Draw graph with sql variables

Question

I used Google graph to draw a graph by below code, but the problem is that my data is on a sql table, how can I put my table variables into this graph?

<html>
<head>
<script type="text/javascript" src="https://www.google.com/jsapi"></script>
<script type="text/javascript">
google.load("visualization", "1", {packages:["corechart"]});
  google.setOnLoadCallback(drawChart);
  function drawChart() {
    var data = google.visualization.arrayToDataTable([
      ['Year', 'Sales', 'Expenses'],
      ['2004',  1000,      400],
      ['2005',  1170,      460],
      ['2006',  660,       1120],
      ['2007',  1030,      540]
    ]);

    var options = {
      title: 'Company Performance'
    };

    var chart = new google.visualization.LineChart(document.getElementById('chart_div'));
    chart.draw(data, options);
  }
</script>
</head>
<body>
<div id="chart_div" style="width: 900px; height: 500px;"></div>
</body>
</html>

You must SELECT the data on the php file where you prepare this page you posted and then echo the data as the format you show in your sample code. — Snow Blind
– Snow Blind, Commented Jul 29, 2013 at 11:13
You first need to extract the data from the database - the recommend approach for PHP in this day and age is to just PHP Data Objects — John Parker
– John Parker, Commented Jul 29, 2013 at 11:15
@SnowBlind number of rows are not specified, every time it changed, how to echo them? — Sunny Leone
– Sunny Leone, Commented Jul 29, 2013 at 11:27
@r.vengadesh I have table named result with two column: attempt and login, every time that visitor loged in one row adds — Sunny Leone
– Sunny Leone, Commented Jul 29, 2013 at 11:29

r.vengadesh · Accepted Answer · 2013-07-29 12:51:32Z

0

<?php


$result = mysql_query("SELECT year,sales,expenses from tablename");
$value=array();
while($r = mysql_fetch_assoc($result)) {
$year=$r['year'];
$sales=$r['sales'];
$expenses=$r['expenses'];
$val="[".$year.",".$sales.",".$expenses."]";
array_push($value,$val );
}
$final_value = implode(",", $value);

?>

put this $final_value into google chart

function drawChart() {
    var data = google.visualization.arrayToDataTable([
      <?php echo $final_value?>
    ]);

    var options = {
      title: 'Company Performance'
    };

    var chart = new google.visualization.LineChart(document.getElementById('chart_div'));
    chart.draw(data, options);
  }

edited Jul 29, 2013 at 12:51

answered Jul 29, 2013 at 11:53

r.vengadesh

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2 Comments

Sunny Leone Over a year ago

what is $array in $final_value = implode(",", $array); ?

r.vengadesh Over a year ago

sorry change $array into $ value @SunnyLeone

Pandiyan Cool · Accepted Answer · 2013-07-29 11:15:05Z

0

Retrieve your data From tables by querying database and convert it to JSON string. Then pass your json string to your variable

var data=/*your json string*/

answered Jul 29, 2013 at 11:15

Pandiyan Cool

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2 Comments

Sunny Leone Over a year ago

I am new to PHP and SQL, could you help me more, how to do this?

Pandiyan Cool Over a year ago

You can get some ideas by visiting posts like this stackoverflow.com/questions/10045341/…

Snow Blind · Accepted Answer · 2013-07-29 12:29:30Z

0

$db = new mysqli('localhost', 'user', 'pass', 'demo');

if($db->connect_errno > 0){
    die('Unable to connect to database [' . $db->connect_error . ']');
}

$sql = "SELECT attempt, login FROM result";

if(!$result = $db->query($sql)){
    die('There was an error running the query [' . $db->error . ']');
}

echo "var data = google.visualization.arrayToDataTable([['Attempt', 'Login']";

while($row = $result->fetch_assoc()){
    echo ",['".$row['attempt'] ."','".$row['attempt']."']";
}

echo "]);";

$result->free();

$db->close();

edited Jul 29, 2013 at 12:29

answered Jul 29, 2013 at 11:50

Snow Blind

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2 Comments

Sunny Leone Over a year ago

there is error on echo "var data = google.visualization.arrayToDataTable([['Attempt', 'Login']";

Snow Blind Over a year ago

@SunnyLeone Editted the answer, please try again.

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