如何从Oracle数据库的CHAR列获取数据/值的大小？

Question

I have a table in Oracle DB with column CHAR(11 ).

I have inserted the value 'abcd ' to it. (Note there is single space after abcd)

When I do LENGTH(column) - it returns 11 instead of 5.

How do I get the data size instead of column size?

Reason I need it because I am reading the value from table1 having CHAR datatype and inserting the data to table2 having VARCHAR datatype .

So in this case, it adds 'abcd ' (abcd with trailing 7 spaces) to table2 instead of 'abcd ' (with single space).

How do I avoid this? If I can get hold of data size from CHAR column then I can do that.

Please let me know how it can be achieved?

Answer 1

If you want a variable length column then use VARCHAR2 and not CHAR .

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A CHAR column is a fixed length string that will be right-padded with space (ASCII 32) characters. So, if you want the length of a CHAR(11) then it will always be 11 characters regardless of whether you tried to insert a 1-character or an 11-character string as it will be right padded.

If you want to ignore the right padding then RTRIM the string before measuring the length (however this will also remove any intentional trailing white spaces as well as the unintentional ones).

SELECT LENGTH( RTRIM( column_name ) ) FROM table_name

For example:

Oracle Setup :

CREATE TABLE table_name ( column_name CHAR(11) );

INSERT INTO table_name ( column_name )
  SELECT 'abcd' FROM DUAL UNION ALL
  SELECT ' abc' FROM DUAL;

Query :

SELECT LENGTH( column_name ),
       LENGTH( TRIM( column_name ) ),
       LENGTH( RTRIM( column_name ) )
FROM   table_name

Output :

\nLENGTH(COLUMN_NAME) |  LENGTH(TRIM(COLUMN_NAME)) |  LENGTH(RTRIM(COLUMN_NAME)) \n------------------: |  ------------------------: |  -------------------------: \n                 11 |  4 |  4 \n                 11 |  3 |  4 \n

db<>fiddle here

As you can see, LENGTH on its own includes the entire padded string. Using TRIM gives the wrong answer as it also removes leading white-space as well as trailing white-space. RTRIM gives the correct length.

Answer 2

Please try with trim .

select trim(column_name) from table;