Using % to get all including null or empty values

Question

dear all;
Please find below the following sample data
create table t1
ID varchar2(200),
time_create date
insert into t1
    (id, time_create)
  values
    ('A', sysdate);
  insert into t1
    (id, time_create)
  values
    ('B', sysdate);
    insert into t1
    (id, time_create)
  values
    (null, sysdate);
    I have the following sql statement below
    select * from t1
    where t1.id like decode(:id, 'ALL', '%', :id);now, I have a situation where if the user input is ALL, I would like to get all results including the null values or space values, how can I modify the decode statement to do that.
This is the output I want below for ALL
ID  TIME_CREATE
A   5/23/2011 11:14:23 PM
B   5/23/2011 11:14:24 PM
     5/23/2011  11:14:25 PMAll help is appreciated. Thank you.

Answer 1

Both work in exactly the same manner.
The OR condition says either boolean expression needs to be true for the row to be returned, so when :id = 'ALL' (and bind variable is set to 'ALL')- all rows are returned regardless.
When id = :ID, only values that are equal to your bind variable will be returned.
You could modify to this to provide even more flexibility
select * from t1
where :ID = 'ALL'
OR like :ID;
Scott