function data_init(&$options, $memberInfo, &$args) {
//A3 Aqui filtraremos para q solo salga la data con un campo del loogeado
//
if(!$_POST['FilterField'][1] && !$_GET['FilterField'][1]){
/*
In the call below, we want to display records of the customers
table where the value of the 7th field is equal to 'New'.
*/
addFilter(1, 'and', 50, 'equal-to', 'tratante1');
}
return TRUE;
}
Using the example as blueprint and only changed it for the field 50 to be the word tratante.
Its the only one filter
Sorry to bother, but cant figure it out
a3
basic filter not working
-
- Veteran Member
- Posts: 104
- Joined: 2018-12-10 21:52
Re: basic filter not working
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 '<=> 'tratante1')) LIMIT 0, 50' at line 1
is what i get
is what i get
-
- Veteran Member
- Posts: 104
- Joined: 2018-12-10 21:52
Re: basic filter not working
Anyone?
Nor a PRO in php or Javascript but I think I know my way around...
Any help will be BLESS for me...
a3
Nor a PRO in php or Javascript but I think I know my way around...
Any help will be BLESS for me...
a3
-
- Veteran Member
- Posts: 104
- Joined: 2018-12-10 21:52
Re: basic filter not working
I havent figure it out yet, i know its maybe cause im missing something, but im lost at this point.
Anyone can help?
a3
Anyone can help?
a3
-
- Veteran Member
- Posts: 104
- Joined: 2018-12-10 21:52
Re: basic filter not working -->SOLVED<--
SOLVED! As a Good practice I post my solutions
The problem was all my FAULT!
The instruction was CLEAR: “ The filter index. The first filter has an index of 1, the second filter 2, ... etc.â€
People, COUNT APPGINI INDEX, not your MySQL INDEX, I was missing by a number, and that was it…
SOLVED!
The problem was all my FAULT!
The instruction was CLEAR: “ The filter index. The first filter has an index of 1, the second filter 2, ... etc.â€
People, COUNT APPGINI INDEX, not your MySQL INDEX, I was missing by a number, and that was it…
SOLVED!