How to create array using javascript with a database mysql?

Question

anyone can help me about this problem thank you in advance this is a google chart I need to change the task and Hours per day dynamically, so basically need a database that can connect to the database and will put them on array how can I do that.

 <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([
              ['Task', 'Hours per Day'],
              ['Work',     11],
              ['Eat',      2],
              ['Commute',  2],
              ['Watch TV', 2],
              ['Sleep',    7]
            ]);

            var options = {
              title: 'My Daily Activities'
            };

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

I can connect to the database I only need how to create a array for javascript basically I need this code to change into dynamic: var data = google.visualization.arrayToDataTable([ ['Task', 'Hours per Day'], ['Work', 11], ['Eat', 2], ['Commute', 2], ['Watch TV', 2], ['Sleep', 7] ]); — Hanz
– Hanz, Commented Apr 5, 2013 at 8:37

S.Visser · Accepted Answer · 2013-04-05 08:51:55Z

3

First you need to design a database structure. For example:

** Activities ** 
| ID |   Task   | Hours  | 
| 1  |   Work   |   11   |
| 2  |   Eat    |   2    |
| 3  | Cummute  |   2    |
| 4  | Watch TV |   2    | 
| 5  |  Sleep   |   7    |

Now you need to make a database connection. ( easiest way, there a much better ways to do this, this is just an example )

<?php
 $link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
 if (!$link) {
     die('Not connected : ' . mysql_error());
 } 

 // make foo the current db
 $db_selected = mysql_select_db('your_db_name', $link);
 if (!$db_selected) {
     die ('Can\'t use database : ' . mysql_error());
 }
?>

You need to build this code dynamic:

   var data = google.visualization.arrayToDataTable([
      ['Task', 'Hours per Day'],
      ['Work',     11],
      ['Eat',      2],
      ['Commute',  2],
      ['Watch TV', 2],
      ['Sleep',    7]
    ]);

In this example, the title is static

   var data = google.visualization.arrayToDataTable([
      ['Task', 'Hours per Day'],
      <?php
         $query = mysql_query("SELECT * FROM Activities");
         $count = mysql_row_count($query);
         $i = 0;
         $last = ',';
         while ($row = mysql_fetch_array($result)) {
          $i++;
          if($i == $count) { $last = ''; )
          echo "['". $row['Task'] .", ". $row['Hours'] ." '] " . $last
         }
       ?>
    ]);

edited Apr 5, 2013 at 8:51

answered Apr 5, 2013 at 8:42

S.Visser

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7 Comments

Kumar Saurabh Sinha Over a year ago

dont use mysql functions always use mysqli functions.

S.Visser Over a year ago

@SaurabhSinha thats why I say in it "Now you need to make a database connection. ( easiest way )" I'm not suggesting its the best way. if you want to do it the best way use PDO...

Kumar Saurabh Sinha Over a year ago

you need not have $count = mysql_row_count($query); and then check that if condition for the count. the while keyword will only run for the rows returning true.

Kumar Saurabh Sinha Over a year ago

if($i == $count) { $last = ','; ) this line will produce an error. why do we need to have that last paranthesis

Hanz Over a year ago

Thank you Visser I will try your code now. I will send a feedback later and thank you to all who has a suggestion.

|

Kumar Saurabh Sinha · Accepted Answer · 2013-04-05 08:34:56Z

-1

try the following:

Make DB Connect with mysqli_connect and query the DB for the resultset

Assign the PHP variable as mentioned below:

          ['Task', 'Hours per Day'],
          ['Work',     <?php echo $work; ?>],
          ['Eat',      <?php echo $eat; ?>],
          ['Commute',  <?php echo $commute; ?>],
          ['Watch TV', <?php echo $watchTv; ?>],
          ['Sleep',    <?php echo $sleep; ?>]

The variables :

$work
$eat
$commute
$watchTv
$sleep

are the PHP variables which contains corresponding values from DB

I hope this would help.

answered Apr 5, 2013 at 8:34

Kumar Saurabh Sinha

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5 Comments

Kumar Saurabh Sinha Over a year ago

@S.Visser: why not dynamic, could you please explain

Hanz Over a year ago

I need the "Task" and "Hours per day" to be dynamic and it comes to the database anyone can update the code.

Kumar Saurabh Sinha Over a year ago

both the tasklist and their value will be dynamic?

S.Visser Over a year ago

Saurabh, what I he goes sporting for one hour? then he need the edit the code to add the task sport to it. Not really dynamic.

Hanz Over a year ago

yup for example I added a new "task" equal to Listening and "Hours" equal to 3 and saved to the database how can I output that.

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