简体   繁体   English

一对多关系codeigniter / datamapper

[英]One To Many relationship codeigniter/datamapper

I need help on this plz : 我需要这个plz的帮助:

2 tables 2桌

Service : has many Sub_Service Sub_Service : has one Service 服务:有很多Sub_Service Sub_Service:有一个Service

My code : 我的代码:

    $services = new Service();
    $services->where('org_id', $org_id)->get();
    $services->sub_service->get();

    foreach($services as $service)
    {
        echo $service->title;

        foreach($service->sub_service as $sub_service)
        {
                  echo $sub_service->title;
        }
    } 

But it doesn't work, if I want to access sub_service i have to take it out of the first loop and do somthing like 但这是行不通的,如果我想访问sub_service,我必须将其从第一个循环中取出并做类似的事情

foreach($services->sub_service as $sub_service)
            {
                   echo $sub_service->title;
            } 

But this is not what I want, i want to get an array like this one : 但这不是我想要的,我想要一个像这样的数组:

Service 1 服务1

—Subservice 1 —子服务1

—Subservice 3 —子服务3

Service 2 服务2

Service 3 服务3

—Subservice 2 —子服务2

Info i'm using DM 1.8 and CI2. 信息我正在使用DM 1.8和CI2。

Thx for ur help 感谢您的帮助

Try using the 'all' property available in datamapper objects 尝试使用datamapper对象中可用的'all'属性

foreach($services->all as $service)
{
    echo $service->title;

    foreach($service->sub_service->all as $sub_service)
    {
              echo $sub_service->title;
    }
} 

You have to use the include_join_fields method of datamapper in order to create the relationship as you want it. 您必须使用datamapper的include_join_fields方法来创建所需的关系。 In your controller, include this: 在您的控制器中,包括以下内容:

$service = new Service();
$data = $service->get(); 
$options['records'] = $data;

and then in your view you can do this: 然后您可以执行以下操作:

    if(isset($records)){

    foreach($records as $service){

            echo $service->title;
        $subservice = $service->subservice->include_join_fields()->get();

        foreach($subservise as $subservice){
            if(!is_null($subservice->title)){  
                echo $subservice->title;
            }
        }

    }

This will give you what you want. 这会给你你想要的。

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM