[英]Query joining in sql server 2005
我有兩個查詢,例如:
SELECT PC_COMP_CODE,
PC_SL_LDGR_CODE,
PC_SL_ACNO ACCOUNT,
COUNT(PC_CHEQUE_NO) CHQS,
SUM(CONVERT(FLOAT, PC_AMOUNT)) CHQ_AMT
FROM GLAS_PDC_CHEQUES
WHERE PC_COMP_CODE = '1'
AND PC_DISCD IS NULL
GROUP BY PC_SL_LDGR_CODE,
PC_SL_ACNO ,PC_COMP_CODE
ORDER BY PC_SL_ACNO
--------------------------------------------------
SELECT COAD_PTY_FULL_NAME,PC_COMP_CODE, PC_SL_LDGR_CODE, PC_SL_ACNO, PC_DEPT_NO, PC_DOC_TYPE, PC_CHEQUE_NO, PC_BANK_AC_NO FROM GLAS_PTY_ADDRESS,GLAS_SBLGR_MASTERS,GLAS_PDC_CHEQUES WHERE
COAD_COMP_CODE = '1' AND SLMA_COMP_CODE = COAD_COMP_CODE AND SLMA_ADDR_ID = COAD_ADDR_ID
AND SLMA_LDGRCTL_CODE = PC_SL_LDGR_CODE AND PC_COMP_CODE=SLMA_COMP_CODE
AND SLMA_ACNO = PC_SL_ACNO
AND SLMA_LDGRCTL_YEAR = DBO.GLAS_VALIDATIONS_GET_OPEN_YEAR(PC_COMP_CODE)
如果我僅執行第一個查詢,就會得到5條記錄...
如果我加入上述兩個查詢,如:
SELECT
PC_COMP_CODE,
PC_SL_LDGR_CODE,
PC_SL_ACNO ACCOUNT,
COUNT(PC_CHEQUE_NO) CHQS,
SUM(CONVERT(FLOAT, PC_AMOUNT)) CHQ_AMT,
COAD_PTY_FULL_NAME
FROM GLAS_PDC_CHEQUES
LEFT OUTER JOIN GLAS_SBLGR_MASTERS
ON( SLMA_COMP_CODE=PC_COMP_CODE AND
SLMA_LDGRCTL_CODE = PC_SL_LDGR_CODE AND SLMA_ACNO = PC_SL_ACNO )
LEFT OUTER JOIN GLAS_PTY_ADDRESS ON( SLMA_COMP_CODE = COAD_COMP_CODE AND SLMA_ADDR_ID = COAD_ADDR_ID)
WHERE PC_COMP_CODE = '1'
AND PC_DISCD IS NULL AND SLMA_LDGRCTL_YEAR = DBO.GLAS_VALIDATIONS_GET_OPEN_YEAR(PC_COMP_CODE)
GROUP BY PC_SL_LDGR_CODE,
PC_SL_ACNO ,PC_COMP_CODE,COAD_PTY_FULL_NAME
ORDER BY PC_SL_ACNO
那我只得到2條記錄。
加入后,我需要顯示5條記錄.....我該怎么做?
我在第一個查詢中添加了所有字段,以使其更具可讀性,並更有可能使人理解事物。 不幸的是,我當時放棄了。
結果如下,但不管它有什么價值,我都會在黑暗中開槍,說你想加入UNION 。
SELECT id, first_name, account, COUNT(pants) pants, SUM(CONVERT(FLOAT, cost)) amount
FROM table1
WHERE id = '1' AND discount IS NULL GROUP BY first_name, account ,id
ORDER BY account
SELECT full_name,id, first_name, account, department, coffee_pref, pants, acct_no
FROM table2,table3,table1
WHERE secret_code = '1'
AND public_code = secret_code
AND addr_id = phone_id
AND nickname = first_name
AND id=public_code
AND movie_rating = account
AND album_year = DBO.get_random_year(id)
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