How to do a conditional where clause with where in PL/SQL within Procedure - sql

Question

I have a pretty simple Stored Procedure that I am in trouble to do because i'm new to SQL and PL/SQL. I Have a table with a name column that is a varchar(55).
I discovered that if the user executes my procedure with an empty string as a paramter the LIKE statment brings all rows from TABLE1
SELECT *
FROM TABLE1
WHERE COLUMN LIKE VARIABLE || '%'
AND...
So I tried to change the query so if the VARIABLE is passed with a empty string it can still perform other conditions in the where statment.
SELECT *
FROM TABLE1
WHERE (VARIABLE <> '' AND COLUMN LIKE VARIABLE || '%')
AND...
But now wherever I pass as variable ('', NULL, 'anystring') I get no rows returned.
How can I build a query that validates if the variable is different of empty string and if it is it performs the LIKE statment with the variable correctly?

Answer 1

If I understand you correctly, it is not difficult thing to do. You can use conditional WHERE clause using CASE WHEN. So your query will support different scenarios, something like this:
SELECT *
FROM TABLE1
WHERE (CASE WHEN variable IS NULL AND column IS NULL THEN 1
WHEN variable LIKE '%' AND column LIKE variable||'%' THEN 1
ELSE 0
END) = 1
AND...
Basically, it checks if the variable = '' then it will compare the column against ''. Otherwise, it will compare it against variable||'%'.
Notice, Oracle treats empty string of the type VARCHAR as NULL (this does not apply to CHAR). So, in the first scenario we compare against NULL.

Answer 2

Hello Just a thought for this we can use Dynamic sql too. If you may try this approach. Hope it helps.
CREATE OR REPLACE PROCEDURE SPS_TEST_OUT(
p_input_in IN VARCHAR2
)
AS
lv_sql LONG;
lv_where VARCHAR2(100);
BEGIN
lv_where:= CASE WHEN p_input_in IS NULL OR p_input_in = '' THEN
''
ELSE
' AND COLUMN1 LIKE '''||p_input_in||'''%'
END;
lv_sql:='SELECT * FROM TABLE
WHERE 1 = 1
' ||lv_where;
dbms_output.put_line(lv_sql);
END;