jquery combine two mysql values - php

Question

I use jquery (JQwidgets)to display values in a grid. I have a database with multiple columns and lets say i have in column A the value "http://www.google.com" and in column B the value "Google".
What i want is to display "Google" in the grid and when people click on it it opens the url.
Right now i have this code wich gets the variables from the database.
var source =
{
datatype: "json",
datafields: [
{ name: 'sitename', type: 'string'},
{ name: 'url', type: 'string'},
],
url: 'http://www.site.com/data.php',
cache: false
};
After that the values are displayed using this...
$("#jqxgrid").jqxGrid(
{
pagesize : 25,
pagesizeoptions: ['25', '50', '100'],
width : 1170,
theme:"black",
source: dataAdapter,
pageable: true,
autoheight: true,
selectionmode: 'multiplecellsextended',
columns: [
{ text: 'Website', datafield: 'website' , width: 700, cellsrenderer: linkrenderer},
]
});
So what i cannot get to work is to combine the two values. What i tried is this...
var website = function (row, datafield, value) {
return 'sitename' + 'url';
}

Answer 1

You haven't set a linkrenderer function.
Above it, add
var linkrenderer = function(row, column, value, defaultHtml, columnSettings, rowData) {
return '<a href="' + rowData.url + '">' + value + '</div>';
}
The last column rowData is a JSON object of everything already rendered, so maybe add the columns
columns: [{text: 'URL', datafield: 'url' , width: 100}, {text: 'Website', datafield: 'sitename' , width: 700, cellsrenderer: linkrenderer}]
Hiding the URL column
$("#jqxgrid").jqxGrid('hidecolumn', 'url');