Oracle PL/SQL : remove “space characters” from a string

Question

In my Oracle 10g database I would like to remove \"space characters\" (spaces, tabs, carriage returns...) from the values of a table field.

Is TRANSLATE() t

Answer 1

I'd go for regexp_replace, although I'm not 100% sure this is usable in PL/SQL

my_value := regexp_replace(my_value, '[[:space:]]*',''); 

Answer 2

To replace one or more white space characters by a single blank you should use {2,} instead of *, otherwise you would insert a blank between all non-blank characters.

REGEXP_REPLACE( my_value, '[[:space:]]{2,}', ' ' )

Answer 3

select regexp_replace('This is a test   ' || chr(9) || ' foo ', '[[:space:]]', '') from dual;

REGEXP_REPLACE
--------------
Thisisatestfoo

Answer 4

Since you're comfortable with regular expressions, you probably want to use the REGEXP_REPLACE function. If you want to eliminate anything that matches the [:space:] POSIX class

REGEXP_REPLACE( my_value, '[[:space:]]', '' )


SQL> ed
Wrote file afiedt.buf

  1  select '|' ||
  2         regexp_replace( 'foo ' || chr(9), '[[:space:]]', '' ) ||
  3         '|'
  4*   from dual
SQL> /

'|'||
-----
|foo|

If you want to leave one space in place for every set of continuous space characters, just add the + to the regular expression and use a space as the replacement character.

with x as (
  select 'abc 123  234     5' str
    from dual
)
select regexp_replace( str, '[[:space:]]+', ' ' )
  from x

Answer 5

To remove any whitespaces you could use:

myValue := replace(replace(replace(replace(replace(replace(myValue, chr(32)), chr(9)),  chr(10)), chr(11)), chr(12)), chr(13));

Example: remove all whitespaces in a table:

update myTable t
    set t.myValue = replace(replace(replace(replace(replace(replace(t.myValue, chr(32)), chr(9)), chr(10)), chr(11)), chr(12)), chr(13))
where
    length(t.myValue) > length(replace(replace(replace(replace(replace(replace(t.myValue, chr(32)), chr(9)), chr(10)), chr(11)), chr(12)), chr(13)));

or

update myTable t
    set t.myValue = replace(replace(replace(replace(replace(replace(t.myValue, chr(32)), chr(9)), chr(10)), chr(11)), chr(12)), chr(13))
where
    t.myValue like '% %'

Answer 6

Shorter version of:

REGEXP_REPLACE( my_value, '[[:space:]]', '' )

Would be:

REGEXP_REPLACE( my_value, '\s')

Neither of the above statements will remove "null" characters.

To remove "nulls" encase the statement with a replace

Like so:

REPLACE(REGEXP_REPLACE( my_value, '\s'), CHR(0))