用於測試字符串值是否包含回車符的SQL查詢

Question

試圖找出測試VARCHAR列值是否以回車結束的正確方法。試過，但它不工作，數據庫是Oracle 11g ...

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

Answer 1

查找包含非打印字符，如回車或垂直選項卡或行尾值您可以使用regexp_like功能。在你的情況下，顯示行的最後一個特定列的字符串值包含回車，可以使用類似的查詢。

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

Demo

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答到的意見

Trim功能，默認情況下，刪除開頭和結尾的空格，它不會刪除線的回車或結束個字符。讓我們進行一個簡單的測試：

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> 

Answer 2

嘗試

SELECT name from myTable where name like '%'||chr(10) or name like '%'||chr(13)