CURSOR WITH PARAMETER

Hey bright guys !! you sould know this

CURSOR report_cursor(v_namesearch VARCHAR2) IS
SELECT name, address
FROM emp AND name LIKE 'v_namesearch%';

There seems to be something wrong with the way I have defined the above cursor. Looks like, since v_namesearch is
a variable (& not a constant) , the singe quotes seems to be messing up with what I am trying to accomplish.

Shall appreciate any help.. as I seem to be stuck with this
issue >> >>

Thank you & Merry X'mas
-NK

CURSOR report_cursor(v_namesearch VARCHAR2) IS
SELECT name, address
FROM emp WHERE name LIKE 'v_namesearch%';

Pardon my typing the
the cursor declaration should read as above

thank,
nk

CURSOR report_cursor(v_namesearch VARCHAR2) IS
SELECT name, address
FROM emp WHERE name LIKE v_namesearch || '%';

Anything inside a quote is taken as a literal. Therefore, you cannot embed variables inside quotes.

Hope this helps,

- Chris

THANKS Chris..
But this doesn't seem to work at first glance. I will dig deeper!!
In the meanwhile let me share my views.

the clause
LIKE v_namesearch || '%';
seems to be the problem. May be because v_namesearch is a string variable, it also needs a single quotes. But how??

Any ideas.

nk

This should work:

LIKE ' '%'||v_namesearch||'%' ';

Try:

LIKE ''''|| '%'||v_namesearch||'%'||'''' OR
LIKE '''%'||v_namesearch||'%''';

Basically, '' represents " ' " .