SQL To remove new line of column length more than 4000(CLOB Datatype)

Question

I have a column with data such as -

COL                  STR
1                    
                    Testing data entered. This is not true but I will like to enter more data

2                   
                    New column like Noida Limtd.Able to work with or without supervision
                    Able to lead a team and  accomplished given tasks
                    Able to assemble and disassemble finals and undercarriage.

The STR column is in clob format and has a newline as the first line in the text. To remove the newline I am using -

 translate(STR, ' '||chr(10)||chr(11)||chr(13), ' ') r

The above code is working for most data but some of the text in STR column is > 4000 and I am getting the error -

ORA-22835: Buffer too small for CLOB to CHAR or BLOB to RAW conversion  translate

I can reduce the STR column length to 4000 and translate but SUBSTR is not working.

Answer 1

Can you get away with using REPLACE? It has CLOB handling where TRANSLATE does not, eg

SQL> create table t ( x clob );

Table created.

SQL> insert into t values ( '123' ||rpad('x', 5000,'x'));

1 row created.

SQL> update t set x = x || x;

1 row updated.

SQL> update t set x = x || x;

1 row updated.

SQL> update t set x = x || x;

1 row updated.

SQL> update t set x = x || x;

1 row updated.

SQL> update t set x = x || x;

1 row updated.

SQL> commit;

Commit complete.

SQL> select length(x) from t;

 LENGTH(X)
----------
    160096

SQL>
SQL> select translate(x,'123','456') from t;
select translate(x,'123','456') from t
                 *
ERROR at line 1:
ORA-22835: Buffer too small for CLOB to CHAR or BLOB to RAW conversion (actual: 160096, maximum: 32767)


SQL>
SQL> select replace(x,'123','456') from t;

REPLACE(X,'123','456')
--------------------------------------------------------------------------------
456xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx