SQL查询以测试字符串值是否包含回车符

Question

试图找出测试VARCHAR列值是否以回车结尾的正确方法。试过这个，但是不起作用，数据库是Oracle 11g ...

select name from myTable where name LIKE '%\r' OR name like '%\n'

Answer 1

要查找包含不可打印字符(例如回车符或垂直制表符或行尾)的值，可以使用regexp_like函数。在您要显示行的情况下，特定列的字符串值最后包含回车符的行，可以使用类似的查询。

select *
  from your_table_name
 where regexp_like(trim(string_column), '[[:space:]]$')

Demo

对评论的回答
Trim函数默认情况下会删除前导和尾随空格，并且不会删除 回车或 行尾字符。让我们进行一个简单的测试:

SQL> create table Test_Table(
  2    id number,
  3    col1 varchar2(101)
  4  );

Table created

SQL> insert into Test_Table (id, col1)
  2    values(1, 'Simple string');

1 row inserted

SQL> commit;

Commit complete

SQL> insert into Test_Table (id, col1)
  2    values(1, 'Simple string with carriage return at the end' || chr(13));

1 row inserted

SQL> commit;

Commit complete

SQL> insert into Test_Table (id, col1)
  2    values(1, '   Simple string with carriage return at the end leading and trailing spaces' || chr(13)||'   ');

1 row inserted

SQL> commit;

Commit complete

SQL> insert into Test_Table (id, col1)
  2    values(1, '  Simple string leading and trailing spaces  ');

1 row inserted

SQL> commit;

Commit complete

SQL> select *
  2    from test_table;

        ID COL1
--------------------------------------------------------------------------------
         1 Simple string
         1 Simple string with carriage return at the end
         1    Simple string with carriage return at the end leading and trailing spaces
         1   Simple string leading and trailing spaces

SQL> 
SQL> select *
  2    from test_table
  3   where regexp_like(trim(col1), '[[:space:]]$')
  4  ;

        ID COL1
----------------------------------------------------------------------------------
         1 Simple string with carriage return at the end
         1    Simple string with carriage return at the end leading and trailing spaces

SQL>