努力使用以下不起作用的代碼（$ThisYear 引數）
$ThisYear = date("Y");
$ThisYear = strtotime($ThisYear);
$sql2 = 'SELECT
distinct
DATE_FORMAT(TransDate, "%M %Y") as MonthYear
, DATE_FORMAT(TransDate, "%Y") as TransYear
, DATE_FORMAT(TransDate, "%m") as TransMonth
from Leave
where YEAR(TransDate) = **"$ThisYear"**
group by DATE_FORMAT(TransDate, "%M %Y")
order BY YEAR(TransDate), MONTH(TransDate) desc';
與之相比，這是有效的：
$ThisYear = date("Y");
$sql2 = 'SELECT
distinct
DATE_FORMAT(TransDate, "%M %Y") as MonthYear
, DATE_FORMAT(TransDate, "%Y") as TransYear
, DATE_FORMAT(TransDate, "%m") as TransMonth
from Leave
where YEAR(TransDate) = **2022**
group by DATE_FORMAT(TransDate, "%M %Y")
order BY YEAR(TransDate), MONTH(TransDate) desc';
uj5u.com熱心網友回復：
根據 PHP 檔案，strtotime()將任何英文文本日期時間描述決議為Unix timestamp。
為了：
- 您的變數已轉換為數字（2022 年的 1666484520 Unix 時間戳）。
- 服務器將年份數與您的變數 = 1666484520 進行比較。
這是正確的查詢和變數。
$ThisYear = date("Y");
$sql2 = 'SELECT
distinct
DATE_FORMAT(TransDate, "%M %Y") as MonthYear
, DATE_FORMAT(TransDate, "%Y") as TransYear
, DATE_FORMAT(TransDate, "%m") as TransMonth
from Leave
where YEAR(TransDate) = $ThisYear
group by DATE_FORMAT(TransDate, "%M %Y")
order BY YEAR(TransDate), MONTH(TransDate) desc';
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標籤：phpmysql日期
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