附加< options>以< select>使用AJAX&的PHP [英] Append <options> to <select> using AJAX & PHP

问题描述

我的数据库中有两个表:

I have two tables in my database:

1..动物类型 2.品种

animal_type:

type_id  |    name
------------------------
1        |  Dog
2        |  Cat
3        |  etc...

品种

ID   |  type_id |  name 
---------------------------
1    |    1     | Labrador
2    |    1     | Putbull
3    |    2     | Persian
etc.....

在我的首页上，我想显示带有动物类型的选择下拉菜单，并根据动物类型选择品种.

On my front page I want to show select dropdown menu with animal types and breeds select based on animal type.

1.选择动物类型

<select id="animal_type">
   <option value="<?php $type->id">Dog</option>
   <option value="<?php $type->id">Cat</option>
</select>

2.如果用户选择类型无效，则选择基于类型的品种列表

<!-- user select Dog type!  fetch all dog breeds -->
<select id="breeds" disabled>
   <option value="<?php $type->id">Labrador</option>
   <option value="<?php $type->id">Pitbull</option>
</select>

所以我想基于选择的女巫类型从我的控制器加载所有品种.我试图用Ajax来解决这个问题，但对他来说并不是很好.我尝试这样做，不知道如何添加新选项来选择下拉菜单.

So I want to load all breeds from my controller based on witch type is selected. I try to slove this with ajax but am not realy good with him. I try this and dont know how to append new option to select dropdown.

脚本:

$(document).ready(function() {

   // $("#breeds").attr('disabled', true);

    // check if selected
    if($("#animal_type").find('option:selected').val() == 0) {
        $("#breeds").attr('disabled', true);
    }

    $('#animal_type').change(function(){
        // get value of selected animal type
        var selected_animal_type = $(this).find('option:selected').val();

        $.ajax({
            url : "url-to-site/index.php/account/getBreedsByTypeID/1",
            type:'POST',
            dataType: 'json',
            success: function(response) {
                $("#breeds").attr('disabled', false);

                //alert(response); // show [object, Object]

                var $select = $('#breeds');

                $select.find('option').remove();
                $.each(response,function(key, value)
                {
                    $select.append('<option value=' + key + '>' + value + '</option>'); // return empty
                });
             }
        });
    });
 });

JSON返回我的自定义品种名称

[{"breed_name":"Nema\u010dki prepeli\u010dar"},{"breed_name":"Irski vodeni \u0161panijel"},{"breed_name":"Barbe (Barbet)"},{"breed_name":"Lagoto 
My controller `mysite.com/index.php/account/getBreedsByTypeID/1`

此网址返回以下编码的JSON

This url returns the following encoded JSON

$breeds = $this->animal_breed->findAllBreedsByType($id); // Model @return array

return json_encode($breeds);

那么如何根据类型将结果附加到我的选择中呢?

So how can I append that result to my select based on type?

您能提供一个解决此问题的示例吗?谢谢.

Could you offer an example to solve this problem? Thanks.

推荐答案

$.ajax({
            url : "url-to-site/index.php/account/getBreedsByTypeID/1",
            type:'POST',
            dataType: 'json',
            success: function(response) {
                $("#breeds").attr('disabled', false);
                $.each(response,function(key, value)
                {
                    $("#breeds").append('<option value=' + key + '>' + value + '</option>');
                });
             }
        });

这篇关于附加< options>以< select>使用AJAX&的PHP的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！