[英]Codeigniter active record query count
in my project i have 4 tables. 在我的项目中,我有4张桌子。 One called
Calls
, Projects
, Products
and Users
. 一种叫做
Calls
, Projects
, Products
和Users
。
Calls has 2 FK ( project_id
and user_id
) and Projects has one FK ( product_id
). 呼叫有2个FK(
project_id
和user_id
),而Projects有一个FK( product_id
)。
I'm trying to query the table Calls and I want to know the amount of calls for each product has been received from each operator. 我正在尝试查询表Calls ,我想知道已经从每个运营商处收到了每种产品的呼叫数量。
I'm trying to query something like this: 我正在尝试查询如下内容:
[{"name":"User1","Project1":"120","Project2":"10,"Project3":"140...}
{"name":"User2","Project1":"80","Project2":"60,"Project3":"14...}]
I am guessing when you mean operator you mean user 我猜你是什么意思
$this->db->select('count(P.product_id) as total_product ,CL.project_id, CL.user_id, P.project_name')
->from('Calls CL')
->join('Projects P', 'P.project_id=CL.project_id')
->group_by('CL.project_id, CL.user_id')
->get();
Now your can write a php script to loop through your result to make the json 现在,您可以编写一个php脚本循环遍历结果以制作json
$index = 0;
$user_data = array();
$user_key = array();
foreach ($result as $row) {
if(!isset($user_key[$row['user_id']])){
$user_key[$row['user_id']] = $index;
$index++;
}
$user_data[$user_key[$row['user_id']]]['user_name'] = $row['user_name'];
$user_data[$user_key[$row['user_id']]][$row['project_name']] = $row['total_product'];
}
I did not tested the code but it should work 我没有测试代码,但应该可以
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