从日期字符串 PHP MySQL 开始的每个新年 [重复]

Question

如何在PHP 或MySQL 中创建一个结果，例如显示2005 年，并对所有结果求和

示例输入

-------------------------
Date       \    Quantiti
-------------------------
12-05-2005 \      5
-------------------------
23-08-2005 \      8
-------------------------
11-02-2006 \      4
-------------------------
26-09-2006 \      2
-------------------------

示例输出

---------------------------
year    \    Total
---------------------------
2005    \     13
---------------------------
2006    \     6
---------------------------

<?php
$sql = "SELECT * FROM mytable ORDER BY date DESC"; 
$result = $conn->query($sql);

if ($result->num_rows > 0) {

    while($row = $result->fetch_assoc()) {
        $date = $row["date"];
        $quantiti = $row["quantiti"];

        // here i need code for each new year 
    }
    } else {


    }
?>

Answer 1

您可以使用GROUP BY YEAR(date) 将年份分组，SUM() 所有具有相同年份的数量。像这样的简单查询应该可以完成这项工作

SELECT SUM(Quantiti) as `total`, 
       YEAR(date) as `year` 
FROM mytable 
GROUP BY YEAR(date)

请注意，根据您的示例输出，返回结果名为 total 和 year。

Answer 2

您可以使用 SUM 来汇总数量，使用 GROUP BY 来按年份对结果进行分组。

满足您需求的 sql 查询是：

SELECT YEAR(Date), SUM(Quantiti) 
FROM mytaple
GROUP BY YEAR(Date)

所以你的代码是：

<?php
$sql = "SELECT YEAR(Date) as y, SUM(Quantiti) as q FROM mytaple GROUP BY YEAR(Date)"; 
$result = $conn->query($sql);

if ($result->num_rows > 0) {

    while($row = $result->fetch_assoc()) {
        $date[] = $row["y"];
        $quantiti[] = $row["q"];
    }
for($i<0; $i<count($date);$i++) {
        echo $date[$i]."/".$quantiti[$i];
    }
    } else {


    }

    ?>

Answer 3

Group by With year and Month 将实现这一点

按年（日期）、月（日期）分组