問題描述
視圖:learning_view.php
view : learning_view.php
這是我從數據庫中填充的第一個下拉列表.
Here is the first dropdown which I am populating from database.
<select name = 'category' id = 'category'>
<option value="">-- Select Category --</option>
<?php foreach($category as $item){ ?>
<option value="<?php echo $item->id_cat; ?>"><?php echo $item->name; ?></option>
<?php } ?>
</select>
<br><br>
我想要的是填充另一個依賴于第一個下拉列表的下拉列表.為此,我使用了 jQuery ajax 帖子.
What I want is to populate another dropdown which is dependent on the first dropdown. For that I have used the jQuery ajax post.
第二個下拉列表:
<select name = 'type' id = 'type'>
<option value="">-- Select Type --</option>
<?php foreach($type as $item){ ?>
<option value="<?php echo $item->id_type; ?>"><?php echo $item->name; ?></option>
<?php } ?>
</select>
<br><br>
ajax 帖子:
jQuery(document).ready(function(){
$("#category").change(function() {
var category_id = {"category_id" : $('#category').val()};
console.log(category_id);
$.ajax({
type: "POST",
data: category_id,
url: "<?= base_url() ?>learning/dependent_dropdown",
success: function(data){
$.each(data, function(i, data){
console.log(data.name);
console.log(data.id_type)
});
}
});
});
});
控制器:learning.php
controller : learning.php
public function dependent_dropdown()
{
if(isset($_POST['category_id']))
{
$this->output
->set_content_type("application/json")
->set_output(json_encode($this->learning_model->getType($_POST['category_id'])));
}
}
數據來自我檢查過的ajax帖子后的數據庫
The data is coming from the database after ajax post which I checked by
console.log(data.name);
console.log(data.id_type)
在控制臺中.
但無法弄清楚如何使用我視圖的第二個下拉列表中的數據.
But couldn't able to figure out how to use the data in the second dropdown of my view.
我的意思是如何使用我在 ajax 發布后收到的數據填充第二個下拉列表.
I mean how can i populate the second dropdown with the data i have received after ajax post.
推薦答案
我通過修改ajax帖子的success函數找到了解決我的問題的方法:
I found a solution to my problem by modifying the success function of the ajax post:
success: function(data) {
$.each(data, function(i, data) {
$('#type').append("<option value='" + data.id_type + "'>" + data.name + "</option>");
});
}
將值附加到下拉列表中.
Which append the value into the drop down.
<select name="type" id="type">
<option value="">-- Select Type --</option>
</select>
我只是在ajax帖子的success函數中給了select塊的id,并附加了值.它可以工作,但有一個問題,即當我更改第一個下拉菜單的選擇時會出現新值,但為上一個選擇顯示的值并沒有消失.
I just gave the id of the select block into the success function of the ajax post and appended the value. It works but there is a problem which is when I change the selection of the first dropdown new value appears but the values which were showing for the previous selection doesn't go away.
這篇關于codeigniter - 依賴于 jquery 和 ajax post 的下拉列表的文章就介紹到這了,希望我們推薦的答案對大家有所幫助,也希望大家多多支持html5模板網!