Retrieving database column using JSON [migrated]
Posted
by
arokia
on Programmers
See other posts from Programmers
or by arokia
Published on 2013-06-25T13:39:46Z
Indexed on
2013/06/25
16:28 UTC
Read the original article
Hit count: 242
I have a database consist of 4 columns (id-symbol-name-contractnumber). All 4 columns with their data are being displayed on the user interface using JSON.
There is a function which is responisble to add new column to the database e.g (countrycode).
The coulmn is added successfully to the database BUT not able to show the new added coulmn in the user interface.
Below is my code that is displaying the columns.
Can you help me?
table.php $(document).ready(function () { // prepare the data var theme = getDemoTheme();
var source =
{
datatype: "json",
datafields: [
{ name: 'id' },
{ name: 'symbol' },
{ name: 'name' },
{ name: 'contractnumber' }
],
url: 'data.php',
filter: function()
{
// update the grid and send a request to the server.
$("#jqxgrid").jqxGrid('updatebounddata', 'filter');
},
cache: false
};
var dataAdapter = new $.jqx.dataAdapter(source);
// initialize jqxGrid
$("#jqxgrid").jqxGrid(
{
source: dataAdapter,
width: 670,
theme: theme,
showfilterrow: true,
filterable: true,
columns: [
{ text: 'id', datafield: 'id', width: 200 },
{ text: 'symbol', datafield: 'symbol', width: 200 },
{ text: 'name', datafield: 'name', width: 100 },
{ text: 'contractnumber', filtertype: 'list', datafield: 'contractnumber' }
]
});
});
data.php
<?php
#Include the db.php file
include('db.php');
$query = "SELECT * FROM pricelist";
$result = mysql_query($query) or die("SQL Error 1: " . mysql_error());
$orders = array();
// get data and store in a json array
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$pricelist[] = array(
'id' => $row['id'],
'symbol' => $row['symbol'],
'name' => $row['name'],
'contractnumber' => $row['contractnumber']
);
}
echo json_encode($pricelist);
?>
© Programmers or respective owner