Why left outer join works like inner join on ABAP CDS view and a custom table? - abap

Question

I have created an ABAP CDS view using ACDOCA, BSID, MARA, MAKT and such tables.
Now my CDS view contains list of customers along with their outstanding (FI) documents.
Against these customers, FI documents and certain more parameters, I am maintaining certain data in a custom table.
Now using left outer join I want to connect this CDS view with custom table.
The problem now is that this join is working like inner join, so if my custom table is blank my output is displaying without values whereas it should be displaying CDS view as it is, and custom table fields as blank.
Now based on similar issues on web I did research and found some useful points:
all conditions on right table are included at most in ON join condition
tried to include null values from right side custom table as well.
tried using just one parameter on join condition to see the result (still working as inner join)
Here is my complete select statement, where:
ZPD1 is my ABAP CDS view of ACDOCA, BSDI ... tables
ZPD_HSL is an ABAP CDS view for customer wise GL amount calculation.
ZFI_PDRATE_MAP is my custom table which contains rate for customer / FI docs combination.
What could be a misstep in this statement?
SELECT
FROM ZPD1( PBUKRS = #BUKRS-LOW ,
POSTFROM = #COL_DATE-LOW ,
POSTTO = #COL_DATE-HIGH ) AS A
LEFT OUTER JOIN ZFI_PDRATE_MAP AS F
ON A~PROD_CATEGORY = F~PROD_CATEGORY
AND A~OD_DATE GE F~POST_DT_FROM
AND A~OD_DATE LE F~POST_DT_TO
AND F~COL_DT_FROM GE #COL_DATE-LOW
AND F~COL_DT_TO LE #COL_DATE-HIGH
AND F~SALES_OFFICE = #VKBUR-LOW
AND CASE WHEN A~ARREAR_DAYS <= 0 THEN 'NOD'
WHEN A~ARREAR_DAYS > 0 THEN 'OD'
END = F~OD
INNER JOIN KNA1 AS H
ON H~KUNNR = A~KUNNR
INNER JOIN TVKBT AS I
ON I~VKBUR = F~SALES_OFFICE
AND I~SPRAS = #SY-LANGU
LEFT OUTER JOIN SKAT AS J
ON J~SAKNR = A~HKONT
AND J~SPRAS = 'E'
AND J~KTOPL = '1000'
LEFT OUTER JOIN MAKT AS K
ON K~MATNR = A~MATNR
AND K~SPRAS = 'E'
LEFT OUTER JOIN T001W AS L
ON L~WERKS = A~WERKS
FIELDS
A~RBUKRS AS BUKRS,
A~KUNNR,
H~NAME1,
F~SALES_OFFICE,
I~BEZEI,
A~BELNR AS COL_BELNR,
A~GJAHR AS COL_GJAHR,
A~HSL AS COL_AMT,
A~BUDAT AS COL_DATE,
A~KEY_DATE AS KEY_OS_DATE,
A~OD_DOC AS OS_BELNR,
A~OD_YR AS OS_GJAHR,
A~BLART ,
A~OD_DATE AS OS_DATE,
CASE WHEN A~SHKZG = 'H' THEN CAST( A~WRBTR * -1 AS CURR( 12, 2 ) )
ELSE A~WRBTR
END AS OS_AMT ,
A~DUE_DATE,
A~ARREAR_DAYS,
CASE WHEN A~ARREAR_DAYS <= 0 THEN 'NOD'
WHEN A~ARREAR_DAYS > 0 THEN 'OD'
ELSE ' '
END AS OD_IND,
A~HKONT,
J~TXT50,
A~PROD_CATEGORY,
A~MATNR,
K~MAKTX,
A~WERKS ,
L~NAME1,
A~BWTAR,
A~PROFIT_CENTRE AS PRCTR,
COALESCE( F~PD_RATE , 0 ) AS RATE,
0 AS PD_AMT,
CASE WHEN A~ARREAR_DAYS > 0 THEN
CAST(
DIVISION(
( CAST( A~WRBTR * F~PD_RATE AS CURR( 15 ,2 ) ) ) ,
100 ,
3 )
AS CURR( 15, 2 ) )
* -1
ELSE
DIVISION(
( CAST( A~WRBTR * F~PD_RATE AS CURR( 15 , 2 ) ) ) ,
100 ,
3 )
END AS ITEM_PD_AMT, "DISCOUNT_RATE,
0 AS ITEM_PD_AMT1,
A~MAIN_TYPE
WHERE A~KUNNR IN #KUNNR
AND NOT EXISTS (
SELECT *
FROM ZFI_PAYMENT_DISC
WHERE BELNR = A~BELNR
AND GJAHR = A~GJAHR
AND BUKRS = A~RBUKRS
AND SALES_OFFICE = #VKBUR-LOW
AND A~KUNNR IN #KUNNR
AND OS_BELNR EQ A~OD_DOC )
AND A~ARREAR_DAYS <= #OD
AND ( F~COL_AMT_FROM LE (
SELECT SUM( COLLECT )
FROM ZPD_HSL( BUKRS = #BUKRS-LOW ,
BUDATFROM = #COL_DATE-LOW ,
BUDATO = #COL_DATE-HIGH )
WHERE KUNNR = A~KUNNR )
AND F~COL_AMT_TO GE (
SELECT SUM( COLLECT )
FROM ZPD_HSL( BUKRS = #BUKRS-LOW ,
BUDATFROM = #COL_DATE-LOW ,
BUDATO = #COL_DATE-HIGH )
WHERE KUNNR = A~KUNNR )
)
ORDER BY A~KUNNR, A~BELNR ,A~DUE_DATE ASCENDING
INTO TABLE #PROV_POST.

Answer 1

This is a mere guess, as your complex setup makes it hard to reproduce a working example.
My assumption is that the INNER JOIN TVKBT AS I ON I~VKBUR = F~SALES_OFFICE is getting in the way. It inner-joins to your custom table, aliased as F, and might lead to an overreduction of records if the join order is interpreted wrong.
I'd recommend to remove that join and see whether that fixes the basic join result.
In any way, I would recommend to add parentheses and reorder the join clauses to make clear what the "left" side of the equation is, and to remove any ambiguity, both for the interpreter, and the reader.