與聯合許多表讓許多價值

Question

問題這是我的腳本：

select c.rendering_id as prov_number, c.begin_date_of_service as date_of_service, 
    c.practice_id as group_number, v.enc_nbr as invoice, p.person_nbr as patient, 
    v.enc_nbr as invoice_number, c.charge_id as transaction_number, 
    t.med_rec_nbr as primary_mrn, p.last_name, p.first_name, 
    z.payer_id as orig_fsc_number, z.payer_id as curr_fsc_number, 
    c.location_id as location_number, c.closing_date as posting_date, 
    c.quantity as service_units, c.amt as charge_amount, 
    c.cpt4_code_id as procedure_code, r.description as procedure_name, 
    x.tran_code_id as pay_code_number, ISNULL([modifier_1],'') as modifier_code_1, 
    ISNULL([modifier_2],'') as modifier_code_2, ISNULL([modifier_3],'') as modifier_code_3, 
    ISNULL ([icd9cm_code_id],'') as dx_code_1, ISNULL ([icd9cm_code_id_2],'') as dx_code_2, 
    ISNULL ([icd9cm_code_id_3],'') as dx_code_3, ISNULL ([icd9cm_code_id_4],'') as dx_code_4 

from charges c, person p, patient t, patient_encounter v, encounter_payer z, cpt4_code_mstr r, transactions x 

where c.person_id = p.person_id 
    and c.person_id = t.person_id 
    and c.person_id = v.person_id 
    and c.person_id = z.person_id 
    and c.cpt4_code_id = r.cpt4_code_id 
    and c.person_id = x.person_id 
    and c.practice_id = '0001' 
    and c.closing_date >= GetDate() - 7 

我應該得到大約14K行，但這個我得到一對夫婦十萬。我覺得應該有一個內部連接來糾正它，但我已閱讀了一堆帖子，似乎可以得到它的工作。它是迄今爲止我在SQL中做過的最大的一次拉。

任何幫助將大大幫助。

Answer 1

不知道更多關於數據結構和外鍵關係，這個答案只是教育炒作。但在回答之前，您需要學習正確的JOIN語法。您的查詢應該如下所示：

from charges c join 
     person p 
     on . . . . 

也就是說，您的問題可能是您在同一時間沿多個維度加入。儘管沒有明確說明，但我猜測一個人可能有多個患者遇到，例如A，B和C.一個人也可能有多個費用，比如說10,11和12.

您的查詢會產生九個在這種情況下行，每個組合一個。

換句話說，你需要確定：

1. 確認表之間的連接鍵。被稱爲交易的表是否真正加入了使用person_id的遭遇和成本？
2. 找出你在哪裏得到交叉產品，並將其拆分成兩個子查詢，然後將它們恰當地結合在一起。

我建議你開始與前兩個表，看看你是否得到預期的行數：

select * 
from charges c join 
    person p 
    on c.person_id = p.person_id 
where c.practice_id = '0001' and 
    c.closing_date >= GetDate() - 7 

然後同時建立查詢一個表得到的結果你要。

最後一個注意事項是，當使用表別名時，我發現使用喚醒表的別名更清晰。 「C」收費非常好。考慮一下像patient_encounters的「pe」等等。

Answer 2

它應該是這樣的，或者您可以使用左連接

select c.rendering_id as prov_number, c.begin_date_of_service as date_of_service, 
c.practice_id as group_number, v.enc_nbr as invoice, p.person_nbr as patient, 
v.enc_nbr as invoice_number, c.charge_id as transaction_number, 
t.med_rec_nbr as primary_mrn, p.last_name, p.first_name, 
z.payer_id as orig_fsc_number, z.payer_id as curr_fsc_number, 
c.location_id as location_number, c.closing_date as posting_date, 
c.quantity as service_units, c.amt as charge_amount, 
c.cpt4_code_id as procedure_code, r.description as procedure_name, 
x.tran_code_id as pay_code_number, ISNULL([modifier_1],'') as modifier_code_1, 
ISNULL([modifier_2],'') as modifier_code_2, ISNULL([modifier_3],'') as modifier_code_3, 
ISNULL ([icd9cm_code_id],'') as dx_code_1, ISNULL ([icd9cm_code_id_2],'') as dx_code_2, 
ISNULL ([icd9cm_code_id_3],'') as dx_code_3, ISNULL ([icd9cm_code_id_4],'') as dx_code_4 

from charges c 
inner join person p on c.person_id = p.person_id 
inner join patient t on c.person_id = t.person_id 
inner join patient_encounter v on c.person_id = v.person_id 
inner join encounter_payer z on c.person_id = z.person_id 
inner join cpt4_code_mstr r on c.cpt4_code_id = r.cpt4_code_id 
inner join transactions x on c.person_id = x.person_id 

where c.practice_id = '0001' 
and c.closing_date >= GetDate() - 7 

Answer 3

現在，你對此有何評論一個內部聯接的時間和執行下面的查詢，看看其中哪些連接是造成一對多的關係...當計數給出14 K附近時，這意味着註釋表導致了1對多關係。

否則最好的方法是根據這些表上的唯一鍵，主鍵和FK找到關係​​。

select 
count(c.person_id) 
from charges c 
inner join person p on c.person_id = p.person_id 
inner join patient t on c.person_id = t.person_id 
inner join patient_encounter v on c.person_id = v.person_id 
inner join encounter_payer z on c.person_id = z.person_id 
inner join cpt4_code_mstr r on c.cpt4_code_id = r.cpt4_code_id 
inner join transactions x on c.person_id = x.person_id 

where c.practice_id = '0001' 
and c.closing_date >= GetDate() - 7 

您可以嘗試

select count(*) from <tablename> group by person_id having count(*) > 1 

，並重覆上述查詢所有表這會給你什麼樣的收費表和其他表之間關係的想法。 Offcourse對cpt4_code_mstr表使用cpt4_code_id，但通過名稱，它看起來像是這張表是主表，所以它對於收費表中的每個cpt4-code_id值將具有一個標號值。

我希望它能幫到