PHP日期打开和关闭脚本 [英] PHP Date Open and Close Script

问题描述

我有一个任务，使用php创建一个脚本，以在正确的时间显示打开和关闭。到目前为止，我有时间正确地工作，如果在这个时间里这个业务在一周内开放了7天，这将是很好的。然而，该项目的风景是商业开放时间为上午7:00 - 下午5:30，然后星期六上午7:00至下午1:00开放，星期日休息。我以为我可以使用一个日期功能，因为显示0-6，如果

I have a task to create a script using php to display open and closed during the correct times. So far I have the time working correcty and this would be fine if the business was open during this time for 7 days a week. However the scenerio for the project is the business is open mon-fri 7:00am - 5:30 pm then open saturdays 7:00am to 1:00pm and closed sundays. I thought I could use a date function w since is displays 0-6 and call if

if($date >= 0 && $date < 6)

但是没有工作。这是我迄今为止的代码

but that didn't work. Here is the code I have so far

<?php

date_default_timezone_set('America/Chicago');
$open = "700";
$close = "1730";
$time = date('Gi');
$day = date('w');

if ($time >= $open && $time <= $close) {
    echo "We are Open";
} else {
    echo "We are closed";
}

?>

推荐答案

如果您不使用数据库，您可以硬编码每周的每一天都有一些简单易懂的格式：

If you're not using a database you can hardcode each day of the week in some easily parsable format:

$schedule[0] = "700-1730";
$schedule[1] = "700-1730";
$schedule[2] = "700-1730";
$schedule[3] = "700-1730";
$schedule[4] = "700-1730";

$schedule[5] = "700-1300";
$schedule[6] = "0";

$today = $schedule[date('w')];

list($open, $close) = explode('-', $schedule);

$now = (int) date('Gi');

$state = 'Open';

if ($today[0] == 0 || $now < (int) $today[0] || $now > (int) $today[1]) {
  $state = 'Closed';
}

只是写了代码，没有测试它。

Just wrote the code, didn't test it yet.

祝你好运！

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