在 Select 标签上选择一个选项并在重新加载/刷新时将其保留在那里

Question

Select我正在尝试在您选择并单击其中的位置后创建一个标签option，它将加载到标签href所option具有的值。加载后，option您选择的将在标签上突出显示select。

到目前为止，从我搜索过的内容来看，我只有：

HTML:

    <select id="jumpid" onchange="jump(this)" >

        <option value="index.php?ID=1">Val 1</option>

        <option value="index.php?ID=2">Val 2</option>

        <option value="index.php?ID=3">Val 3</option>

        <option value="index.php?ID=4">Val 4</option>

        <option value="index.php?ID=5">Val 5</option>

        <option value="index.php?ID=6">Val 6</option>

        <option value="index.php?ID=7">Val 7</option>

    </select>

记者：

<script>

    function jump(selectedLink){

    var chpSelect = selectedLink.value;

    location.href=document.getElementById("jumpid").value;

}

</script>

例如，当我选择某些内容时Val 3，浏览器将重新加载到index.php?ID=3它的内容。但select标签仍然显示Val 1，我希望它留在Val 3。

Answer 1

它很粗糙，但它会起作用：<?php$id=0;if(isset($_GET['ID']){&nbsp; &nbsp; $id = $_GET['ID'];}?>&nbsp; &nbsp;&nbsp;<select id="jumpid" onchange="jump(this)" >&nbsp; &nbsp; &nbsp; &nbsp; <option value="index.php?ID=1" <?php if($id==1){echo "selected";}?>>Val 1</option>&nbsp; &nbsp; &nbsp; &nbsp; <option value="index.php?ID=2" <?php if($id==2){echo "selected";}?>>Val 2</option>&nbsp; &nbsp; &nbsp; &nbsp; <option value="index.php?ID=3" <?php if($id==3){echo "selected";}?>>Val 3</option>&nbsp; &nbsp; &nbsp; &nbsp; <option value="index.php?ID=4" <?php if($id==4){echo "selected";}?>>Val 4</option>&nbsp; &nbsp; &nbsp; &nbsp; <option value="index.php?ID=5" <?php if($id==5){echo "selected";}?>>Val 5</option>&nbsp; &nbsp; &nbsp; &nbsp; <option value="index.php?ID=6" <?php if($id==6){echo "selected";}?>>Val 6</option>&nbsp; &nbsp; &nbsp; &nbsp; <option value="index.php?ID=7" <?php if($id==7){echo "selected";}?>>Val 7</option>&nbsp; &nbsp; </select>