Question

如何让这一天变成浅红色。其他日子都是绿色的。我尝试过类似的东西，但缺少一些东西。有人可以帮忙吗？如果您阅读以下代码，您将了解我想要做什么。我错过了一部分，但我不知道哪部分。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>

Answer 1

<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在$days[$day]中添加了空格，因为它无法与之比较 '22 ' == '22'

今日日历以红色显示

1 个答案: