Add data from a database to an array

Question

I need to write a script that is able to search for a particular result, put it into a graph. I've already developed the graph. However, when hovered over the section of the graph, there would be a list of the data that is in that section.

Example:
25% of the pie chart is boys. When hovered over the boys, it will show a list of the names of the boys. In other words, i would need to be able to retrieve the multiple data from the database, put it in a list and echo out the results.

so far I've managed to code this:

$data = array();<br>
    while ($row = mysqli_fetch_array($result))<br>
    {<br>
    $data[]= $row;<br>
}<br>
echo "$data";

However I do not know what I'm echoing out. Please help me

$data= $row; for data assign data to an array . please can you provide the data structure of your chart and database — Anik Islam Abhi
– Anik Islam Abhi, Commented Nov 18, 2014 at 8:34
Uhm. on Hover you can call (with javascript) an ajax call that will ask a php page information about that. — Marco Mura
– Marco Mura, Commented Nov 18, 2014 at 8:35
I don't think it's a good idea to make a call to the database every time someone hovers over a pie chart. It's better if you get the results on page load and then show them using JS when someone hovers their mouse over the pie chart. — Alternatex
– Alternatex, Commented Nov 18, 2014 at 8:41
I'm sorry, I dont have enough reputation to be able to input an image. >< @AnikIslamAbhi gwt-google-apis.googlecode.com/svn/wiki/SimpleVizQuery-1.png — crypatix
– crypatix, Commented Nov 18, 2014 at 8:58
Something like that, however, when hovered, it would show the list of names retrieved from the database @AnikIslamAbhi — crypatix
– crypatix, Commented Nov 18, 2014 at 8:58

Yang Yuan · Accepted Answer · 2014-11-18 08:39:26Z

1

i guess this is what you want do to...

$data = array();
while ($row = mysqli_fetch_array($result)) {
    array_push($data, $row);
}
print_r($data); // dump

answered Nov 18, 2014 at 8:39

Yang Yuan

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2 Comments

crypatix Over a year ago

Thank you so much, but I just want 1 section of the data. I tried that code by using echo $data[0]['gene_id'] But it only showed 1 of the results. should i also put it in a loop?

Yang Yuan Over a year ago

maybe... array_push($data, $row['gene_id']); or just 'SELECT gene_id from ...' in your SQL

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