Oracle查詢 - 加入用逗號分隔的數據

Question

表名：crm_mrdetails

id | mr_name | me_email  | mr_mobile | mr_doctor| 
----------------------------------------------------- 
1 | John |abc@gmail.com | 1234555555 | ,1,2,3 | 

表名：crm_mr_doctor

id | dr_name  | specialization| 
---------------------------------- 
1 | Abhishek | cordiologist | 
2 | Krishnan | Physician  | 
3 | Krishnan | Nurse   | 

在mrdetails.mr_doctor連接值是mr_doctor.id外鍵。我需要加入他們來產生這樣的輸出：

id | mr_name | me_email  |Doctor_Specialization| 
------------------------------------------------- 
1 | John |abc@gmail.com |cordiologist,Physician,Nurse| 

我是Oracle的新手，我使用Oracle 12C。任何幫助非常感謝。

Answer 1

請根據您的要求更改列名稱。

CREATE OR REPLACE Function ReplaceSpec 
    (String_Inside IN Varchar2) 
    Return Varchar2 Is 

     outputString Varchar2(5000); 
     tempOutputString crm_doc.specialization%TYPE; 

    Begin 

     FOR i in 1..(LENGTH(String_Inside)-LENGTH(REPLACE(String_Inside,',',''))+1) 
     LOOP 

      Select specialization into tempOutputString From crm_doc 
      Where id = PARSING_STRING(String_Inside,i); 

      If i != 1 Then 
       outputString := outputString || ','; 
      end if; 
      outputString := outputString || tempOutputString; 

     END LOOP; 

     Return outputString; 


    End; 
/

Parsing_String函數幫助拆分逗號分隔值。

CREATE OR REPLACE Function PARSING_STRING 
(String_Inside IN Varchar2, Position_No IN Number) 
Return Varchar2 Is 
    OurEnd Number; Beginn Number; 
Begin 

If Position_No < 1 Then 
Return Null; 
End If; 

OurEnd := Instr(String_Inside, ',', 1, Position_No); 

If OurEnd = 0 Then 
    OurEnd := Length(String_Inside) + 1; 
End If; 

If Position_No = 1 Then 
    Beginn := 1; 
Else 
    Beginn := Instr(String_Inside, ',', 1, Position_No-1) + 1; 
End If; 

Return Substr(String_Inside, Beginn, OurEnd-Beginn); 

End; 
/

請注意，我只給出了一個基本函數來獲得您的輸出。您可能需要添加一些例外等

Eg. When the doc_id [mr_doctor] is empty, what to do. 

使用

select t1.*,ReplaceSpec(doc_id) from crm_details t1 

如果您mr_doctor數據總是用逗號使用開始：

Select t1.*,ReplaceSpec(Substr(doc_id,2)) from crm_details t1 

Answer 2

該解決方案使用正則表達式來拆分mr_doctor列轉換成ID然後加入到mr_doctor表中; specialization列連接在一起以產生所需的輸出。

select mrdet.id, 
     mrdet.mr_name, 
     mrdet.me_email, 
     listagg(mrdoc.specialization, ',') 
        within group (order by mrdoc.specialization) as doctor_specialization 
from mr_details mrdet 
    join (
     select distinct id, 
       regexp_substr(mr_doctor, '(,?)([0-9]+)(,?)', 1, level, null, 2) as dr_id 
     from mr_details 
     connect by level <= regexp_count(mr_doctor, '(,?)([0-9]+)') 
     ) as mrids 
    on mrids.id = mrdet.id 
    join mr_doctor mrdoc 
     on mrids.dr_id = mr_doc.id 
group by mrdet.id, 
     mrdet.mr_name, 
     mrdet.me_email 
/

儘管數據模型很脆弱，但此解決方案仍具有相當的彈性。如果字符串中包含太多逗號或空格，它將返回結果。它會忽略字母或其他不是數字的值。如果提取的號碼與mr_doctor表中的ID不匹配，則不會投擲。顯然這些結果是不可信的，但這是輕快數據模型價格的一部分。

能否請你解釋下：(,?)([0-9]+)(,?)

的模式匹配零個或一個逗號其次一個或多個數字其次零或一個逗號。也許匹配模式中的(,?)不是絕對必要的。但是，如果沒有它們，該字符串2 3 4將匹配與此字符串2,3,4相同的三個ID。也許這是正確的，但事實並非如此。當外鍵存儲在CSV列中而不是通過適當的約束強制執行時，「正確」的含義甚至意味着什麼？

Answer 3

您必須將mr_doctor列中的數據拆分成行，連接表crm_mrdoctor，然後使用listagg()。 如何分割數據？Splitting string into multiple rows in Oracle

select t.id, max(mr_name) mr_name, 
     listagg(specialization, ', ') within group (order by rn) specs 
    from (
    select id, mr_name, levels.column_value rn, 
      trim(regexp_substr(mr_doctor, '[^,]+', 1, levels.column_value)) as did 
     from crm_mrdetails t, 
      table(cast(multiset(select level 
           from dual 
           connect by level <= 
            length(regexp_replace(t.mr_doctor, '[^,]+')) + 1) 
         as sys.odcinumberlist)) levels) t 
    left join crm_mr_doctor d on t.did = d.id 
    group by t.id 

演示和結果：

with crm_mrdetails (id, mr_name, mr_doctor) as (
    select 1, 'John', ',1,2,3' from dual union all 
    select 2, 'Anne', ',4,2,6,5' from dual union all 
    select 3, 'Dave', ',4'  from dual), 
crm_mr_doctor (id, dr_name, specialization) as (
    select 1, 'Abhishek', 'cordiologist' from dual union all 
    select 2, 'Krishnan', 'Physician' from dual union all 
    select 3, 'Krishnan', 'Nurse'  from dual union all 
    select 4, 'Krishnan', 'Onkologist' from dual union all 
    select 5, 'Krishnan', 'Surgeon'  from dual union all 
    select 6, 'Krishnan', 'Nurse'  from dual 
    ) 
select t.id, max(mr_name) mr_name, 
     listagg(specialization, ', ') within group (order by rn) specs 
    from (
    select id, mr_name, levels.column_value rn, 
      trim(regexp_substr(mr_doctor, '[^,]+', 1, levels.column_value)) as did 
     from crm_mrdetails t, 
      table(cast(multiset(select level 
           from dual 
           connect by level <= 
            length(regexp_replace(t.mr_doctor, '[^,]+')) + 1) 
         as sys.odcinumberlist)) levels) t 
    left join crm_mr_doctor d on t.did = d.id 
    group by t.id 

輸出：

ID MR_NAME SPECS 
------ ------- ------------------------------------- 
    1 John cordiologist, Physician, Nurse 
    2 Anne Onkologist, Physician, Nurse, Surgeon 
    3 Dave Onkologist 

Answer 4

這件怎麼樣？我沒有測試過，所以可能會有語法錯誤。

select id,mr_name,me_email,listagg(specialization,',') within group (order by specialization) as Doctor_Specialization 
from 
(select dtls.id,dtls.mr_name,dtls.me_email,dr.specialization 
from crm_mrdetails dtls, 
crm_mr_doctor dr 
where INSTR(','||dtls.mr_doctor||',' , ','||dr.id||',') > 0 
) group by id,mr_name,me_email;