Creating a line graph taking data from database - php

Question

I tried to create a simple bar chart in D3 taking data from database. The idea is to fetch year and value from the database using the selection buttons of parameters and country and display it in the form of simple bar chart. I have two selection buttons and a submit button. The code for that is like this :
<form action ="data.php" method ="post">
<tr>
<th valign="bottom">Select country:</th>
<td >
<select name="country" class="">
<option value= "NPL">Nepal</option>
<option value= "IND">India</option>
</select>
</td>
<th valign="bottom">Select parameters:</th>
<td >
<select name="indices" class="">
<option value= "foodfinal">Vulnerability on food</option>
<option value= "waterfinal">Vulnerability water </option>
</select>
</td>
<td valign="top">
<input type="submit" class="action" />
</td>
</tr>
</form>
Foodfinal and waterfinal are the names of tables in the database called 'Climate'. NPL and IND are the countrycodes. My PHP code ( data.php ) is like this :
<?php
$username = "root";
$password = "";
$host = "localhost";
$database="climate";
$country="AFG";
$country=$_POST["country"];
$indices=$_POST["indices"];
$server = mysql_connect($host, $username, $password);
$connection = mysql_select_db($database, $server);
$myquery = "SELECT `year`, `values` FROM `$indices` WHERE `countrycode`= '$country'";
$query = mysql_query($myquery);
if ( ! $query ) {
echo mysql_error();
die;
}
$data = array();
for ($x = 0; $x < mysql_num_rows($query); $x++) {
$data[] = mysql_fetch_assoc($query);
}
echo json_encode($data);
include("linegraph.html");
mysql_close($server);
?>
My problem is that linegraph.html had been working well before the use of the selection forms but this time, it only displays the output of data.php which looks like this.
{"year":"1995","values":"0.525999"},{"year":"1996","values":"0.570037"},{"year":"1997","values":"0.563966"},{"year":"1998","values":"0.513896"},{"year":"1999","values":"0.5346"},{"year":"2000","values":"0.552691"},{"year":"2001","values":"0.545319"},{"year":"2004","values":"0.543972"},{"year":"2005","values":"0.543299"},{"year":"2006","values":"0.546682"},{"year":"2007","values":"0.550066"},{"year":"2008","values":"0.549449"},{"year":"2009","values":"0.548832"},{"year":"2010","values":"0.548215"},{"year":"2011","values":"0.547536"},{"year":"2012","values":"0.547536"}]
Would you please help me.

Answer 1

change this
include("linegraph.html");
To ( as a best practice thing if the "include" is required to display the page just use require )
require("linegraph.html");
And add at the top of the php
ini_set('display_errors', 1);
You will probably find your page missing ( at the very least your life will be easier with error reporting turned on )
oh and change this part too,
for ($x = 0; $x < mysql_num_rows($query); $x++) {
$data[] = mysql_fetch_assoc($query);
}
To this
$data = array();
while ( false !== ( $row = mysql_fetch_assoc($query))){
$data[] = $row;
}
It probably won't fix your problem ( hard to do without the other html page ) but its the preferred way to do this type of loop.
Update: Here you can't pass both html and json the way you are ( just dawned on me ) you have to pick one. Your problem is a structural one, as you cant do both things at the same time, where does the json data go, it just floats on top of the html? Sorry but it wont work like that, either you already have the html and fill it in with the returned json data, or you build the html with the data already in it and send it back.
change this
echo json_encode($data);
include("linegraph.html");
to ( now you are sending valid json / with the html but you'll have to assemble it client side )
ob_start();
include("linegraph.html");
$html = ob_get_clean();
$responce = array( 'data'=>$data, 'html'=>$html );
echo json_encode($responce);
At the very least your output will change. Hard to say how to do it ( again ) without the other page.
Ok,,,, when I asked to echo something at the bottom did you do this?
echo json_encode($data);
include("linegraph.html");
mysql_close($server);
echo "works";
?> //id remove this end tag it's not needed and can cause problems if there is space ( line returns ) after it. Because it will make you json which you may or may not need invalid.
Oyyy. this is .. uhh
if ( ! $query ) {
echo mysql_error();
die;
}
you can just do like this.
if ( ! $query ) {
die ('Query error' . mysql_error());
}