Question

我的full outer join子句中出现无效的标识符错误。我认为这可能与我的select语句有关，但不确定在哪里。以下是查询：

select  h.laborlev2nm "Labor Acct", h.laborlev2dsc "Dept Name",
    to_char(b.pay_date, 'MM/DD/YYYY') as "PDate", 
    to_char(d.start_date, 'MM/DD/YYYY') as "SDate", 
    to_char(d.end_date, 'MM/DD/YYYY') as "EDate",
    c.personnum "ID", c.fullnm "FullName",
    d.pay_type "Pay Type", f.name "Paycode", f.paycode_id "Paycode ID", 
    j.dags_objectcode "DAGS", NULL AS "Raw Hours", 


'Regular' AS "Retro Paycode Name",
101 AS "Retro Paycode ID", 0 AS "Computed Hours", 0 AS "Retro Hours", 

CASE 
    WHEN f.name = 'CH to Salaried Reg Pay Adj' THEN li.amount
    WHEN f.name = 'Change in BU Reg Pay Adj' THEN li.amount
    WHEN f.name = 'FTE Change Reg Pay Adj' THEN li.amount
    WHEN f.name = 'LOA Overpayment Adjustment' THEN li.amount
    WHEN f.name = 'LOA Underpayment Adjustment' THEN li.amount
    WHEN f.name = 'New Hire Reg Pay Adj' THEN li.amount
    WHEN f.name = 'Overpayment Adjustment' THEN li.amount
    WHEN f.name = 'Retro Adjustment' THEN li.amount
    WHEN f.name = 'Salaried to CH Reg Pay Adj' THEN li.amount
    WHEN f.name = 'Termination Pay Adj' THEN li.amount
    WHEN f.name = 'Underpayment Adjustment' THEN li.amount
    ELSE NULL
END AS "Retro Dollars",


b.gross_pay "Gross", c.ftepct "FTE", 'Regular' "Reg/Cas",  bu.external_key "BU",  
d.pay_adjustment_id "Pay Adj ID", d.hourly_rate "H Rate", d.pay_type "P type"
--jc.jobtitledsc "Job Title"

from 
TKCSOWNER.KSS_EMP_JOB_INFO a , TKCSOWNER.KSS_PAYROLL b,  TKCSOWNER.PERSON c, 
TKCSOWNER.KSS_PAYROLL_DETAIL d, TKCSOWNER.KSS_PAYCODE_MAP f, TKCSOWNER.KSS_BARGAINING_UNIT BU,
TKCSOWNER.LABORACCT h, TKCSOWNER.KSS_DAGSPAY_OBJCD_MAP j

full outer join TKCSOWNER.KSS_PAY_ADJ_LINE_ITEM li on (d.pay_adjustment_id = li.pay_adjustment_id)

该错误表明我的d.pay_adjustment_id是最后一行代码的无效标识符。

感谢您提前获取任何建议

Answer 1

正如在评论和接受的答案中所讨论的那样，你绝不应该混合使用Oracle＆＃34;逗号语法＆＃34;和SQL标准（ANSI）语法在同一查询中连接。这不是非法的，但语法规则是如此复杂，你不可能让它们正确 - 而且，没有理由这样做。

针对您的具体问题：

该错误与上次加入外部连接无关。如果将其更改为内部连接，则会出现相同的错误。相反：当您混合使用Oracle语法和ANSI语法时，首先使用ANSI语法进行连接，就像括在括号中一样。您是全外连接（或实际上是ANSI连接）别名j和li的表。此时，只有名称j和li在范围内;此时d不可见。这就是它告诉你的错误信息。

您可以非常轻松地重新创建相同的问题：

SQL> select 1 from dual a, dual b inner join dual c on a.dummy = c.dummy;
select 1 from dual a, dual b inner join dual c on a.dummy = c.dummy
                                                  *
ERROR at line 1:
ORA-00904: "A"."DUMMY": invalid identifier

相反：

SQL> select 1 from dual a, dual b inner join dual c on b.dummy = c.dummy;

         1
----------
         1

您已经在接受的答案中使用正确的方法。

Answer 2

当表格用逗号分隔并且通过where子句包含其他任何内容以限制效果时，您正在创建一个CROSS JOIN。所以每个逗号都有一个CROSS JOIN。

这就是我理解您的查询的方式，它涉及多表笛卡尔积：

SELECT h.laborlev2nm "Labor Acct"
       , h.laborlev2dsc "Dept Name"
       , to_char(b.pay_date, 'MM/DD/YYYY') AS "PDate"
       , to_char(d.start_date, 'MM/DD/YYYY') AS "SDate"
       , to_char(d.end_date, 'MM/DD/YYYY') AS "EDate"
       , c.personnum "ID"
       , c.fullnm "FullName"
       , d.pay_type "Pay Type"
       , f.NAME "Paycode"
       , f.paycode_id "Paycode ID"
       , j.dags_objectcode "DAGS"
       , NULL AS "Raw Hours"
       , 'Regular' AS "Retro Paycode Name"
       , 101 AS "Retro Paycode ID"
       , 0 AS "Computed Hours"
       , 0 AS "Retro Hours"
       , CASE WHEN f.NAME = 'CH to Salaried Reg Pay Adj' THEN li.amount WHEN f.NAME = 'Change in BU Reg Pay Adj' THEN li.amount WHEN f.NAME = 'FTE Change Reg Pay Adj' THEN li.amount WHEN f.NAME = 'LOA Overpayment Adjustment' THEN li.amount WHEN f.NAME = 'LOA Underpayment Adjustment' THEN li.amount WHEN f.NAME = 'New Hire Reg Pay Adj' THEN li.amount WHEN f.NAME = 'Overpayment Adjustment' THEN li.amount WHEN f.NAME = 'Retro Adjustment' THEN li.amount WHEN f.NAME = 'Salaried to CH Reg Pay Adj' THEN li.amount WHEN f.NAME = 'Termination Pay Adj' THEN li.amount WHEN f.NAME = 'Underpayment Adjustment' THEN li.amount ELSE NULL END AS "Retro Dollars"
       , b.gross_pay "Gross"
       , c.ftepct "FTE"
       , 'Regular' "Reg/Cas"
       , bu.external_key "BU"
       , d.pay_adjustment_id "Pay Adj ID"
       , d.hourly_rate "H Rate"
       , d.pay_type "P type"
      --jc.jobtitledsc "Job Title"
FROM TKCSOWNER.KSS_EMP_JOB_INFO a
CROSS JOIN TKCSOWNER.KSS_PAYROLL b
CROSS JOIN TKCSOWNER.PERSON c
CROSS JOIN TKCSOWNER.KSS_PAYROLL_DETAIL d
CROSS JOIN TKCSOWNER.KSS_PAYCODE_MAP f
CROSS JOIN TKCSOWNER.KSS_BARGAINING_UNIT BU
CROSS JOIN TKCSOWNER.LABORACCT h
CROSS JOIN TKCSOWNER.KSS_DAGSPAY_OBJCD_MAP j
FULL JOIN TKCSOWNER.KSS_PAY_ADJ_LINE_ITEM li ON (d.pay_adjustment_id = li.pay_adjustment_id)

Oracle SQL全外连接无效标识符

2 个答案: