Creating mysql Function with parameters

Question

delimeter //
DROP function IF EXISTS get_seq_next//
create function get_seq_next(
IN sequence_ref_ varchar(30)
) returns int(11) unsigned
BEGIN
    DECLARE seq_val_  int(11) unsigned;
    LOCK TABLE ga_seq_tab WRITE;
    select sequence_no into seq_val_ from ga_seq_tab where sequence_ref=sequence_ref_;
    if not seq_val_ is null then
        update ga_seq_tab set sequence_no=sequence_no+1 where sequence_ref=sequence_ref_;
    end if
    UNLOCK TABLES;
    return seq_val;
END //
DELIMETER ;

I'm trying to create a function but it keeps saying I have syntax errors and I am not sure what is wrong with it

please post the error. one problem is that you've misspelled delimiter — mechanical_meat
– mechanical_meat, Commented Jun 13, 2012 at 20:58
#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'IN sequence_ref_ varchar(30) ) returns int(11) unsigned BEGIN DECLARE seq_va' at line 2 (I don't have the delimiter fields there when posting it as I am doing so through phpmyadmin (field that specifies it for me)) — James
– James, Commented Jun 14, 2012 at 12:34
I removed the "IN" and I got further, I guess all parameters are IN parameters in a function — James
– James, Commented Jun 14, 2012 at 12:44

Al3x · Accepted Answer · 2014-04-23 23:14:49Z

1

Try removing the IN reserved word in the parameters list.

answered Apr 23, 2014 at 23:14

Al3x

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