Question

我正在检查列长度和精度，如果精度高于数据库，那么我应该执行alter system。

这里出现的错误是(COL_COUNT <> 0 AND TO_NUMBER(OLD_LENGTH) < 6,2 )

修改

DECLARE COL_COUNT NUMBER;
OLD_LENGTH number(6);
 old_prec number(6);
BEGIN
COL_COUNT:= 0;
SELECT DATA_LENGTH, DATA_PRECISION INTO OLD_LENGTH , old_prec  FROM USER_TAB_COLS WHERE TABLE_NAME='EX_EMPLOYEE' AND COLUMN_NAME='ID';
SELECT COUNT (1) INTO COL_COUNT FROM USER_TAB_COLUMNS WHERE TABLE_NAME='EX_EMPLOYEE'AND COLUMN_NAME='ID';
IF (COL_COUNT <> 0 AND to_number(OLD_LENGTH) < 6.2 ) THEN
EXECUTE IMMEDIATE ('ALTER TABLE EX_EMPLOYEE MODIFY ID NUMERIC(6,2)');
END IF;
END;
/

Answer 1

这不是您应该如何比较data_length和data_precision的值。将它们作为正确的数字进行比较，不要将其转换为字符串：

DECLARE 
  old_length NUMBER;
  old_prec   NUMBER;
BEGIN

  SELECT data_length, coalesce(data_precision,0)
     INTO old_length, old_prec 
  FROM user_tab_cols 
  WHERE table_name='EX_EMPLOYEE' 
    AND column_name='ID';

  IF (old_length <> 6 and old_prec <> 2) THEN
    EXECUTE IMMEDIATE ('ALTER TABLE EX_EMPLOYEE MODIFY ID NUMERIC(6,2)');
  END IF;

END;
/

请注意，第一次选择后的SELECT COUNT(*) 是无用的。如果没有这样的列，则第一个SELECT data_length, data_precision INTO将抛出ORA-01403: no data found异常，如果列存在，代码将永远不会到达检查。

Answer 2

Oracle使用小数周期。逗号把它丢掉了。尝试：

  DECLARE COL_COUNT NUMBER;
OLD_LENGTH varchar2(50);
BEGIN
COL_COUNT:= 0;
SELECT DATA_LENGTH||'.'||DATA_PRECISION INTO OLD_LENGTH  FROM USER_TAB_COLS WHERE TABLE_NAME='EX_EMPLOYEE' AND COLUMN_NAME='ID';
SELECT COUNT (1) INTO COL_COUNT FROM USER_TAB_COLUMNS WHERE TABLE_NAME='EX_EMPLOYEE'AND COLUMN_NAME='ID';
IF (COL_COUNT <> 0 AND to_number(OLD_LENGTH) < 6.2 ) THEN
EXECUTE IMMEDIATE ('ALTER TABLE EX_EMPLOYEE MODIFY ID NUMERIC(6,2)');
END IF;
END;

Answer 3

尝试使用“。”分隔符而不是逗号： select to_number('6.2') from dual; --works select to_number('6,2') from dual; --returns error

检查变量时oracle表达错误类型

3 个答案: