php - 如何从 mysql 表字段格式中选择不同的年月值 2013-01-01

Question

我有如下数据表

"id","date_occurred","country","town","quantity"
"1","2012-06-01","England","Andover","82"
"2","2012-06-01","England","Colchester","569"
"3","2012-06-01","England","Farnham","1"
"4","2012-06-01","England","England","4"
"5","2012-06-01","England","America","13"
"6","2012-06-01","America","England","114"
"7","2012-06-02","England","Andover","4"
"8","2012-06-02","England","Colchester","207"
"9","2012-06-02","America","England","14"
"10","2012-06-03","England","Andover","3"
"11","2012-06-03","England","Colchester","72"
"12","2012-06-03","England","America","1"
"13","2012-06-03","America","England","15"
"14","2012-07-04","England","Andover","1"
"15","2012-07-04","England","Colchester","309"
"16","2012-07-04","England","America","4"
"17","2012-07-04","America","England","11"
"18","2012-08-05","England","Andover","2"
"19","2012-08-05","England","Colchester","319"
"20","2012-08-05","England","Farnham","1"
"21","2012-08-05","England","America","4"
"22","2012-08-05","America","England","25"
"23","2012-08-06","England","Andover","93"
"24","2013-06-06","England","Colchester","542"
"25","2013-06-06","England","Farnham","1"
"26","2013-06-06","England","England","4"
"27","2013-06-06","England","America","7"
"28","2013-06-06","America","England","115"

当我运行下面的独立查询时，我会生成我需要的结果。

给定月份查询每个国家/地区每天的总销售额:

SELECT date_occurred, country, SUM(quantity)
FROM tableA 
WHERE YEAR(date_occurred) = 2013 AND MONTH(date_occurred) = 6
GROUP BY date_occurred, country

给定月份每个国家/地区每天的总销售额

2012-06-01 England 669
2012-06-01 America 114
2012-06-02 England 211 

我创建了一个 ajax 下拉列表来使用三个选择框查询数据库

问题

我的 Ajax 脚本产生 3 个变量
$日期_o $国家 $镇

通过 $_POST

如何单独使用变量来创建上面的查询？

我尝试过的

下面的查询使用了 $country 变量并且有效。

 SELECT date_occurred, country, SUM(quantity) AS quantity
 FROM TableA
 WHERE YEAR(date_occurred) =2013
 AND MONTH(date_occurred) = 6 AND country ='$country'
 GROUP BY date_occurred , country = '$country'

但是我已经尝试了很多排列来仅使用 $_POSTED 值来生成查询但都没有成功。

** 所需的输出如下所示。 **

 SELECT $date_o, $country, SUM(quantity) AS quantity
 FROM TableA
 WHERE YEAR($date_o) = year_part_of_$date_o
 AND MONTH($date_o) = month_part_of_$date_o AND country ='$country'
 GROUP BY $date_o , country = '$country'

我想我需要从 $date_o 变量中去除月份和年份位，但无法解决。欢迎任何简单的答案或建议!提前致谢。

Answer 1

$year = substr($date_o, 0, 4);
$month = substr($date_o, 5, 2);

如果您不需要月份中的前导零

$month = ltrim(substr($date_o, 5, 2), '0');