dongshuzhuo5659 2018-05-22 12:15
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今日日历以红色显示

How do I make this day light red. The other days are green. I've tried something like this but something is missing. Can someone help?
You'll understand what I want to do if you read the code below. I'm missing a part but I don't know what part. Please help me.

    <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))."
";
            $daysnumber=$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>
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1条回答 默认 最新

  • douqu8828 2018-05-22 13:20
    关注
    <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>
    

    You are makeing mistake here:

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

    added space in $days[$day] because of it not able to compare with '22 ' == '22'

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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