Reusable scripts - Part1

Find the letter occurrences in the statement:

The below query is finding the number of occurrence letter “a” in the string.

select length('Find a letter in a string') - length(replace('Find a letter in a string', 'a','')) as "Number of occurrence" from dual ;

Number of occurrence
--------------------
2


Remove the alphabetic letters in the given string:

declare
l_start number;
len number;
i number;
p_str varchar2(1000) := 'aa123yksjds45';
p_str1 varchar2(1000);
res varchar2(1000);
begin
i:=0;
len :=length(p_str);
WHILE i <= len
loop
len := ASCII(substr(p_str,i,1));
IF len > 47 AND len < 58 then
i:= i+1;
else
p_str := replace(p_str,substr(p_str,i,1),'');
END IF;
end loop;
dbms_output.put_line(p_str);
end;

Result:

12345

Divide the given comma separated words:

declare v_comma_position number;
v_len number;
p_comma_val varchar2(1000) := 'value1,value2,value3,value4,value5'; p_comma_tmp varchar2(1000);
v_result varchar2(1000);
begin p_comma_val := p_comma_val ',';
v_len :=length(p_comma_val);
WHILE v_len >= 0
loop
v_comma_position := instr(p_comma_val,',')+1;
p_comma_tmp := '';v_result := '';
p_comma_tmp := p_comma_val;
p_comma_val:='';
v_result := substr(p_comma_tmp,1,v_comma_position-2);
p_comma_val := substr(p_comma_tmp,v_comma_position,v_len);
v_len := v_len - v_comma_position;
dbms_output.put_line( v_result);
end loop;
end;

Result:

value1
value2
value3
value4
value5