首页 > 解决方案 > SQL 错误:ORA-00904:“GENDER”:无效标识符 00904。00000 -“%s:无效标识符”

问题描述

CREATE TABLE COM_MANAGEMENT (
    TITLE VARCHAR(20),
    name_a VARCHAR(20),
    POST VARCHAR(20),
    DATE_OF_BIRTH DATE,
    JOINING_DATE DATE,
    YEAR_OF_EXPERIENCE NUMBER(4),
    PHONE NUMBER(15),
    E_MAIL VARCHAR(60),
    ADDRESS VARCHAR(20),
    GENDER VARCHAR(10)
);

如果我想在这个 SQL 中插入数据,输出给我 SQL 错误:

ORA-00904: "GENDER": 无效标识符 00904. 00000 - "%s: invalid identifier" 这是错误

INSERT INTO COM_MANAGEMENT VALUES (TITLE('FRIENDS SHOP'),NAME_A('FOZLE'),POST('MD'), DATE_OF_BIRTH(1999-JAN-01), JOINING_DATE(1999-JAN-20),YEAR_OF_EXPERIENCE(4),PHONE(017554122114),
E_MAIL('APPSLF@GMAIL.COM'),ADDRESS('FDSJDF'), GENDER(MALE));

标签: oracle

解决方案


我认为您可能会在这里跨不同平台混合和匹配您的 SQL 语法。无论如何,这是一个快速修复

SQL>
SQL> CREATE TABLE COM_MANAGEMENT (
  2      TITLE VARCHAR(20),
  3      name_a VARCHAR(20),
  4      POST VARCHAR(20),
  5      DATE_OF_BIRTH DATE,
  6      JOINING_DATE DATE,
  7      YEAR_OF_EXPERIENCE NUMBER(4),
  8      PHONE NUMBER(15),
  9      E_MAIL VARCHAR(60),
 10      ADDRESS VARCHAR(20),
 11      GENDER VARCHAR(10)
 12  );

Table created.

SQL>
SQL> INSERT INTO COM_MANAGEMENT VALUES (TITLE('FRIENDS SHOP'),NAME_A('FOZLE'),POST('MD'), DATE_OF_BIRTH(1999-JAN-01),
  2  JOINING_DATE(1999-JAN-20),YEAR_OF_EXPERIENCE(4),PHONE(017554122114),E_MAIL('APPSLF@GMAIL.COM'),ADDRESS('FDSJDF'), GENDER(MALE
));
JOINING_DATE(1999-JAN-20),YEAR_OF_EXPERIENCE(4),PHONE(017554122114),E_MAIL('APPSLF@GMAIL.COM'),ADDRESS('FDSJDF'), GENDER(MALE))
                                                                                                                  *
ERROR at line 2:
ORA-00904: "GENDER": invalid identifier


SQL>
SQL>
SQL> INSERT INTO COM_MANAGEMENT
  2    (TITLE
  3   ,NAME_A
  4   ,POST
  5   ,DATE_OF_BIRTH
  6   ,JOINING_DATE
  7   ,YEAR_OF_EXPERIENCE
  8   ,PHONE
  9   ,E_MAIL
 10   ,ADDRESS
 11   ,GENDER
 12  )
 13  VALUES (
 14  'FRIENDS SHOP',
 15  'FOZLE',
 16  'MD',
 17  date '1999-01-01',
 18  date '1999-01-20',
 19  4,
 20  '017554122114',
 21  'APPSLF@GMAIL.COM',
 22  'FDSJDF',
 23  'MALE');

1 row created.

SQL>
SQL>

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