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probably because of this... It's better to use the same column names.

As for your problem with joins on icon_ids, I wouldn't bother. Create an array of icons from the icon table $res = $pdo->query("SELECT icon_id, icon_code FROM icon"); $icons = array_column($res->fetchAll(), 'icon_code', 'icon_id'); then use the $card['Icon'] to access key to this array when outputting $icon_code = $icons[$card['Icon']];

As the data I was given has no data for RAGUL's empno in any tables (except usertable) I cannot verify any of my theories on what may be wrong. Also for branch 2 I have only this... +---------+---------------+---------------+----------------+-----------------+------------------+----------------+-----------------+ | month | name | CustomerTotal | CustomerActual | ProductionTotal | ProductionActual | MarketingTotal | MarketingActual | +---------+---------------+---------------+----------------+-----------------+------------------+----------------+-----------------+ | 2024-03 | Palanikumar B | | | | | 1 | | | 2024-02 | NAVEENKUMAR P | | | | | 1 | | +---------+---------------+---------------+----------------+-----------------+------------------+----------------+-----------------+ For branch 5... +---------+------------------+---------------+----------------+-----------------+------------------+----------------+-----------------+ | month | name | CustomerTotal | CustomerActual | ProductionTotal | ProductionActual | MarketingTotal | MarketingActual | +---------+------------------+---------------+----------------+-----------------+------------------+----------------+-----------------+ | 2024-02 | Nethaji.MK | | | | | 1 | | | 2024-02 | T.Anandakrishnan | | | | | 1 | | | 2024-02 | Sivanraj.A | | | | | 1 | | | 2024-02 | kannan.r | | | | | 1 | | | 2024-01 | thayyanayaki | 5 | 5 | 2 | 2 | | | | 2024-02 | thayyanayaki | 5 | 5 | 2 | 2 | | | | 2024-03 | thayyanayaki | 5 | 5 | 2 | 2 | 1 | | | 2024-04 | thayyanayaki | 5 | 3 | 2 | 1 | | | | 2024-01 | Vignesh.M | | | | | 1 | 1 | | 2024-02 | Vignesh.M | | | | | 1 | | | 2024-02 | N.Manikandan | | | | | 1 | | | 2024-02 | Dinesh.C | | | | | 1 | | +---------+------------------+---------------+----------------+-----------------+------------------+----------------+-----------------+ The previous queries (your and mine) would only report on whatever empno, branch and month values existed in the production table (the one that isn't LEFT JOINED). This version removes that reliance ... SELECT u.Month , Name , CustomerTotal , CustomerActual , ProductionTotal , ProductionActual , MarketingTotal , MarketingActual FROM ( WITH RECURSIVE allmonths (id, month) as ( SELECT 1, '2024-01' UNION ALL SELECT id+1, concat('2024-', lpad(id+1,2,'0')) FROM allmonths WHERE id < 4 ) SELECT empNo as empId , fname as name , month , ? as branch FROM usertable, allmonths -- WHERE role IN (4, 5) ) u LEFT JOIN ( SELECT count(*) as CustomerTotal , sum(VisitType = 'No Due' OR VisitDate != '') as CustomerActual , month , empid , branch FROM customerdata GROUP BY month,branch,empid ) ca ON u.month = ca.month AND u.empid = ca.empid AND u.branch = ca.branch LEFT JOIN ( SELECT count(*) as ProductionTotal , sum(MCubicmeter OR MHourmeter) as ProductionActual , month , empid , branch FROM production GROUP BY month, branch, empid ) pa ON u.month = pa.month AND u.empid = pa.empid AND u.branch = pa.branch LEFT JOIN ( SELECT count(*) as MarketingTotal , month , empid , branch FROM marketing_target GROUP BY month, branch, empid ) mt ON u.month = mt.month AND u.empid = mt.empid AND u.branch = mt.branch LEFT JOIN ( SELECT count(*) as MarketingActual , month , empid FROM marketing_data GROUP BY month, empid ) ma ON u.month = ma.month AND u.empid = ma.empid AND u.branch = mt.branch HAVING CustomerTotal OR CustomerActual OR ProductionTotal OR ProductionActual OR MarketingTotal OR MarketingActual ORDER BY u.empid, u.month;