Konvertieren Sie Roh-SQL mit JOINed-Unterabfragen in den Abfrage-Builder von CodeIgniter [geschlossen] ⇐ Php

[Anonymous]

Ist es möglich, die folgende Abfrage mit den Abfrage-Builder-Methoden von CodeIgniter zu generieren?

Code: Select all

SELECT
dept,
COUNT(nik) * 3 * (
SELECT
COUNT(DISTINCT(tanggal)) AS jday
FROM `tab_detail_absen`
WHERE tanggal BETWEEN '2021-05-31' AND '2021-06-04'
) AS jabs
FROM `tab_member_meet`
GROUP BY dept
) a
LEFT JOIN (
SELECT
LEFT(kode_org,3) AS dept,
COUNT(status) AS jth
FROM `tab_detail_absen`
WHERE
status = 'T'
AND keterangan = ''
AND tanggal BETWEEN '2021-05-31' AND '2021-06-04'
GROUP BY dept
) b ON a.dept = b.dept
LEFT JOIN (
SELECT
LEFT(kode_org,3) AS dept,
COUNT(nik) AS jentry
FROM tab_detail_absen
WHERE tanggal BETWEEN '2021-05-31' and '2021-06-04'
GROUP BY left(kode_org,3)
) c ON a.dept = c.dept