Codeigniter，如何以 json 格式显示选择查询结果？答案

【问题标题】：Codeigniter, How to display select query result in json format?Codeigniter，如何以 json 格式显示选择查询结果？
【发布时间】：2017-01-07 12:00:43
【问题描述】：

我有两张桌子表 1：dr_country

table2：dr_city

我必须从两个表中获取数据。我所需的输出格式应如下所示：

[{"id":"1","country_name":"Australia","country_code":"61","iso_code":"AUS","city":
[{"id":"1","city_name":"sydney"},{"id":"2","city_name":"melbourne"},{"id":"3","city_name":"perth"},{"id":"4","city_name":"brisbane"}]},
{"id":"2","country_name":"Bangladesh","country_code":"880","iso_code":"BGD","city":
[{"id":"5","city_name":"dhaka"},{"id":"6","city_name":"chittagong"}]}]

我的模型方法如下：

public function countryAction()
{
    $this->db->select("*");
    $this->db->from('dr_country');
    $this->db->join('dr_city', 'dr_country.id = dr_city.country_id','left');

    $result = $this->db->get()->result_array();
   if($result)
   {
        print_r(json_encode($result));

    }
    else
    {
         $detail = array(
            'status'=>'unsucess',   
           );

          echo  json_encode($detail); 
    }
}

它产生如下输出：

[{"id":"1","country_name":"Australia","country_code":"61","iso_code":"AUS","city_name":"sydney","country_id":"1"},{"id":"2","country_name":"Australia","country_code":"61","iso_code":"AUS","city_name":"melbourne","country_id":"1"},{"id":"3","country_name":"Australia","country_code":"61","iso_code":"AUS","city_name":"perth","country_id":"1"},{"id":"4","country_name":"Australia","country_code":"61","iso_code":"AUS","city_name":"brisbane","country_id":"1"},{"id":"5","country_name":"Bangladesh","country_code":"880","iso_code":"BGD","city_name":"dhaka","country_id":"2"},{"id":"6","country_name":"Bangladesh","country_code":"880","iso_code":"BGD","city_name":"chittagong","country_id":"2"}]

因此，为了获得所需的 json 格式，我应该在模型函数中进行哪些更改，我是新手，提前致谢。

【问题讨论】：

dr_country.country_name,dr_city.city_name 分组

标签： php mysql json codeigniter

【解决方案1】：

在你的模型中试试这个

$countries = $this->db->get('dr_country')->result();
foreach ($countries as $key => $country) {
    $country->city = $this->db->get_where('dr_city', array('country_id' => $country->id))->result();
}
print_r(json_encode($countries));

【讨论】：

print_r(json_encode($country));我只为所有国家/地区获得一个城市（相同的悉尼）
[{"id":"1","country_name":"Australia","country_code":"61","iso_code":"AUS","city":[{"city_id ":"1","city_name":"sydney","country_id":"1"}]},{"id":"2","country_name":"Bangladesh","country_code":"880", "iso_code":"BGD","city":[{"city_id":"1","city_name":"sydney","country_id":"1"}]}] 这就是我得到的 city(sydney ) 对所有重复

【解决方案2】：

请尝试以下步骤：

获取所有国家 - 从 dr_country 中选择 *
遍历国家并获取每个国家/地区的城市

foreach($countries as $c) { $countries[$city_id] = select * from dr_city where country_id = $c; }
1. json 编码

【讨论】：

【解决方案3】：

您的预期输出永远不会与直接数据库一起出现。因为数据库以 TABLE 格式提供输出。

如果您想要您的预期输出，那么您应该执行 2 个 SQL 查询。 i) 获取国家 ii) 获取城市并制作一个国家数组的foreach循环并查找与国家相关的所有城市并制作输出数组。

【讨论】：

【解决方案4】：

$countries = $this->db->get('dr_country')->result();
foreach ($countries as $key => $country) {
    $country->city = $this->db->get_where('dr_city', array('country_id' => 
$country->id))->result();
    $countryfinal[]=$country;
}
print_r(json_encode($countryfinal));

【讨论】：

【解决方案5】：

    $countries = $this->db->get('dr_country')->result();
foreach ($countries as $key => $country) {
    $country->city = $this->db->get_where('dr_city', array('country_id' => 
$country->id))->result();
    $countryfinal[]=$country;
}
print_r(json_encode($countryfinal));

【讨论】：