Re: DataGrid Problems

mark — Wed May 21, 2008 10:48 pm

Author: mark

Posted: Wed May 21, 2008 10:48 pm

PinkCloud, please open a bug report on pear.php.net about this issue.

Re: DataGrid Problems

vanila2000 — Wed May 21, 2008 7:29 am

Author: vanila2000

Posted: Wed May 21, 2008 7:29 am

No. unfortunately, this post has not received any reply at all and I had to switch to simple php programming for the report ...

Re: DataGrid Problems

PinkCloud — Tue May 20, 2008 10:42 am

Author: PinkCloud

Posted: Tue May 20, 2008 10:42 am

Hi there,

I have been trying to attempt to exactly the same thing as you, but when i try to change the label the data does not print.

I know this post is quite old now, but has anyone got any answers on how to do this?

Re: DataGrid Problems

vanila2000 — Wed Jun 28, 2006 11:18 am

Author: vanila2000

Posted: Wed Jun 28, 2006 11:18 am

Can anyone please help me out??? it is really urgent .. Thanks in adv.

DataGrid Problems

vanila2000 — Fri Jun 23, 2006 1:39 pm

Author: vanila2000

Posted: Fri Jun 23, 2006 1:39 pm

Hi,
I downloaded the example script given for datagrid. my script is like this:
<html>
<style type="text/css">

</style>
<?php
/* Includes */
require_once "PEAR.php";
define("DB_DATAOBJECT_NO_OVERLOAD",true); /* This is needed for some buggy versions of PHP4 */
require_once "DB/DataObject.php";
require_once "Structures/DataGrid.php";

/* Database and DataObject setup */
$options = &PEAR::getStaticProperty("DB_DataObject","options");
$options["database"] = "oci8://user:passwd@host/dbname"; (i am giving my own user here)
$options["proxy"] = "full";
DB_DataObject::debugLevel(0);

class DataObject_Fruits extends DB_DataObject
{
var $__table = "fruits";
var $id;
var $name;
var $stock;
var $price;
}

/* Instantiate */
$dataobject = new DataObject_Fruits();
$dataobject->keys("id");
$datagrid =& new Structures_DataGrid(10); /* 10 rows per table */

/* Required in order to use the "fields" and "labels" options */
//global $datagridOptions;
$options["generate_columns"] = true;

/* The fields we want to display */
$options["fields"] = array ("id","name", "stock", "price");

/* Translate the fields names into user-friendly labels */
$options["labels"] = array (
"id"=>"Product Id",
"name" => "Product Name",
"stock" => "Quantity in Stock",
"price" => "Price (€)"
);
//print_r($datagridOptions);
echo "<p align=\"center\" > size=\"4\"> FRUIT WISE REPORT <br>";
/* Pass these options at binding time */
$datagrid->bind($dataobject,$print_r);
/* Get a reference to the Renderer object */
$renderer =& $datagrid->getRenderer();

/* For the <table> element : */
$renderer->setTableAttribute("class", "fruits");

/* For every odd <tr> elements */
$renderer->setTableOddRowAttributes(array ("class" => "odd"));

$pagingHtml = $renderer->getPaging();
echo "<p class=\"paging\">Pages : $pagingHtml</p>";

/* And render */
$datagrid->render();
?>

Here, whenever i try to change the labels in the output, the data is not printed. That is in the bind statement like this:

$datagrid->bind($dataobject,$datagridOptions);

However, if i give

$datagrid->bind($dataobject,$print_r);
I am getting the data with the table field names as the labels.

I want to change the lables. How do i do it? Even though in the code i have defined the labels, it is not printed. Please tell me how to do it.

PEAR Forum

Re: DataGrid Problems

Re: DataGrid Problems

Re: DataGrid Problems

Re: DataGrid Problems

DataGrid Problems