mysql实现九九乘法表

如果想完全实现需要
这个库
下面是这个库的地址。
https://blog.csdn.net/GongmissYan/article/details/102937816
其实没有也无所谓只有 一 这一列结果就行了
要是完全想先实现还是建议建库的

1.首先随便找一列1-9的行

，在加九列分别为1，2，3，4，5，6，7，8，9 这些常数会出现如下结果
代码如下

SELECT 
  `employee_id` - 99 AS 一,1,2,3,4,5,6,7,8,9
FROM
  `employees` 
ORDER BY `employee_id` 
LIMIT 0, 9 ;

然后将1，2，3，4，5，6，7，8，9

分别于 一 相乘
实现效果如下

代码如下

SELECT em.一*em.1,em.一*em.2,em.一*em.3,em.一*em.4,em.一*em.5,em.一*em.6,em.一*em.7,em.一*em.8,em.一*em.9 FROM (SELECT 
  `employee_id` - 99 AS 一,1,2,3,4,5,6,7,8,9
FROM
  `employees` 
ORDER BY `employee_id` 
LIMIT 0, 9)AS em  

三、在相乘的列前面加1，2，3，4，5，6，7，8，9

这九列

实现效果如下

代码如下：

SELECT * FROM (
SELECT em.一 AS 零,em.1,em.一*em.1 AS 一,em.2,em.一*em.2 AS 二,em.3,em.一*em.3 AS 三,em.4,em.一*em.4 AS 四,em.5,em.一*em.5 AS 五,em.6,em.一*em.6 AS 六,em.7,em.一*em.7 AS 七,em.8,em.一*em.8 AS 八,em.9,em.一*em.9 AS 九 FROM (SELECT 
  `employee_id` - 99 AS 一,1,2,3,4,5,6,7,8,9
FROM
  `employees` 
ORDER BY `employee_id` 
LIMIT 0, 9)AS em )AS em1

四使用if(expr1,expr2,expr3)将多余的去掉即可

实现如下：

代码如下：

SELECT eml.一,IF(eml.2*eml.2<=eml.二,eml.二,"  ") 二,IF(eml.3*eml.3<=eml.三,eml.三,"  ") 三,IF(eml.4*eml.4<=eml.四,eml.四,"  ") 四,IF(eml.5*eml.5<=eml.五,eml.五,"  ") 五,IF(eml.6*eml.6<=eml.六,eml.六,"  ") 六,IF(eml.7*eml.7<=eml.七,eml.七,"  ") 七,IF(eml.8*eml.8<=eml.八,eml.八,"  ") 八,IF(eml.9*eml.9<=eml.九,eml.九,"  ") 九 FROM (
SELECT em.一 AS 零,em.1,em.一*em.1 AS 一,em.2,em.一*em.2 AS 二,em.3,em.一*em.3 AS 三,em.4,em.一*em.4 AS 四,em.5,em.一*em.5 AS 五,em.6,em.一*em.6 AS 六,em.7,em.一*em.7 AS 七,em.8,em.一*em.8 AS 八,em.9,em.一*em.9 AS 九 FROM (SELECT 
  `employee_id` - 99 AS 一,1,2,3,4,5,6,7,8,9
FROM
  `employees` 
ORDER BY `employee_id` 
LIMIT 0, 9)AS em )AS eml

一下是上传的图片

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