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如何在单个数组中显示更多数量的数组结果?

[英]How to display more number of array results in a single array?

I have two tables: login and follow`. 我有两个表:login and follow`。

  • Table name: login 表名: login
    Fields: id,email,username,imageurl 栏位: id,email,username,imageurl

  • Table name: follow 表名: follow
    Fields: id:user_id:follow_id 栏位: id:user_id:follow_id

It's like a Twitter follower's concept. 这就像Twitter追随者的概念。 I want to get the details of myfollower name and also myfollower 's following person's name. 我想获取myfollower名称的详细信息,以及myfollower的跟随者的姓名。

For that I have written the coding as like below. 为此,我编写了如下代码。

public function follw ()
{
    if( $this->input->get("userid")  )

     {          

    extract($this->input->get()); 

    $followers_list = array();

    $follower = array();

    $query = $this->db->query('select follow_id from follow where user_id = '.$userid.'')->result();

    foreach($query as $row)

    { 
        $follower['follower_id'] = $row->follow_id;

        if($follower['follower_id'] == "")
        {
            echo "hi";
        }

        else
            {
        $query3 = $this->db->query('select username from login where id = '.$follower['follower_id'].'')->result();

        foreach($query3 as $row3)

            {
                $follower['followuser'] = $row3->username;
            }

        $query1 = $this->db->query('select follow_id from follow where user_id = '.$follower['follower_id'].'')->result();

        foreach($query1 as $row1)

            {
                $follower['follow_id'] = $row1->follow_id;

            if($follower['follow_id'] == "")

            {
                echo "jeeva";
                }


            else
            {
        $query2 = $this->db->query('select username from login where id = '.$follower['follow_id'].'')->result();

        foreach($query2 as $row2)

            {
                $follower['username'] = $row2->username;


            }

            }//second for each in else loop
            }//first foreach in else loop

            }//main else
        $followers_list[] = $follower;

        }

        $str = json_encode($followers_list);
    echo stripslashes($str);
}

else
{
    echo '[{"status":"Failure  - Error Occured - Not Enough Details provided"}]';

    }
}

I get the output like this: 我得到这样的输出:

[{"follower_id":"12","followuser":"janmejoy","follow_id":"24","username":"sarvana"},{"follower_id":"10","followuser":"jeeva","follow_id":"23","username":"selva"},{"follower_id":"6","followuser":"raj","follow_id":"17","username":"jeeva"},{"follower_id":"23","followuser":"selva","follow_id":"22","username":"guru"}]

This output displays myfollower 's name and myfollower 's following person name, but the problem is it displays only one member of myfollower 's following person name. 此输出显示myfollower的名称和myfollower的跟随人的名字,但是问题在于它仅显示myfollower的跟随人的名字的一个成员。

However, I want to the output like this: 但是,我想要这样的输出:

[{"follower_id":"12","followuser":"janmejoy",{"follow_id":"24","username":"sarvana",follow_id":"13","username":"jai",follow_id":"9","username":"raj"}},{"follower_id":"10","followuser":"jeeva","follow_id":"23","username":"selva"},{"follower_id":"6","followuser":"raj","follow_id":"17","username":"jeeva"},{"follower_id":"23","followuser":"selva","follow_id":"22","username":"guru"}]

[{"follower_id":"12","followuser":"janmejoy",{"follow_id":"24","username":"sarvana",follow_id":"13","username":"jai",follow_id":"9","username":"raj"}},{"follower_id":"10","followuser":"jeeva","follow_id":"23","username":"selva"},{"follower_id":"6","followuser":"raj","follow_id":"17","username":"jeeva"},{"follower_id":"23","followuser":"selva","follow_id":"22","username":"guru"}] [{“ follower_id”:“ 12”,“ followuser”:“ janmejoy”,{“ follow_id”:“ 24”,“用户名”:“ sarvana”,follow_id“:” 13“,”用户名“:” jai“, follow_id“:” 9“,”用户名“:” raj“}},{” follower_id“:” 10“,” followuser“:” jeeva“,” follow_id“:” 23“,”用户名“:” selva“} ,{“ follower_id”:“ 6”,“ followuser”:“ raj”,“ follow_id”:“ 17”,“用户名”:“ jeeva”},{“ follower_id”:“ 23”,“ followuser”:“ selva “,” follow_id“:” 22“,”用户名“:”专家“}]

This code is invalid JSON variable. 此代码是无效的JSON变量。 i can't solve this question for you. 我不能为你解决这个问题。 Please update your question. 请更新您的问题。

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