jquery autosuggest add heading before items - php

Question

I am using jquery autosuggest and would like to add a default heading just before all the items appear. For example when the user starts his search the first drop down item would say 'We Suggest:' followed by the list of items. Here is what I currently have
$term=$_GET["term"];
$query=mysql_query("SELECT DISTINCT region FROM business where region like '$term%' limit 10 ");
$json=array();
while($region=mysql_fetch_array($query)){
$json[]=array(
'value'=> $region["region"],
'label'=>$region["region"]
);
}
echo json_encode($json);
I have a reasonable understanding of php and jquery but not json
Thanks for any help

Answer 1

NEW ANSWER:
I may have misread your question before. Perhaps you are just asking for this:
$( "#mylist" ).autocomplete({
source: myjsonarray,
open: function( event, ui ) {
$('.ui-autocomplete').find('li:first').prepend("<div>We Suggest:</div>")
}
});
Access the "open" event, add a div before the first list item.
OLD ANSWER:
//make an empty container
$myresults=array();
//manually create your first item
$myfirstitem=array("value"=>"We suggest","label"=>"We suggest")
//put your first item in the container
array_push($myresults,$myfirstitem);
// fetch your data
while $result=mysql_query("SELECT phrase from mytable"){
//put each item in the container
array_push($myresults,array(
"value"=>$result,
"label"=>$result
));
}
//$myresults is now full of phrases.
///Uncomment this to check it if you want.
//var_dump($myresults);
//encode and echo your result
echo json_encode($myresults);
The above code illustrates one way to do what you want to do. With that said, I'd probably not put "We Suggest" as an actual option... if you're presenting a list of baseball cards, it wouldn't make sense for the user to choose "we suggest" as their favorite baseball card...
Also, don't let json bother you- it's just a "notation". This example is very simple- think of it and use it as any other array.
You can also do this more efficiently- you don't need to return a label and a value... but that's another conversation.
Also, it's important that you do not use mysql_* functions. Look into mysqli and pdo.
Good luck, I hope that helped.