Comments by harristars

harristars 27-May-19 7:10am View

my trigger is not updating. right now i dont have an error but its not updating the column that it is suppose to update.what could be the problem.

DELIMITER //
CREATE TRIGGER `after_update_debt_defaulters` AFTER INSERT ON `defaulters`
 FOR EACH ROW BEGIN
       IF ( EXISTS (
            SELECT DISTINCT `KRAPIN` FROM defaulters WHERE `Debt_Amount` = '0'
           )   
       )
        THEN
            UPDATE users 
            SET users.debtor = '0'
            WHERE KRAPIN = NEW.KRAPIN;
        END IF;
END

harristars 27-May-19 6:04am View

i have included the DELIMITER // at the beginning and also at the END // and now i am getting the another
#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'WHERE KRAPIN = NEW.KRAPIN;
END IF;
END' at line 11

harristars 27-May-19 5:54am View

i have removed but its giving me this error
#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near '' at line 10

harristars 27-May-19 5:40am View

even after removing semicolon i get the same error.
#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'AFTER insert on defaulters
FOR EACH ROW
BEGIN
IF ( EXISTS (
' at line 2

harristars 27-May-19 5:20am View

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'AFTER insert on defaulters
FOR EACH ROW
BEGIN
IF ( EXISTS (
' at line 2

harristars 13-May-19 23:30pm View

Undefined variable: transaction_id (0)

when i uncomment my declaration it is giving me the above error

// $transaction_id = transaction::find ($req->transaction_id);

harristars 2-May-19 21:27pm View

I want it to check if mypin has balance from a table called default ers. So if my has balance it show you the balance and prevent you from logining in.

harristars 2-May-19 2:03am View

syntax error, unexpected 'if' (T_IF), expecting ']'

harristars 2-May-19 1:32am View

Deleted

i have update the code this is what i seeing

 syntax error, unexpected 'if' (T_IF), expecting ']'

harristars 10-Apr-19 7:02am View

i have fixed

harristars 16-Oct-18 7:17am View

plus i am testing all this in local environment this localhost (development server) on local machine

harristars 16-Oct-18 7:15am View

i have already allowed .htaccess in this file httpd.conf .because .htaccess is always disabled in
<Directory />
#Allowoverride none
Allowoverride All

harristars 16-Oct-18 7:12am View

idont think so because when .php is allowed its working.http://localhost/sites/website/plan_right_website%c2%a0/controller/home/home.php

harristars 15-Oct-18 8:45am View

inside controller/home/

there is a file called home.php

harristars 15-Oct-18 7:41am View

i have updated the question just check

harristars 31-Jul-18 0:09am View

I want to know exactly what is the reason why the datatable is not not filtering single record.

harristars 25-Jul-18 11:09am View

My problem is not sql injection. In fact I have no problem with my db it is doing exactly what is required to do and I have a file that is taking care of security related issue. I think my question is very clear. It's not db related.it about a JavaScript code that is ment to populate data into input box once you select the data in option box.

harristars 23-Jul-18 14:32pm View

yah.but i will share with you my code on git

harristars 13-Jul-18 6:39am View

thanks men it have worked

harristars 13-Jul-18 5:37am View

this the error i am get
/editmaster.php?item_id=Notice: Undefined variable: row in C:\xampp\htdocs\sites\admin_panel\student_center_system\controller\admin\viewitems.php on line 49

harristars 13-Jul-18 5:31am View

yah the previous error now is gone.The problem the big problem is nothing is being displayed on the editmaster.php when i click edit button on viewitems.php.

harristars 13-Jul-18 4:46am View

when i select edit nothing is showing in the form

harristars 13-Jul-18 4:22am View

yah but it did not work

harristars 20-Jun-18 7:00am View

yah

harristars 10-May-18 8:47am View

When I right click "SystemFunction" Visualization Studio does not suggest anything.

harristars 23-Mar-18 6:27am View

can you please let me see the code you test and its working

harristars 23-Mar-18 6:25am View

CREATE TRIGGER product_id
AFTER INSERT
ON test

BEGIN

-- variable declarations

-- trigger code
String.Format("PC{0}//POS", id.ToString(D2));

END;
the above code i have tried it several tmes and its not working

harristars 19-Mar-18 22:02pm View

Yah and it did not work.I have tried many times but still it gives me Mysql said : #1303 -can't create a trigger from within another stored routine.I wish if u could give me an example that is working that would be better

harristars 7-Mar-18 9:17am View

I am still stuck .I am using phpmyadmin.every time i write the trigger I am getting this error
Mysql said : #1303 -can't create a trigger from within another stored routine

harristars 7-Mar-18 9:01am View

@John Simmons I am still stuck on this code..how can I call the above static method on your code.

harristars 23-Feb-18 9:54am View

i have tried it's not working

harristars 23-Feb-18 9:50am View

 One or more errors have occurred while processing your request:
The following query has failed: "CREATE DEFINER=`root`@`localhost` TRIGGER `product_code` AFTER INSERT ON `inventory_table` FOR EACH ROW CREATE TRIGGER product_code AFTER INSERT ON test FOR EACH ROW BEGIN UPDATE test set textid = concat('PC', LPAD(NEW.id, 5, '0'), '/POS') WHERE id=NEW.id; END"

MySQL said: #1303 - Can't create a TRIGGER from within another stored routine

The above code is the error i am getting when i execute the code for trigger

harristars 23-Feb-18 5:50am View

private void Genarate_productcode()
{
product_code.Text = IncrementID2(" ", 9);

}
is this the correct way to call the function generating the product_code

harristars 21-Feb-18 6:44am View

Oooh I have let me try again and go through it

harristars 21-Feb-18 3:38am View

if the number doesn't parse correctly, set it to zero
number = 0;

on this line there is a problem

harristars 21-Feb-18 3:20am View

CREATE TRIGGER product_id
AFTER INSERT
ON test

BEGIN

-- variable declarations

-- trigger code
String.Format("PC{0}//POS", id.ToString(D2));

END;

this where i am hanging i got no idea to be sincere.i have never written a mysql trigger and i am ready to learn with this one

harristars 19-Feb-18 8:55am View

still i can't get it .i am been looking to various examples of writing a trigger but still my is not working

harristars 14-Feb-18 17:30pm View

explain to me with an example please

harristars 14-Feb-18 17:21pm View

in trigger do i to set the Definer or Definition after selecting the table .or how do i set it

harristars 14-Feb-18 17:14pm View

i am using xampp for MySQL database

harristars 14-Feb-18 17:14pm View

i am actually at the database.i know how to set the auto increment but this one where by i am suppose to have my product_id look like this PC01/POS. i am unable

harristars 14-Feb-18 15:27pm View

i have tried i am unable to go about it .i have studied the link plus the code u provide..but where exactly should i place the MYSQL Trigger in the code

harristars 14-Feb-18 10:42am View

let me try this too

harristars 14-Feb-18 10:33am View

let me try it and see if it will work

Comments by harristars (Top 44 by date)