I'm currently working on a MySQL query, but I've been having a bit of trouble with it and am looking for help. I have been struggling with it for a while now and have searched for the solution from the Internet and from different coding forums, but, alas, still nothing.
我目前正在研究MySQL查询，但我遇到了一些麻烦，我正在寻求帮助。我一直在努力研究它并从互联网和不同的编码论坛寻找解决方案，但是，唉，仍然没有。
Basically, I have a MySQL table as follows:
基本上，我有一个MySQL表如下：
Table name: 'data'
'id' SMALLINT(5) auto_increment
'value_1' SMALLINT(5)
'value_2' SMALLINT(5)
The column 'value_2' has duplicate values and, so, I need to find the distinct values to process the results. So I created a MySQL query to do that.
列'value_2'具有重复值，因此，我需要找到不同的值来处理结果。所以我创建了一个MySQL查询来做到这一点。
SELECT DISTINCT value_1 FROM data WHERE value_2!="0"
The problem I'm having is that I want to order the results I just had by not the id nor by any other field, but by the number of rows that had those certain distinct values in that column ('value_1').
我遇到的问题是，我想订购我刚刚拥有的结果，而不是id或任何其他字段，而是通过在该列中具有那些特定不同值的行数（'value_1'）。
Could anyone please assist me?
有人可以帮助我吗？
2 个解决方案
#1
3
What you'll need is GROUP BY and COUNT.
你需要的是GROUP BY和COUNT。
SELECT `value_1`, COUNT(`value_1`) AS `count` FROM `data` WHERE `value_2` != '0' GROUP BY `value_1` ORDER BY `count`;
For more information, see: http://dev.mysql.com/doc/refman/5.1/en/counting-rows.html.
有关更多信息，请参阅：http：//dev.mysql.com/doc/refman/5.1/en/counting-rows.html。
#2
0
SELECT DISTINCT value_1, COUNT(*) FROM data GROUP BY value_1 ORDER BY 2 DESC;
#1
3
What you'll need is GROUP BY and COUNT.
你需要的是GROUP BY和COUNT。
SELECT `value_1`, COUNT(`value_1`) AS `count` FROM `data` WHERE `value_2` != '0' GROUP BY `value_1` ORDER BY `count`;
For more information, see: http://dev.mysql.com/doc/refman/5.1/en/counting-rows.html.
有关更多信息，请参阅：http：//dev.mysql.com/doc/refman/5.1/en/counting-rows.html。
#2
0
SELECT DISTINCT value_1, COUNT(*) FROM data GROUP BY value_1 ORDER BY 2 DESC;