奇妙な結果に与えたグループの連結方式は、私は、このSQLクエリを持って
vacancy_id,org_id,name,description,number_required,occupancy_kind,website,offer,logo,banner,address_country,address_city,address_postal_code,address_line_1,address_line_2,vacancy_visibility_start_date,vacancy_visibility_end_date,engagement,interests,skills,date_from
"2","1","test123","aze<sdgqswdfg","1","1","","blabla",NULL,"12049394_10208129537615226_4853636504350654671_n.jpg","Belgie","Brussel","1000","Brusselsestraat 15",NULL,"2016-09-02 00:00:00","2016-09-19 00:00:00","3","13,6,1","4,3","2016-09-13 00:00:00"
"3","1","blablabla","lkpjoip","1","2","","blabla",NULL,NULL,"Belgie","Antwerpen","2000","Antwerpsestraat 16",NULL,"2016-09-02 00:00:00","2016-09-29 00:00:00","3","28","7,8,5","2016-09-01 00:00:00"
"4","1","hahaha","14556dsf","1","3","","blabla",NULL,NULL,"Belgie","Mechelen","2800","Mechelsesteenweg 17",NULL,"2016-09-02 00:00:00","2016-09-28 00:00:00","3",NULL,NULL,"2016-09-26 00:00:00"
"5","1","omggg","45sdfdj5","1","1","","blabla",NULL,NULL,"Belgie","Gent","3000","Gentsesteenweg 18",NULL,"2016-09-02 00:00:00","2016-09-30 00:00:00","3","17,11","4,1","2016-09-19 00:00:00"
"6","1","this is a test","wauhiufdsq","1","2","","blabla",NULL,NULL,"Belgie","Luik","4000","Luikseweg 19",NULL,"2016-09-02 00:00:00","2016-09-30 00:00:00","3","19,17,22","6","2016-08-10 00:00:00"
空席の関心と空席スキルの表には、単一空室の複数のレコードを含めることができます。例えば。空席3にはすべて3つの行があり、すべての異なるinterest_idがある可能性があります。 group_concatは私の問題をここで解決します。 したがって、このクエリは正常に動作します。
しかし、私が遭遇した2の問題は、以下の通りである:
1)私はこれが唯一の私の代わりに期待二列の1行を返しIDにより興味にHAVINGにフィルタを追加する場合。
SELECT v.*, group_concat(distinct(vi.interest_id)) as interests, group_concat(distinct(vs.skill_id)) as skills, vc.date_from
FROM `vacancies` as v
LEFT JOIN `vacancy_interests` as vi on v.vacancy_id = vi.vacancy_id
LEFT JOIN `vacancy_skills` as vs on v.vacancy_id = vs.vacancy_id
LEFT JOIN `vacancy_calendar` as vc on v.vacancy_id = vc.vacancy_id
WHERE v.vacancy_visibility_end_date >= CURDATE()
GROUP BY v.vacancy_id
HAVING interests IN (17)
これは私に1行しか返しません。つまり、vacanacy_id 5で記録してください。また、vacancy_id = 6も返さなければなりません。
私には最も奇妙なことは、私が技術と同じことをしているのですが(スキルを持っている(4)) 、これは私に正しい結果で複数の行を返します。
2)私はDATE_FROMにフィルタリングする(一緒に持つことに関心やスキルを持つ、私は次のようにします。
SELECT v.*, group_concat(distinct(vi.interest_id)) as interests, group_concat(distinct(vs.skill_id)) as skills, vc.date_from
FROM `vacancies` as v
LEFT JOIN `vacancy_interests` as vi on v.vacancy_id = vi.vacancy_id
LEFT JOIN `vacancy_skills` as vs on v.vacancy_id = vs.vacancy_id
LEFT JOIN `vacancy_calendar` as vc on v.vacancy_id = vc.vacancy_id
WHERE v.vacancy_visibility_end_date >= CURDATE() AND date(vc.date_from) > '2016-09-10'
GROUP BY v.vacancy_id
HAVING skills IN (4)
これだけもしばらくは明らかに、私に欠員数5を返します。空席番号2は、2016年9月10日より大きい日付(2016年9月13日00:00:00)持っている....
私はここで間違って何をしているのですか?
最大30個の最大金利と10個のスキルがあり、空き1つにつき最大3個あります。すぐにこれらをアップグレードする予定はありません。 :) あなたの答えは完全にありがとうございます! – Dennis