我已经运行了这个查询:
SELECT
OWNER,
TABLE_NAME,
COLUMN_NAME,
DATA_TYPE,
DATA_LENGTH,
(CASE
WHEN DATA_PRECISION IS NULL THEN 0
ELSE DATA_PRECISION
END) DATA_PRECISION,
(CASE
WHEN DATA_SCALE IS NULL THEN 0
ELSE DATA_SCALE
END) DATA_SCALE,
NULLABLE,
COLUMN_ID
DEFAULT_LENGTH,
DATA_DEFAULT,
(CASE
WHEN DATA_DEFAULT IS NULL THEN '0'
ELSE DATA_DEFAULT
END) DATA_DEFAULT1
FROM
all_tab_columns
WHERE
table_name LIKE 'TABLE1';
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但它在 column 引发错误DATA_DEFAULT:
ORA-00932:不一致的数据类型:预期的 CHAR 得到了 LONG
00932。00000 - “不一致的数据类型:预期的 %s 得到了 %s”
我该如何解决?
谢谢!
你不能用LONG. Oracle 仍然在数据字典中使用它们是一个 PITA。
您可以使用 XML:
select owner
, table_name
, column_name
, data_type
, data_length
, case
when data_precision is null then 0
else data_precision
end data_precision
, case
when data_scale is null then 0
else data_scale
end data_scale
, nullable
, column_id
, default_length
, case
when default_length is null then '0'
else
extractvalue
( dbms_xmlgen.getxmltype
( 'select data_default from user_tab_columns where table_name = ''' || c.table_name || ''' and column_name = ''' || c.column_name || '''' )
, '//text()' )
end as data_default
from all_tab_columns c
where table_name like 'TABLE1';
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从 12.1 开始,您可以内联编写自己的查找函数:
with
function get_default(tab varchar2, col varchar2) return varchar2
as
dflt varchar2(4000);
begin
select c.data_default into dflt
from user_tab_columns c
where c.table_name = upper(tab)
and c.column_name = upper(col);
return dflt;
end get_default;
select owner
, table_name
, column_name
, data_type
, data_length
, case
when data_precision is null then 0
else data_precision
end data_precision
, case
when data_scale is null then 0
else data_scale
end data_scale
, nullable
, column_id
, default_length
, get_default(c.table_name, c.column_name) as data_default
from all_tab_columns c
where table_name like 'TABLE1%'
/
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或者当然制作一个独立的函数或包函数来做同样的事情。
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