How to group and uniqe each row with ORACLE? - oracle

Question

Anyone please help me, how to uniqe each row
SELECT T3.ID_JS,T5.DIVISI_AREA,T4.NAME_METHODE_REPAIR, T1.URUTAN AS SORT, TO_DATE(T1.TIME_FINISHED_WORK,'DD/MM/YYYY') AS DATE_FINISHED
FROM TB_WORK T1
JOIN TB_INSPECTION T2 ON T1.ID_INSPECTION=T2.ID_INSPECTION
JOIN TB_JOBSHEET T3 ON T2.ID_JS = T3.ID_JS
JOIN TB_METHODE_REPAIR T4 ON T1.ID_METHODE_REPAIR=T4.ID_METHODE_REPAIR
JOIN TB_DIVISI T5 ON T1.ID_DIVISI=T5.ID_DIVISI
where t3.id_js=142414
GROUP BY T3.ID_JS,T5.DIVISI_AREA,T4.NAME_METHODE_REPAIR, T1.URUTAN,TO_DATE(T1.TIME_FINISHED_WORK,'DD/MM/YYYY')
ORDER BY T3.ID_JS, TO_DATE(T1.TIME_FINISHED_WORK,'DD/MM/YYYY') desc
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Answer 1

Those rows are unique, so I'll guess: there are two rows with sort = 3 and you'd want to have only one. If that's so, regarding the fact that date value is the only thing that makes difference, a simple way is to use aggregate function (such as min or max). I used max, but you can change it if you want:
SELECT t3.id_js,
t5.divisi_area,
t4.name_methode_repair,
t1.urutan AS sort,
MAX (TO_DATE (t1.time_finished_work, 'DD/MM/YYYY')) AS date_finished
FROM tb_work t1
JOIN tb_inspection t2 ON t1.id_inspection = t2.id_inspection
JOIN tb_jobsheet t3 ON t2.id_js = t3.id_js
JOIN tb_methode_repair t4
ON t1.id_methode_repair = t4.id_methode_repair
JOIN tb_divisi t5 ON t1.id_divisi = t5.id_divisi
WHERE t3.id_js = 142414
GROUP BY t3.id_js,
t5.divisi_area,
t4.name_methode_repair,
t1.urutan
ORDER BY t3.id_js, date_finished DESC
You'd remove date column from GROUP BY; besides, if you wanted to select distinct rows, why didn't you apply select distinct instead of using group by? Result is the same, but - group by is generally used with aggregates, not to return distinct result.