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php - 日历今天日期为红色

问题描述

我如何让这一天变成红色。其他日子是绿色的。我已经尝试过这样的事情，但缺少一些东西。有人可以帮忙吗？
如果您阅读下面的代码，您就会明白我想要做什么。我错过了一个部分，但我不知道是什么部分。请帮我。

    <table border="1">
        <?php
        date_default_timezone_set('Europe/Stockholm');
        echo"<tr>";
        $week_number = 21;
        $year = 2018;

        if($week_number < 10){
           $week_number = "0".$week_number;
        }

        for($day=1; $day<=7; $day++)
        {
            $days[$day] = date('d', strtotime($year."W".$week_number.$day))."\n";
            $daysnumber=$days[$day];
            $today=date('d');
        if ($daysnumber>=$today){

        echo"<td bgcolor='red'>$daysnumber</td>" ;
        }
        else

        {


        if ($daysnumber<=$today){

        echo"<td bgcolor='green'>$daysnumber</td>" ;
        }
        else

        {


    }

}}
?>
</table>

标签： php

解决方案

<table border="1">
    <?php
        date_default_timezone_set('Europe/Stockholm');
        echo"<tr>";
        $week_number = 21;
        $year = 2018;

        if($week_number < 10):
           $week_number = "0".$week_number;
        endif;

        for($day = 1; $day <= 7; $day++):

            $daysnumber = date('d', strtotime($year."W".$week_number.$day));
            $today = date('d');

            if($daysnumber == $today):
                echo "<td bgcolor='red'>$daysnumber</td>";
            else:
                echo "<td bgcolor='green'>$daysnumber</td>" ;
            endif;


        endfor;
    ?>
</table>

你在这里犯了错误：

$days[$day] = date('d', strtotime($year."W".$week_number.$day))."\n";

\n添加空间，$days[$day]因为它无法与 '22 ' == '22'

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