如何将多个查询合并为一个查询结果

Question

如果您帮助我们以下主题，我将非常有用：

select z1.nature_projet as typeCredit,COUNT(z1.nature_projet) as nbreTotalRejete, SUM(z1.montant_emprunt) as totalMontantRejete                                                 
FROM (
select * from dossier
where (EXTRACT(MONTH FROM date_affectation_banque ) = EXTRACT(MONTH FROM now())) and partner_id=1 and statut_dossier='Envoye'
) z1
where banque_statut_dossier='Rejete'
group by nature_projet**

以下图片描述了结果：

select  z1.nature_projet as typeCredit,COUNT(z1.nature_projet) as nbreTotalAccorde, SUM(z1.montant_emprunt) as totalMontantAccorde                                                          
FROM (
select * from dossier
where (EXTRACT(MONTH FROM date_affectation_banque ) = EXTRACT(MONTH FROM now())) and partner_id=1 and statut_dossier='Envoye'
) z1
where banque_statut_dossier='Accorde'
group by nature_projet

以下图片描述了结果：

SELECT z1.nature_projet as typeCredit,  
COUNT(z1.nature_projet) as nbreTotal, SUM(z1.montant_emprunt) as totalMontant                                                           
FROM (
select * from dossier
where (EXTRACT(MONTH FROM date_affectation_banque ) = EXTRACT(MONTH FROM now())) and partner_id=1 and statut_dossier='Envoye'
) z1
group by nature_projet

以下图片描述了结果：

我想将前两个查询的结果连接到一个，将最后一个查询连接起来，所以结果列如下：

TypeCredit,totalNbre,totalMontant,TotalNbreAccorde,TotalMontantAccorde,TottalNbreRejete,TotalMontantRejete

Answer 1

您可以进行条件聚合。考虑：

select nature_projet, 
    count(*)             filter(where banque_statut_dossier = 'Rejete' ) cnt_rejete,
    count(*)             filter(where banque_statut_dossier = 'Accorde') cnt_accorde,
    count(*)             filter(where banque_statut_dossier = 'Envoye' ) cnt_envoye,
    sum(montant_emprunt) filter(where banque_statut_dossier = 'Rejete' ) montant_rejete,
    sum(montant_emprunt) filter(where banque_statut_dossier = 'Accorde') montant_accorde,
    sum(montant_emprunt) filter(where banque_statut_dossier = 'Envoye' ) montant_envoye
from dossier
where 
    partner_id = 1
    and banque_statut_dossier in ('Rejete', 'Accorde', 'Envoye')
    and date_affectation_banque >= date_trunc('month', current_date) 
    and date_affectation_banque <  date_trunc('month', current_date) + interval '1 month'
group by nature_projet

请注意，我更改了日期过滤逻辑；您似乎打算在当月进行过滤，并且上述表达式使用针对半开间隔的直接过滤，比起依赖日期函数的原始策略更有效。