oracle最新sql回忆性演练1(二)

2015-11-21 02:03:26 · 作者: · 浏览: 14
ct * from bbb; alter table bbb drop column id; select * from example; select * from xxx; --USER_TAB_COLS中记录了用户表的列信息 SELECT USER_TAB_COLS.TABLE_NAME as 表名, USER_TAB_COLS.COLUMN_NAME as 列名, USER_TAB_COLS.DATA_TYPE as 数据类型, USER_TAB_COLS.DATA_LENGTH as 长度, USER_TAB_COLS.NULLABLE as 是否为空, USER_TAB_COLS.COLUMN_ID as 列序号, user_col_comments.comments as 备注 FROM USER_TAB_COLS inner join user_col_comments on user_col_comments.TABLE_NAME = USER_TAB_COLS.TABLE_NAME and user_col_comments.COLUMN_NAME = USER_TAB_COLS.COLUMN_NAME; --关联到表的所有字段信息 select col.column_name, uc.constraint_type, case uc.constraint_type when 'P' then '√' else '' end "PrimaryKey" from user_tab_columns col left join user_cons_columns ucc on ucc.table_name = col.table_name and ucc.column_name = col.column_name left join user_constraints uc on uc.constraint_name = ucc.constraint_name and uc.constraint_type = 'P' where col.table_name = 'example'; --查询某个表中的外键字段名称、所引用表名、所应用字段名 select distinct (col.column_name), r.table_name, r.column_name from user_constraints con, user_cons_columns col, (select t2.table_name, t2.column_name, t1.r_constraint_name from user_constraints t1, user_cons_columns t2 where t1.r_constraint_name = t2.constraint_name and t1.table_name = 'example') r where con.constraint_name = col.constraint_name and con.r_constraint_name = r.r_constraint_name and con.table_name = 'example';