我有以下代碼，當我單擊提交按鈕時，我想獲取引導下拉按鈕上顯示的文本的值，并且我還希望將 POST 發送到 codeigniter 控制器以對其執行某些操作. 我需要幫助。 <form method="post" action="#" id="user-order-form"> <div class="users-order col-md-5"> <label class=""> <strong>ORDER BY </strong></label> <div class="dropdown form-control" style="padding:0px; margin-left: 20px; margin-right:10px"> <button class="btn btn-default dropdown-toggle form-control" type="button" id="user-order-dropdown" name="user-order-dropdown" data-toggle="dropdown"> Surname <span class="caret"></span> </button> <ul class="dropdown-menu" aria-labelled-by="user-order-dropdown" id="capton"> <li><a href="#">Surname </a> </li> <li><a href="#">First Name </a> </li> <li><a href="#">Company Name </a> </li> </ul> </div> <label class="radio-inline"><input type="radio" name="order-radio" checked>ASC</label> <label class="radio-inline"><input type="radio" name="order-radio">DESC</label> <button type="submit" class="btn" name="order-submit-btn" id="order-submit-btn">Order</button> </div> </form>這是顯示所選下拉項的 jquery 代碼： $('.dropdown').each(function (key, dropdown) { var $dropdown = $(dropdown); $dropdown.find('.dropdown-menu a').on('click', function () { $dropdown.find('button').text($(this).text()).append(' <span class="caret"></span>'); }); });
1 回答
慕妹3242003
TA貢獻1824條經驗 獲得超6個贊
要獲取下拉列表的值：您需要創建一個隱藏的輸入字段，并且每次下拉列表的值發生變化時，您都需要更新它。
要獲取POST中的數據controller：您需要在該控制器中創建一個控制器（顯然）和一個函數。然后將該 URL 提供給表單操作屬性。
我已經為你寫了一個參考代碼，并在評論中進行了必要的解釋，看看它是否對你有幫助。
看法
<form method="post" action="<?php echo base_url('some_controller/some_function');?>" id="user-order-form">
<!-- give the action where controller name(some_controller) and function name(some_function) -->
<div class="users-order col-md-5">
<label class=""> <strong>ORDER BY </strong></label>
<div class="dropdown form-control" style="padding:0px; margin-left: 20px; margin-right:10px">
<button class="btn btn-default dropdown-toggle form-control" type="button" id="user-order-dropdown" name="user-order-dropdown" data-toggle="dropdown">
Surname
<span class="caret"></span>
</button>
<!-- make a hidden field, give it a default value(which is initially selected), give it a name(it will be used to get the post data)-->
<input type="hidden" name="dropdown" value="Surname"id="dropdown_input"/>
<ul class="dropdown-menu" aria-labelled-by="user-order-dropdown" id="capton">
<li><a href="#">Surname </a> </li>
<li><a href="#">First Name </a> </li>
<li><a href="#">Company Name </a> </li>
</ul>
</div>
<label class="radio-inline"><input type="radio" name="order-radio" checked>ASC</label>
<label class="radio-inline"><input type="radio" name="order-radio">DESC</label>
<button type="submit" class="btn" name="order-submit-btn" id="order-submit-btn">Order</button>
</div>
</form>
查詢
$('.dropdown').each(function (key, dropdown) {
var $dropdown = $(dropdown);
$dropdown.find('.dropdown-menu a').on('click', function () {
$dropdown.find('button').text($(this).text()).append(' <span class="caret"></span>');
$('#dropdown_input').val($(this).text()); // change the value of hidden input field
});
});
控制器（Some_controller.php）
function some_function(){
$dropdown = $this->input->post('dropdown'); // get dropdown value
$order_radio = $this->input->post('order-radio'); // get radio value
/* Do whatever you want to with that data here(Eg - Save in DB) -- your logic */
}
- 1 回答
- 0 關注
- 103 瀏覽
添加回答
舉報
0/150
提交
取消