Codeigniter 查询未返回左侧的所有记录 table。只有returns条符合条件的记录

Question

仅查询 returns 加入的 table 中的匹配记录。我究竟做错了什么？我正在尝试显示来自代理 table 的所有记录，无论贷款 table 中是否有匹配的记录。也许我想要实现的目标的逻辑是错误的。

    $select = "agents.person_id, CONCAT(people.first_name, ' ', people.last_name) as last_name, SUM(loans.referral_amount) as referral_amount, COUNT( DISTINCT loans.customer_id ) as no_customers";
            $this->db->select($select, false);
            $this->db->from('agents');
            $this->db->join('people', 'agents.person_id=people.person_id', 'LEFT');
            $this->db->join('loans', 'agents.person_id=loans.referral_agent_id AND loans.loan_status = "paid" AND loans.delete_flag = 0', 'LEFT');
            $this->db->where('agents.deleted', 0);
    return $this->db->get();



Table: agents

+-----------+---------+--+
| person_id | deleted |  |
+-----------+---------+--+
|         1 |       0 |  |
|         2 |       0 |  |
|         3 |       1 |  |
|         4 |       0 |  |
+-----------+---------+--+

Table: loans

|   | loan_status | referral_amount | referral_agent_id | delete_flag|customer_id |
|---|-------------|-----------------|-------------------|-------------|-------------|
|   | paid        | 10              | 1                 | 0           | 2           |
|   | pending     | 20              | 1                 | 0           | 2           |
|   | approved    | 30              | 3                 | 1           | 1           |


Table: people

| person_id | first_name | last_name |
|-----------|------------|-----------|
| 1         | Test       | Ken       |
| 2         | Lorem      | Ipsum     |
| 3         | Stack      | Over      |

The result I am getting

| name     | referral amount | no of customers |
|----------|-----------------|-----------------|
| Test Ken | 10              | 1               |

What I am expecting

| name        | referral amount | no of customers |
|-------------|-----------------|-----------------|
| Test Ken    | 10              | 1               |
| Lorem Ipsum | null            | null            |
| Stack Over  | null            | null            |

Answer 1

首先，据我所知，您将“标准查询”与“查询生成器”混合使用，最好只使用一个（最好是查询生成器，以防您切换到另一个数据库引擎） .

虽然 this is valid, you could try to make the joins work first. I think is working now but to be sure please, debug your query print the result and fix it according to documentation、

，但在第二个 Join 中，您也在进行“AND”比较

$select = "agents.person_id, CONCAT(people.first_name, ' ', people.last_name) as last_name, SUM(loans.referral_amount) as referral_amount, COUNT( DISTINCT loans.customer_id ) as no_customers, people.phone_number";

$this->db->select($select, false);
$this->db->from('agents');
$this->db->join('people', 'agents.person_id=people.person_id', 'LEFT');
$this->db->join('loans', 'agents.person_id=loans.referral_agent_id', 'LEFT'); 
$this->db->where('loans.loan_status', 'paid');
$this->db->where('loans.delete_flag', 0);
$this->db->where('agents.deleted', 0);
$this->db->get();

print_r($this->db->last_query());

（最终建议：无论何时使用 Join，请始终在查询中使用符号 database.column，这样更容易理解并避免在两个数据库具有同名列的情况下出错）

Answer 2

$select = "agents.person_id, CONCAT(people.first_name, ' ', people.last_name) as last_name, SUM(loans.referral_amount) as referral_amount, COUNT( DISTINCT loans.customer_id ) as no_customers, people.phone_number";

$this->db->select($select, false);
$this->db->from('agents');
$this->db->join('people', 'agents.person_id = people.person_id', 'LEFT');
$this->db->join('loans', 'agents.person_id = loans.referral_agent_id', 'LEFT');
$this->db->where('loans.loan_status', "paid");
$this->db->where('loans.delete_flag', 0);
$this->db->where('agents.deleted', 0);

$result = $this->db->get();

Answer 3

想通了

$select = "c19_agents.person_id, referral_sum.user_referral_amount, CONCAT(c19_people.first_name, ' ', c19_people.last_name) as agents_name, customer_count.no_customers, referral_sum.user_referral_amount, phone_number";
        $this->db->select($select, false);
        $this->db->from('agents');
        $this->db->join('people', 'agents.person_id=people.person_id', 'LEFT');
        $this->db->join('(SELECT c19_loans.referral_agent_id, SUM(c19_loans.referral_amount)
                   as user_referral_amount
                   FROM c19_loans
                       WHERE c19_loans.delete_flag = 0 AND c19_loans.loan_status = "paid"
                       GROUP BY c19_loans.referral_agent_id) referral_sum', 'c19_agents.person_id = referral_sum.referral_agent_id', 'LEFT');
        $this->db->join('(SELECT c19_loans.referral_agent_id, COUNT( DISTINCT c19_loans.customer_id)
                    as no_customers
                    FROM c19_loans
                        WHERE c19_loans.delete_flag = 0 AND customer_id > 0
                        GROUP by c19_loans.referral_agent_id) customer_count', 'c19_agents.person_id = customer_count.referral_agent_id', 'LEFT');
        $this->db->where('agents.deleted', 0);

Codeigniter 查询未返回左侧的所有记录 table。只有returns条符合条件的记录

Codeigniter query is not returning all records from the left table. Only returns the matching records

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mysql

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