如何找到明天是 "saturday" 还是不在 PHP？

Question

这是我的代码。我正在使用 strtotime 函数查找第二天是星期六。但是我的代码结果总是运行 else 部分。我哪里错了。

$current_date = date('Y-m-d');

$date = strtotime($current_date); 

$weekDay = date('w', strtotime('+1 day',$date));

if(($weekDay == 'Saturday'))
    echo "Tomorrow is Saturday.";
else
    echo "Tomorrow is not Saturday.";

Answer 1

它会给你 6 作为输出。

$current_date = date('Y-m-d');

$date = strtotime($current_date); 

$weekDay = date('w', strtotime('+1 day',$date));

if(($weekDay == 6))
    echo "Tomorrow is Saturday.";
else
    echo "Tomorrow is not Saturday.";

Answer 2

使用'l'代替"w"

$current_date = date('Y-m-d');

$date = strtotime($current_date); 

$weekDay = date('l', strtotime('+1 day',$date));
if(($weekDay == 'Saturday'))
    echo "Tomorrow is Saturday.";
else
    echo "Tomorrow is not Saturday.";

Answer 3

它会给你 6 作为输出。您可以更改“6”而不是 "saturday"。

$date = time(); //Current date 
$weekDay = date('w', strtotime('+1 day',$date));

if(($weekDay == 6)) //Check if the day is saturday or not.
    echo "Tomorrow is saturday.";
else
    echo "Tomorrow is not saturday.";

结果：

明天是星期六。

NOTE :
0 => Sunday
1 => Monday
2 => Tuesday
3 => Wednesday
4 => Thursday
5 => Friday
6 => Saturday

Answer 4

在 date() 函数中使用 w 参数有一个非常简单的解决方案

if (date('w', strtotime("+1 day")) == 6) {
 // tomorrow is saturday
}

或更多objective解决方案

if ((new DateTime())->modify('+1 day')->format('w') == 6) {
   // tomorrow is saturday
}

w - returns 星期几
6 - 代表星期六

为了缩短你可以使用三元运算符

printf("Tomorrow is%s saturday", date('w', strtotime("+1 day")) == 6 ? '' : " not");

如何找到明天是 "saturday" 还是不在 PHP？

How to find Tomorrow day is "saturday" or not in PHP?

php

strtotime