Question

我想从网络服务中提取一条记录并在div中很好地显示它，就像在this example中一样。

当有人选择“天数”时，它应该过滤结果。

这是我到目前为止所做的：

<script>
function itinerary(name)
{
    var last = document.getElementById(selected);
    selected = name;
    var current = document.getElementById(selected);

    last.style.display = "none";
    current.style.display = "";

    document.getElementById("selectItinerary").blur();
    }
</script>
<?php
    ini_set("display_errors", 1);

    $days = "http://services.gerbertours.com/tours.asmx/Days?type=2&cityID=1146";
    $ch1 = curl_init($days); 
    curl_setopt($ch1, CURLOPT_HEADER, 0); 
    curl_setopt($ch1, CURLOPT_BINARYTRANSFER, 1);
    curl_setopt($ch1, CURLOPT_RETURNTRANSFER, 1);       // make sure we get the response back 
    $xml1 = curl_exec($ch1);        // execute the post 
    curl_close($ch1);               // close our session 
    $var1 = simplexml_load_string($xml1);
    $configData1 = json_decode($var1, true);

    $days = 0;
    foreach ($configData1 as $key=>$val) {
        if ($val > $days) {
            $days = $val;
        }
    }
    //echo $days;

    //$days = "http://services.gerbertours.com/tours.asmx/Days?type=2&cityID=1146";
    $url = "http://services.gerbertours.com/tours.asmx/Sample?type=2&CityID=1146&days=$days";
    $ch = curl_init($url); 
    curl_setopt($ch, CURLOPT_HEADER, 0); 
    curl_setopt($ch, CURLOPT_BINARYTRANSFER, 1);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);        // make sure we get the response back 
    $xml = curl_exec($ch);      // execute the post 
    curl_close($ch);                // close our session 
    $var = simplexml_load_string($xml);

    $configData = json_decode($var, true);

echo '<pre>';
print_r($configData);
echo '</pre>';

foreach ($configData as $row) {
    echo 'Day:' . $row['Day'].'<br>';
    echo 'Time:' .$row['Time'].'<br>';
    echo 'Description:' .$row['Description'].'<br>';
}

?>
    <select id="selectItinerary" onchange="itinerary(this.options[this.selectedIndex].value); this.blur" style="font-size:11px">

<?php
    for ($day = 1; $day <= $days; $day++) {
?>
    <option value="<?php echo $day; ?>"><?php echo $day; ?> Day Tour</option>
<?php

    }
?>
    </select>

<?php
    $dayCount = 1;
    foreach ($configData as $row) {
        $currentDay = $row['Day'];
        if($dayCount==$days);break;
?>

        <div id="<?php echo $dayCount; ?>" class="itinerary">
            <div class="day">
                <?php echo 'Day:' . $row['Day']; ?>
            </div>

            <div class="time">
                <?php echo 'Time:' .$row['Time']; ?>
            </div>

            <div class="description">
                <?php echo 'Description:' .$row['Description']; ?>
            </div>
        </div>

    <?php 
        $dayCount++;
    }
?>

其他一切都没问题。我需要做的就是根据用户选择过滤结果，如果有人选择“3日游”，它应该显示结果中的3个项目。我怎样才能做到这一点？

Answer 1

如果您不使用Jquery，则可以使用以下方式显示/隐藏元素：

document.getElementById('elementId').style.display = 'block'; // to show
document.getElementById('elementId').style.display = 'none';

想法是隐藏所有元素，然后当onchange事件触发时，显示要显示的元素。因此，在您的代码中，您只需要更改

current.style.display = "";

到

current.style.display = "block";

我之前也把它们放在一起，所以你在显示所需的div之前隐藏了你需要隐藏的所有内容：

document.getElementById('1').style.display = 'none';
document.getElementById('2').style.display = 'none';
document.getElementById('3').style.display = 'none';

但是，当然，对于大型项目，我建议使用Jquery或任何Javascript框架来简化操作。查看：Jquery dynamic hide and show for drop down menu

php Foreach只运行有限的时间并在正确的div中插入数据

1 个答案: