在下拉列表更改时重新加载页面并传递该值

Question

我的页面中有一个下拉列表.当我从下拉列表中选择一个选项时,页面应该重新加载并返回到同一页面并且应该传递该下拉列表的值
下面是下拉列表的php代码

<select style="width: 200px;" name="location" onchange="window.location='index.php?id=' + this.value;">
    <option value="All">All</option>
    <option value="Noida Sector 1">Noida Sector 1</option>
    <option value="Noida Sector 2">Noida Sector 2</option>
    <option value="Noida Sector 3">Noida Sector 3</option>
    <option value="Noida Sector 4">Noida Sector 4</option>
    <option value="Noida Sector 5">Noida Sector 5</option>
    <option value="Noida Sector 6">Noida Sector 6</option>
    <option value="Noida Sector 7">Noida Sector 7</option>
</select>

<?php $location=$_POST['id']; ?>

我试图将值传递给$ location.

Answer 1

只需更改您的PHP代码,此代码输出选择选项并突出显示所选值(如下所示)

    <select style="width: 200px;" id="myselect" name="location" onchange="window.location='index.php?id='+this.value+'&pos='+this.selectedIndex;">
        <option value="All">All</option>
        <option value="Noida Sector 1">Noida Sector 1</option>
        <option value="Noida Sector 2">Noida Sector 2</option>
        <option value="Noida Sector 3">Noida Sector 3</option>
        <option value="Noida Sector 4">Noida Sector 4</option>
        <option value="Noida Sector 5">Noida Sector 5</option>
        <option value="Noida Sector 6">Noida Sector 6</option>
        <option value="Noida Sector 7">Noida Sector 7</option>
    </select>

    <?php
    if(isset($_GET['id']))
    {
        $location=$_GET['id'];
        echo $location;
    ?>
    <script>
        var myselect = document.getElementById("myselect");
        myselect.options.selectedIndex = <?php echo $_GET["pos"]; ?>
    </script>
    <?php
    }
    ?>