Mission:
Implement autocomplete of departments(saved in departments table) in employee form field called department.
A user enters a few spellings of department name
That brings up list of the names matching departments
The user select one and that's it.
Platforms
- CakePhp 2.1
- Jquery UI Autocomplete(part of Jquery UI library version 1.8.18)
Database Model
Emplyee (id, first_name,last_name,department_id)
department(id,name)
so in my add.ctp file ajax call is something like
$( "#auto_complete" ).autocomplete({
source: function( request, response ) {
$.ajax({
url: "/employees/showDepartment",
dataType: "json",
data: {
featureClass: "P",
style: "full",
maxRows: 12,
name_startsWith: request.term
},
success: function( data ) {
alert("success--");
response( $.map( data, function( item ) {
//alert(item);
return {
label: item.name,
value: item.id
}
}));
}
});
},
minLength: 2,
select: function( event, ui ) {
log( ui.item ?
"Selected: " + ui.item.label :
"Nothing selected, input was " + this.value);
},
open: function() {
$( this ).removeClass( "ui-corner-all" ).addClass( "ui-corner-top" );
},
close: function() {
$( this ).removeClass( "ui-corner-top" ).addClass( "ui-corner-all" );
}
});
i have a action in my EmployeeController called show_depatment()
public function getAddress() {
$this->autoRender = false;// I do not want to make view against this action.
$this->log($this->params->query['name_startsWith'] , 'debug');
$str = $this->params->query['name_startsWith'];
$this->log($str, 'debug');
$this->layout = 'ajax';
$departments = $this->Employee->Department->find('all', array( 'recursive' => -1,
'conditions'=>array('Department.name LIKE'=>$str.'%'),
'fields'=>array('name', 'id')));
//$this->set('departments',$departments);
$this->log($departments, 'debug');
echo json_encode($departments);
}
I dont want show_department action to have any view so i have made $this->autoRender = false;
but it is not working as expected.
when i debug the response using firebug in response and HTLM section it shows
[{"Department":{"name":"Accounting","id":"4"}}] // when i type "acc" in input field
Question
- How to make it to display in form field.
- echo json_encode($departments); is it right method of sending response in json format?
- when i alert in sucess part of ajax call (alert(item);) it gives error as "undefined"