Question

我正在编写一个PHP程序，它将获取当前工作周（周一除外）的所有日期。我的代码如下：

//Get Year
$sql = "SELECT YEAR(CURDATE()) AS CurYear";
$result = mysqli_query($db, $sql);
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
$curyear = $row['CurYear'];
//Get Week
$sql = "SELECT WEEK(CURDATE()) AS CurWeek";
$result = mysqli_query($db, $sql);
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
$curweek = $row['CurWeek'];
//Get Date of Tuesday
$sql = "SELECT STR_TO_DATE('$curyear$curweek Tuesday', '%X%V %W') AS TueDate";
$result = mysqli_query($db, $sql);
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
$tuedate = $row['TueDate'];

它有效，并将返回本周二的日期，但有更有效的方法这样做，而且在接下来的3天内做这个吗？

Answer 1

如此处所述：How to find the day of week from a date using PHP? 你可以使用date命令：

$time = time(); // present time
$day  = 3; //the day of the week we are looking for 0: sunday, 1 monday and so on
$dayofweek = date('w', $time); //current day week number
$result  = date('Y-m-d', strtotime(($day - $dayofweek).' day', $time));

Answer 2

我不确定您是否需要使用MySQL，但我会直接在PHP中执行此操作。以下是您如何处理它的示例。设置变量$TuesTS后，您可以通过向其添加(24*60*60)或86400来确定第二天，这是一天中的秒数。

<?php
    $ts = time();
    $todayNum = date('N');
    if ( ($todayNum >= 2) && ($todayNum<=5)){
        $offset = $todayNum - 2; // number of days after Tuesday;
        $TuesTS = $ts-($offset*24*60*60);

        echo "Tuesday : ".date('Y-m-d', $TuesTS);
    }
?>

MYSQL - 选择当前工作周的日期

2 个答案: