a logic issue in creating multiple graphs using phpgraphlib

ajoo · March 9, 2015

Hi,

I am creating charts using phpgraphlib - a very easy to use graphing library. once the graph is created in graph.php it just needs to be called from the HTML as follows:-

<html>
    <img src="graph.php" />
</html>

and the graph is plotted. Within graph.php a data array is required which carries the values to be plotted.

Now I need to plot about 10 graphs and in my HTML file I have created a 10 Tabbed layout that should each display one of the 10 graphs. In my Graph.php I have have also created the 10 data arrays from mysql that I require for the 10 graphs. So almost all the work has been done in graph.php and all I need to do is to configure the graph and display the 10 graphs like this.

//configure graph
$graph->addData($dataArray);
$graph->setTitle("Sales by Group");
$graph->setGradient("lime", "green");
$graph->setBarOutlineColor("black");
$graph->createGraph();
?>

But then I need the 10 graphs in 10 files like graph1.php, graph2.php and so on which can be called from their respective tabbed HTML and I need to pass each of these 10 files their data array calculated in graph.php.

If I was to repeat the entire code of graph.php in graph1.php, graph2.php and so on for all 10 files & each of these files had to calculate their own data and invoke the graph it will work but it would be a lot of ( 10 times of ) recalculation of the data arrays. ( All data for the 10 graphs can be calculated from the same single query).

I was just wondering if there can be some method by which I can somehow achieve this without repeating the same process 10 times.

I hope I was able to explain the issue well. Looking for some way out of this.

Thanks all.

Barand · March 9, 2015

If the only thing that changes in each of the 10 graphs is the data then use the same "graph.php" but pass the different data (and maybe a caption) each time. Use something like this

$data_1 = array(1,2,3);
$data_2 = array(5,6,7);

$g1_data = json_encode($data_1);    // convert to a string
$g2_data = json_encode($data_2);    // convert to a string

echo "<img src='graph.php?caption=chart_1&data=$g1_data' />";
echo "<img src='graph.php?caption=chart_2&data=$g2_data' />";

In graph.php the data will be in $_GET['data']

$data_array = json_decode($_GET['data'], 1);  // reconstruct data array  
// create graph

ajoo · March 10, 2015

Hi Guru Barand,

Thanks for the reply. Somehow it fails to display the graph. I have used your files almost as is. I am appending my code below. I do not get any errors either.

index1.php: ( same as yours)

<?php
$data_1 = array(1,2,3,4,5,6,7);
$data_2 = array(5,6,7,7,7,6,5);

$g1_data = json_encode($data_1);    // convert to a string
$g2_data = json_encode($data_2);    // convert to a string

echo "<img src='graph1.php?caption=chart_1&data=$g1_data' />";
// echo "<img src='testGraph1.php?caption=chart_2&data=$g2_data' />";
?>

Graph1.php:

<?php
include('../phpgraphlib-master/phpgraphlib.php');
$mysqli_driver = new mysqli_driver();
$mysqli_driver->report_mode = MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT;	// For error handling as pointed by guru barand

try
{
	$data_array = json_decode($_GET['data'], 1);  // reconstruct data array  
	print_r($data_array);
	$caption = $_GET['caption'];
}
catch (Exception $e) 
{
    echo 'Caught exception: ',  $e->getMessage(), "\n";
}

$graph = new PHPGraphLib(650,350);
$graph->addData($data_array);	
$graph->setLineColor("#ff0000");
$graph->setGradient('Red', 'maroon');	 

$graph->setTitle('caption');
$graph->setBars(true);
$graph->setLine(true);

$graph->setDataValues(true);
$graph->setDataValueColor('maroon');

$graph->createGraph();
?>

I cannot spot the error. Please help.

phpgraphlib was downloaded from : http://www.ebrueggeman.com/phpgraphlib/download

Thanks loads !

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