單擊頁面刷新后，我需要保留下拉列表的選定值。怎么了：從下拉列表中選擇一個項目后，頁面會自行重新加載，并且下拉列表的值會轉到起始位置，盡管該值顯示在 url 的末尾。這是我的代碼：<script>function refreshPage(passValue){ window.location="index.php?cat_page=admin_post&sw_cat=sw_catn="+passValue;}</script><div class="form-group col-md-4"> <label for="inputState">Your health proplem</label> <select name="illness_id_cat" class="form-control" onchange="refreshPage(this.value);" ><?php while(mysqli_stmt_fetch($stmt)){ echo "<option value='{$illness_id_cat}'>{$illness_name_cat}</option>";}?> </select></div>此選項也不起作用：<script>var selectedItem = sessionStorage.getItem("SelectedItem"); $('#dropdown').val(selectedItem);$('#dropdown').change(function() { var dropVal = $(this).val(); sessionStorage.setItem("SelectedItem", dropVal);});</script><div class="form-group col-md-4"> <label for="inputState">Your health proplem</label> <select name="illness_id_cat" class="form-control" id="dropdown" ><?php while(mysqli_stmt_fetch($stmt)){ echo "<option value='{$illness_id_cat}'>{$illness_name_cat}</option>";}?> </select></div>這個我不知道如何實現，什么是“itemname”：<script>var text = localStorage.getItem("itemname");if(text !== null) { $("select option").filter(function() { return this.text == text; }).attr('selected', true);}</script>
1 回答
慕工程0101907
TA貢獻1887條經驗 獲得超5個贊
一種更傳統的方法是使用 PHP，并通過 window.location 命令獲取您傳遞的查詢字符串值：
<script>
function refreshPage(passValue){
window.location="index.php?cat_page=admin_post&sw_cat="+passValue;
}
</script>
<div class="form-group col-md-4">
<label for="inputState">Your health problem</label>
<select name="illness_id_cat" class="form-control" onchange="refreshPage(this.value);" >
<?php
while(mysqli_stmt_fetch($stmt)){
echo "<option";
//get the category value from the querystring and check it against the current category value in the loop. If it matches, pre-set the option as selected
if (isset($_GET["sw_cat"])) {
if ($_GET["sw_cat"] == $illness_id_cat) echo " selected";
}
echo " value='{$illness_id_cat}'>{$illness_name_cat}</option>";
}
?>
</select>
</div>
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