[SOLVED]

SOLVED...thanks 'ryupanqul'

Code:

$rs = mysql_query("SELECT unitID FROM unit ORDER BY case when unitID = 'All' THEN 0 ELSE 1 END, unitID");
$items = array();
while($row = mysql_fetch_object($rs)){

if ($row->unitID == 'ALL') {
$row->selected = true;
}

array_push($items, $row);
}

$result = $items;
echo json_encode ($result);

I have not been able to add the ,"selected":true into that array push item

Code:

$items = array();

while($row = mysql_fetch_object($rs)){
     // PSEDUO Code IF 'All' THEN
     array_push($items, $row + ',"selected":true');  [b]///I'M STUCK HERE...HOW TO CONCAT INTO THIS ITEM[/b]
     //ELSE
    array_push($items, $row);
   }
   
$result = $items;
echo json_encode ($result);

[How do I select the first item in the combox by default WHEN the combobox is popluated from a Database]

I have a combobox:

Code:

<input id="unit_FIELD" class="easyui-combobox"
        name="unit_FIELD" data-options=" url:'get_units.php',
valueField:'unitID',
textField:'unitID'
        ">

I'm populating the ComboBox just fine with the get_units.php code.

Code:

[{"unitID":"ALL"},{"unitID":"001"},{"unitID":"002"},{"unitID":"003"},{"unitID":"004"}]

I need to be able to have a default selected...How do I select the first item in the combox by default WHEN the combobox is popluated from a Database?

-Tom

as soon as I posted i found the error

my POST variables did not match 

Code:

yearFilter_POST

Can someone take a double look, and see why I cannot pass my year FILTEr values over to my PHP page...I think its becuase I;m trying to pass "<, >, =, <="...and that is breaking somehting???

I have two search fields:  Year and Year_Filter

Code:

<input id="year_FIELD" class="easyui-combobox"
        name="year_FIELD" data-options=" data: [
{yearID:'',year:'ALL'},
{yearID:'2013',year:'2013'},
{yearID:'2012',year:'2012'},
{yearID:'2011',year:'2011'},
{yearID:'2010',year:'2010'},
{yearID:'2009',year:'2009'}
],
valueField:'yearID',
textField:'year',
panelHeight:'auto'
        ">

<input id="year_FILTER" class="easyui-combobox"
        name="year_FILTER" data-options=" data: [
{value:'', label:'No Filter'},
{value:'=', label:'Equal'},
{value:'<>', label:'Not Equal'},
{value: '>', label:'Greater Then'},
{value:'<', label:'Less Then'},
{value:'>=', label:'Greater Then OR Equal'}
],
valueField:'value',
textField:'label',
panelHeight:'auto'
        ">

When the search button is clicked...those values are sent the PHP page:

Code:

<script type="text/javascript">
function doSearch(){
$('#tt').datagrid('load',{
                                yearPOST: $('#year_FIELD').combobox('getValue'),
                                yearFilterPOST: $('#year_FILTER').combobox('getValue')
});
}
</script>

My PHP code captures IS NOT CAPTURING THE YEAR_FILTER Value:

Code:

$year_year = isset($_POST['yearPOST']) ? mysql_real_escape_string($_POST['yearPOST']) : '2012';
$year_FILTER = isset($_POST['yearFilter_POST']) ? $_POST['yearFilter_POST'] : '=';

Here is my WHERE statement:

Code:

$where = "year_year $year_FILTER $year_year";