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MYSQL选择PHP_Php_Mysql_Join_Sum - Fatal编程技术网

MYSQL选择PHP

MYSQL选择PHP,php,mysql,join,sum,Php,Mysql,Join,Sum,我有两张表,我想从销售员那里列出每天的结果 一个表包含销售人员信息(id、姓名) 另一个包含当天的销售额(saleschipcell,cell) 加入两个表格后,我每天都会得到结果: salesman John, day -> 01/03/2012 cell -> 15, chip cell ->30 salesman Bob, day -> 01/03/2012 cell -> 5, chip cell ->10 salesman John,

我有两张表,我想从销售员那里列出每天的结果

一个表包含销售人员信息(id、姓名)

另一个包含当天的销售额(saleschipcell,cell)

加入两个表格后,我每天都会得到结果:

salesman John, day -> 01/03/2012   cell -> 15, chip cell ->30

salesman Bob, day  -> 01/03/2012   cell -> 5, chip cell ->10

salesman John, day -> 01/04/2012   cell -> 10, chip cell ->0

salesman Bob, day  -> 01/04/2012   cell -> 10, chip cell ->2
好的。有没有可能像这样从约翰和鲍勃那里得到cell和chipcell的总销售额:

Total of the day , day  -> 01/03/2012   cell -> 20, chip cell ->40

Total of the day , day  -> 01/04/2012   cell -> 20, chip cell ->2

我尝试了
SUM
,但只生成了一个结果,但是SUM从所有销售人员处获取所有结果。

您可以对每个销售人员的结果进行分组,以获取特定人员的总数。例如:

SELECT `day`, SUM(`saleschipcell`) FROM `sales` GROUP BY `salesman_id`;

应该给你每个销售员每天的销售额。

如果你想按天和按人列出,请在“分组依据”中使用两列

SELECT `day`,SUM('cell'), SUM(`chipcell`) FROM `sales` GROUP BY `day`;
SELECT `day`,`salesman_id`,SUM('cell') as cell, SUM(`chipcell`) as chipcell 
FROM `sales` 
GROUP BY `day`, `salesman_id`
ORDER BY `day`

这就是“分组依据”的作用。您可以发布您正在使用的实际查询吗?另请参见选择promotor.id作为promotor id,promotor.promotor作为promotor名称,produtosales.idpromotor作为salepromoto,prosutosales.datesales作为datasell,produtosales.combos作为salescombos从promotor内部加入produtosales作为promotor.id=produtosales.idpromotor,其中promotor.idpddv='33'按promotor.id排序,ProAutoSales.datesales ascI我想从John和bob那里得到一天的总数,我指的是每天不同的芯片销售总数和单元格总数使用您的查询按天分组