Question

我遇到了一个无法找到解决方案的问题。我制定了一个时间表。 现在我想自动转到当天的数据。

现在当你看到它最初显示的星期一的时间表时，我希望它显示当天。

我如何轻松解决这个问题？

这是我的代码：

    <?php
$dagvanweek = $arraydag[date("w")];

$dag = $_GET['dag'];

if ($dag!="maandag" && $dag!="dinsdag" && $dag!="woensdag" && $dag!="donderdag" && $dag!="vrijdag" && $dag!="zaterdag" && $dag!="zondag")


$huidigedag = date("l");
$arraydag = array(
    "zondag",
    "maandag",
    "dinsdag",
    "woensdag",
    "donderdag",
    "vrijdag",
    "zaterdag"
    );
 $huidigedag = $arraydag[date("w")];


$header = "<hr />

Kies een dag: 
<a href='?dag=maandag'>Maandag</a> |
<a href='?dag=dinsdag'>Dinsdag</a> |
<a href='?dag=woensdag'>Woensdag</a> |
<a href='?dag=donderdag>Donderdag</a> |
<a href='?dag=vrijdag'>Vrijdag</a> |
<a href='?dag=zaterdag'>Zaterdag</a> |
<a href='?dag=zondag'>Zondag</a> 

<br />";



$get = mysqli_query($con, "SELECT count(*) AS num FROM event_rooster WHERE dag='$huidigedag' AND `naam`!=''");  
if($get->num_rows == 0) {
    echo 'Geen rijen';
}
else
{
    echo "$header <p><table align='center' cellpadding='1' cellspacing='1' width='400' border='0' cellspacing='3' cellpadding='3'>
    <tr>
    <th>Naam</td>
    <th>Evenement</td>
    <th>Tijd</td>
    </tr>";

    $q1 = mysqli_query($con, "SELECT * FROM event_rooster WHERE `dag`='$dag' AND `naam`!='' ORDER BY `id` ASC ");
    while($r=mysqli_fetch_assoc($q1))
    {

 }




        echo "<tr>
        <td>".stripslashes($r['event'])."</td>
        <td>".stripslashes($r['omschrijving'])."</td>
        <td>".stripslashes($r['tijd'])."</td>
        </tr>";
    }
echo "</table>";
}
?>

抱歉，这是用荷兰语写的。

Maandag = Monday
Dinsdag = Tuesday
Woensdag = Wednesday
Donderdag = Thursday
Vrijdag = Friday
Zaterdag = Saturday
Zondag = Sunday

Answer 1

不清楚你想要什么，但猜测：

$dag = $_GET['dag'];

if(empty($dag)){
    $dag=date("l"); //assumes you set location properly and day name is in dutch
}

自动转到当天的日期

1 个答案: