Преобразовать Select Select, содержащий, где существует, и присоединяйтесь к пунктам к строительству запросов Codeignite ⇐ Php

[Anonymous]

i это mysql Query < /p>

SELECT project.title project_title, category.category, project_details.description description, SUM(project_details.quantity) total_quantity, project_details.unit, SUM(project_details.price) total_price, SUM(project_details.jan) total_jan, SUM(project_details.feb) total_feb, SUM(project_details.mar) total_mar, SUM(project_details.apr) total_apr, SUM(project_details.may) total_may, SUM(project_details.jun) total_jun, SUM(project_details.jul) total_jul, SUM(project_details.aug) total_aug, SUM(project_details.sep) total_sep, SUM(project_details.oct) total_oct, SUM(project_details.nov) total_nov, SUM(project_details.dec) total_dec
FROM project_details
JOIN project ON project_details.project_id = project.id
JOIN category ON project_details.category_id = category.id
JOIN user ON project.user_id = user.id
RIGHT JOIN office ON user.office_id = office.id
WHERE EXISTS
(
SELECT t1.id id1, t2.id id2, t3.id id3, t4.id id4, t5.id id5
FROM user t1
LEFT JOIN user t2 ON t2.reports_to = t1.id
LEFT JOIN user t3 ON t3.reports_to = t2.id
LEFT JOIN user t4 ON t4.reports_to = t3.id
LEFT JOIN user t5 ON t5.reports_to = t4.id
WHERE t1.id = 81
)
AND project.approval_status = 22 AND project.reason_for_rejection IS NULL
GROUP BY project.id, project_details.description
ORDER BY project_details.category_id
< /code>

Однако я не могу найти какую -либо документацию в Codeigniter 3.0, где я могу найти, где существует < /code>. Я уже проверил это https://www.codeigniter.com/userguide3/ ... ilder.html, но есть не.

Это мой код прямо сейчас

$this->db->select('project.title project_title, category.category, project_details.description description, SUM(project_details.quantity) total_quantity, project_details.unit, SUM(project_details.price) total_price, SUM(project_details.jan) total_jan, SUM(project_details.feb) total_feb, SUM(project_details.mar) total_mar, SUM(project_details.apr) total_apr, SUM(project_details.may) total_may, SUM(project_details.jun) total_jun, SUM(project_details.jul) total_jul, SUM(project_details.aug) total_aug, SUM(project_details.sep) total_sep, SUM(project_details.oct) total_oct, SUM(project_details.nov) total_nov, SUM(project_details.dec) total_dec');
$this->db->from('project_details');
$this->db->join('project', 'project_details.project_id = project.id');
$this->db->join('category', 'project_details.category_id = category.id');
$this->db->join('user', 'project.user_id = user.id');
$this->db->join('office', 'user.office_id = office.id', 'right');
// THIS IS WHERE I WANT TO PUT THE WHERE EXISTS
$this->db->where('project.approval_status', $user_id);
$this->db->where('project.reason_for_rejection IS NULL', NULL, TRUE);
$this->db->group_by('project.id');
$this->db->group_by('project_details.description');
$this->db->order_by('project_details.category_id');
$query = $this->db->get();
return $query->result();
< /code>

может кто -нибудь помочь, как я могу достичь этого в Codeigniter? Продвинутая спасибо.

Подробнее здесь: https://stackoverflow.com/questions/360 ... igniters-q