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MySQL按每年的每月获取行

[英]Mysql getting rows by each month in each year

Im trying to find the sum of cost for each month in each year. 我试图找出每年每个月的费用总和。

I have the below table 'logs' 我有下表“日志”

date               |    cost
---------------------------------
Wed, 01 Aug 2016   |    45
Tue, 15 Aug 2016   |    52
Mon, 31 Aug 2016   |    23
Thu, 05 Sep 2016   |    9
Sat, 10 Sep 2016   |    33
Mon, 12 Feb 2017   |    8
Tue, 31 Feb 2017   |    0
Wed, 31 Mar 2017   |    100
Fri, 31 Mar 2017   |    35
Thu, 31 Mar 2017   |    45

This is what im trying to achive, 这就是我想要达到的目标

2017 2017年

Feb 8+0 2月 8 + 0

Mar 100+35+45 三月 100 + 35 + 45


2016 2016年

Aug 45+52+23 8月 45 + 52 + 23

Sep 9+33 9月 9 + 33


Currently im doing this, 目前我正在这样做,

$sqly = "SELECT DISTINCT RIGHT(date,4) as year FROM logs ORDER BY id DESC;";
$resy = mysql_query($sqly);
while($rowy = mysql_fetch_array($resy))
{
   echo $rowy['year'];
}

Result: 结果:

2017 2017年

2016 2016年

I dont know how to proceed further to get the months on each year and sum the total for each month. 我不知道如何进一步获取每年的月份并求和每个月的总数。

This query should work: 此查询应该工作:

SELECT YEAR(STR_TO_DATE(date, '%W, %d %M %Y')) AS year,
       MONTH(STR_TO_DATE(date, '%W, %d %M %Y')) AS month, 
       SUM(cost) 
FROM logs 
GROUP BY year, month;

试试这个查询。

SELECT YEAR(date) AS year, MONTH(date) AS month, COUNT(DISTINCT id) FROM logs GROUP BY year, month
You can use below code to find your answer, Hope this will help you

<?php

$sqly = "SELECT DISTINCT RIGHT(date,4) as year FROM logs ORDER BY id DESC;";
$resy = mysql_query($sqly);
while($rowy = mysql_fetch_array($resy))
{
   echo $rowy['year'];

   $sqly2 = "select SUBSTRING(date, 9, 3) as month_name, Group_concat(cost,' + ') as total_cost from logs where RIGHT(date,4) = '".$rowy['year']."' GROUP BY SUBSTRING(date, 9, 3)";
    $resy2 = mysql_query($sqly2);

    while($rowy2 = mysql_fetch_array($resy2))
{
   echo $rowy2['month_name'] ." ".$rowy2['total_cost'] ;

}
}

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