我的演示数据库中有 2 个表,其中我根据 2 个键连接它们,在这里我想计算匹配结果的不同值,
我的第一张 table
MariaDB [demos]> select * from main_info;
+------+------+-------+-------+----------+
| key1 | key2 | info1 | info2 | date |
+------+------+-------+-------+----------+
| 1 | 1 | 15 | 90 | 20120501 |
| 1 | 2 | 14 | 92 | 20120601 |
| 1 | 3 | 15 | 82 | 20120801 |
| 2 | 1 | 17 | 90 | 20130302 |
| 2 | 2 | 16 | 88 | 20130601 |
+------+------+-------+-------+----------+
5 rows in set (0.00 sec)
还有我的第二张 table
MariaDB [demos]> select * from product1;
+------+------+--------+--------------+
| key1 | key2 | serial | product_data |
+------+------+--------+--------------+
| 1 | 1 | 0 | NaN |
| 1 | 1 | 1 | NaN |
| 1 | 1 | 2 | NaN |
| 1 | 1 | 3 | NaN |
| 1 | 2 | 0 | 12.556 |
| 1 | 2 | 1 | 13.335 |
| 1 | 3 | 1 | NaN |
| 1 | 3 | 2 | 13.556 |
| 1 | 3 | 3 | 14.556 |
| 2 | 1 | 0 | 12.556 |
| 2 | 1 | 1 | 13.553 |
| 2 | 1 | 2 | NaN |
+------+------+--------+--------------+
12 rows in set (0.00 sec)
因此字段序列的不同计数如下,其中序列不为零
MariaDB [demos]> select count(distinct a.key1,a.key2) as serial_count from main_info a,product1 b where a.key1=b.key1 and a.key2=b.key2 and b.serial !=0;
+--------------+
| serial_count |
+--------------+
| 4 |
+--------------+
1 row in set (0.00 sec)
字段product_data的不同计数如下,其中product_data不为NaN
MariaDB [demos]> select count(distinct a.key1,a.key2) as product_count from main_info a,product1 b where a.key1=b.key1 and a.key2=b.key2 and b.product_data !='NaN';
+---------------+
| product_count |
+---------------+
| 3 |
+---------------+
1 row in set (0.00 sec)
在我原来的应用程序中,我的表中有很多列,最后我想使用单个查询得到如下结果,目前我正在使用 PHP 进行多个查询,这需要很长时间,因为例如如果我有 100 列,我必须像上面一样执行函数 100 次,所以请有人指导我加快我的应用程序
+--------------+---------------+
| serial_count |product_count |
+--------------+---------------+
| 4 | 3 |
+--------------+---------------+
以下是表格结构
DROP TABLE IF EXISTS `main_info`;
CREATE TABLE `main_info` (
`key1` int(11) NOT NULL,
`key2` int(11) NOT NULL,
`info1` int(11) NOT NULL,
`info2` int(11) NOT NULL,
`date` int(11) NOT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
LOCK TABLES `main_info` WRITE;
INSERT INTO `main_info` VALUES (1,1,15,90,20120501),(1,2,14,92,20120601),(1,3,15,82,20120801),(2,1,17,90,20130302),(2,2,16,88,20130601);
UNLOCK TABLES;
DROP TABLE IF EXISTS `product1`;
CREATE TABLE `product1` (
`key1` int(11) NOT NULL,
`key2` int(11) NOT NULL,
`serial` int(11) NOT NULL,
`product_data` varchar(1000) DEFAULT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
LOCK TABLES `product1` WRITE;
INSERT INTO `product1` VALUES (1,1,0,'NaN'),(1,1,1,'NaN'),(1,1,2,'NaN'),(1,1,3,'NaN'),(1,2,0,'12.556'),(1,2,1,'13.335'),(1,3,1,'NaN'),(1,3,2,'13.556'),(1,3,3,'14.556'),(2,1,0,'12.556'),(2,1,1,'13.553'),(2,1,2,'NaN');
UNLOCK TABLES;
如果我在终端上运行,我不明白为什么日期类型会改变
$ mysql -u root -p demos -e 'select key1,if(key1 !=0,key1,999.999) as `test1` from main_info'
Enter password:
+------+-------+
| key1 | test1 |
+------+-------+
| 1 | 1.000 |
| 1 | 1.000 |
| 1 | 1.000 |
| 2 | 2.000 |
| 2 | 2.000 |
+------+-------+
如果IF
条件为真,我希望它应该是整数
最佳答案
您可以使用条件聚合在单个查询中完成此操作:
select count(distinct if(b.product_data !='NaN',a.key1, null),
if(b.product_data !='NaN',a.key2, null)) as product_count,
count(distinct if(b.serial !=0,a.key1, null),
if(b.serial !=0,a.key2, null)) as serial_count
from main_info a
inner join product1 b on a.key1=b.key1 and a.key2=b.key2
输出:
product_count serial_count
-----------------------------
3 4
编辑:归功于@Paul Spiegel
可以使用CONCAT
简化查询:
select count(distinct if(b.product_data !='NaN',
CONCAT(a.key1, ',', a.key2),
null)) as product_count,
count(distinct if(b.serial !=0,
CONCAT(a.key1, ',', a.key2),
null)) as serial_count
from main_info a
inner join product1 b on a.key1=b.key1 and a.key2=b.key2
关于php - 如何使用最佳方法获得一行中多个字段的不同计数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35526601/