
+++++++GROUP_CONCAT(CONCAT(post_id,+'(',+points,+')')++SEPARATOR+'+%7C+')+as+point_id_posts,
+++++++SUM(points)+as+totalPoint
FROM+table
GROUP+BY+user_id|code-block|syntax|javascript|3387581|https://www.db-fiddle.com/f/aGHL49ydsUX3LnyAfut4vx/0现场演示|unordered-list-item|3387582|entityMap|0|LINK|mutability|MUTABLE|url|https://www.db-fiddle.com/f/aGHL49ydsUX3LnyAfut4vx/0^0|A|8|J|7|T|E|18|8|1O|8|27|F|2P|E|3J|7|48|1|0|0|0|1G|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|C]|$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1E|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|M|7|1F|8|@]|D|@$9|1G|A|1H|1|1I]]|E|$]]|$1|N|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|O|$P|$5|Q|R|S|E|$T|U]]]]

Run a query where you <code>GROUP BY</code> the <code>user_id</code>. Use <code>GROUP_CONCAT()</code> with <code>CONCAT()</code> to generate your points/posts column. <code>CONCAT()</code> puts together the string to form the <code>postID (points)</code> format, <code>GROUP_CONCAT()</code> puts all the values grouped by that group (in this case the <code>user_id</code>) into one string separated by the separator <code>|</code>.

<pre><code>SELECT user_id, 
 GROUP_CONCAT(CONCAT(post_id, '(', points, ')') SEPARATOR ' | ') as point_id_posts,
 SUM(points) as totalPoint
FROM table
GROUP BY user_id
</code></pre>

<ul>
<li>Live demo at <a href="https://www.db-fiddle.com/f/aGHL49ydsUX3LnyAfut4vx/0" rel="nofollow noreferrer">https://www.db-fiddle.com/f/aGHL49ydsUX3LnyAfut4vx/0</a></li>
</ul>

blocks|key|3387627|text|我想这会对你有帮助|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3387628|SELECT+user_id,GROUP_CONCAT(CONCAT(post_id,'(',point,')')+SEPERATOR+'+%7C+')+as+post_points,SUM(point)+as+total_points+FROM+table_name+GROUP+BY+user_id|code-block|syntax|javascript|3387629|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|$]]

I think This will help you

<pre><code>SELECT user_id,GROUP_CONCAT(CONCAT(post_id,'(',point,')') SEPERATOR ' | ') as post_points,SUM(point) as total_points FROM table_name GROUP BY user_id
</code></pre>

blocks|key|686820|text|查询以实现您的请求|type|unstyled|depth|inlineStyleRanges|entityRanges|data|686821|SELECT+user_id,
+++++++GROUP_CONCAT(+CONCAT(post_id,+'(',+point,+')')+ORDER+BY+post_id+SEPARATOR+'+%7C+'+)+as+post_id_point,
+++++++SUM(point)+
++FROM+tbl
+GROUP+BY+user_id|code-block|syntax|javascript|686822|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|$]]

Query to achieve your requirment

<pre><code>SELECT user_id,
 GROUP_CONCAT( CONCAT(post_id, '(', point, ')') ORDER BY post_id SEPARATOR ' | ' ) as post_id_point,
 SUM(point) 
 FROM tbl
 GROUP BY user_id
</code></pre>

I have a table sql like this

<pre><code>post_id point user_id
5 1.0 1
6 1.5 1
7 0.0 2
8 1.5 3
9 1.0 3
...
</code></pre>

I want to print the data into this ( group the posts by user_id )

<pre><code>user_id post_id(point) total point
1 5(1.0) | 6(1.5) | ... 2.5
2 7(0.0) | | ... 0.0
3 8(1.5) | 9(1.5) | ... 3.0
...
</code></pre>

I tried something like this, but it returned array

<pre><code>SELECT * FROM table WHERE post_id = ?
</code></pre>

I'm sorry if this a silly question but I'm newbie in MySQL

Group data from table in PHP

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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我有一个像这样的表格sqlpost_id    point    user_id5          1.0      16          1.5      17          0.0      28          1.5      39          1.0      3...我想把数据打印到这个(按...

问PHP中表中的数据分组
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