insert into select 要怎么用。报错显示ora-00947：没有足够的值

[u010821061]

t.data_type ,
c.COMMENTS
FROM user_tab_columns t, user_col_comments c
WHERE t.table_name = c.table_name
AND t.column_name = c.column_name
AND t.table_name = 'HDSA0001'));






另外：SELECT t.table_name ,
t.column_name ,
t.data_type ,
c.COMMENTS
FROM user_tab_columns t, user_col_comments c
WHERE t.table_name = c.table_name
AND t.column_name = c.column_name
AND t.table_name = 'HDSA0001' 本身是没有问题的。可以查出来数据

[CT_LXL]

select 查询出来的值得个数和要INSERT的值得个数不匹配

[u010821061]

是匹配的啊。后面查出来4条数据。一共5条数据

[Yole]

seq_Health_Dictionary.nextval 这个是啥东西？

insert into Health_Dictionary(dictionary_id,belong_table_name,column_name,column_data_type,column_describe)

(SELECT  @@identity  as dictionary_id,

t.table_name as belong_table_name,
         t.column_name as column_name,
         t.data_type as column_data_type,
         c.COMMENTS as column_describe
  FROM   user_tab_columns t, user_col_comments c
 WHERE       t.table_name = c.table_name
         AND t.column_name = c.column_name
         AND t.table_name = 'HDSA0001')

[u010821061]

引用 1 楼 zlloct 的回复:

select 查询出来的值得个数和要INSERT的值得个数不匹配

引用 3 楼 u010192842 的回复:

seq_Health_Dictionary.nextval 这个是啥东西？

insert into Health_Dictionary(dictionary_id,belong_table_name,column_name,column_data_type,column_describe)

(SELECT  @@identity  as dictionary_id,

t.table_name as belong_table_name,
         t.column_name as column_name,
         t.data_type as column_data_type,
         c.COMMENTS as column_describe
  FROM   user_tab_columns t, user_col_comments c
 WHERE       t.table_name = c.table_name
         AND t.column_name = c.column_name
         AND t.table_name = 'HDSA0001')

seq_Health_Dictionary.nextval 是一个序列

[Yole]

那也要写到select 里面啊！ 不然语法有问题！

[一颗向往数据的心丶]

insert into table(cols1...) select cols2... from table2 where ... cols1 和 cols2的个数必须相同、

[CT_LXL]

引用 4 楼 u010821061 的回复:

[quote=引用 1 楼 zlloct 的回复:] select 查询出来的值得个数和要INSERT的值得个数不匹配

引用 3 楼 u010192842 的回复:

seq_Health_Dictionary.nextval 这个是啥东西？

insert into Health_Dictionary(dictionary_id,belong_table_name,column_name,column_data_type,column_describe)

(SELECT  @@identity  as dictionary_id,

t.table_name as belong_table_name,
         t.column_name as column_name,
         t.data_type as column_data_type,
         c.COMMENTS as column_describe
  FROM   user_tab_columns t, user_col_comments c
 WHERE       t.table_name = c.table_name
         AND t.column_name = c.column_name
         AND t.table_name = 'HDSA0001')

seq_Health_Dictionary.nextval 是一个序列[/quote] 怀疑这个 @@identity 引起的,换成一个常量试一下吧