[eluser]Unknown[/eluser]
hi,
I'm updating a task manager done in codeigniter.
I created ajax for comunication with controller.
Code:
$('#company_id').on("change",function(e){
//localiza o cliente
$.ajax({
url:'<?=base_url() ?>projects/localizarcliente',
dataType: "text",
cache:false,
type:'GET',
data:'email='+e.val,
beforeSend:function(){
$('#resultado').html('Lendo...');
},
success:function(data){
$('#resultado').html('retornou:'+data);
}
});
But my function localizarcliente return theme the aplication!
Code:
function localizarcliente(){
echo 'teste';
}
This return:
A PHP Error was encountered
Severity: Notice
Message: Undefined variable: widgets
Filename: theme/application.php
Line Number: 106
How i can to make for my code only return my function?
