I am trying to generate a chart based on the example provided by pchart. Here is my code:
<?php
/* Include the pData class */
include("pchart/class/pData.class.php");
/* Create the pData object */
$myData = new pData();
/* Connect to the MySQL database */
$db = mysql_connect("webhost", "user", "pass");
mysql_select_db("database",$db);
/* Build the query that will returns the data to graph */
$Requete = "SELECT * FROM `replies` WHERE `field` LIKE CONCAT ('%', Do you an interest in Green IT, '%')";
$Result = mysql_query($Requete,$db);
$Yes=""; $No=""; $Undecided="";
while($row = mysql_fetch_array($Result));
{
/* Push the results of the query in an array */
$Yes[] = $row["Yes"];
$No[] = $row["No"];
$Undecided[] = $row["Undecided"];
}
/* Save the data in the pData array */
$myData->addPoints($Yes,"Yes");
$myData->addPoints($No,"No");
$myData->addPoints($Undecided,"Undecided");
?>
The error I'm getting is this:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a4728588/public_html/charts.php on line 30
Which points to:
while($row = mysql_fetch_array($Result));
Any ideas on how to fix this so the chart is generated?
Thanks in advance
You have syntax error in your query. CONCAT()
is for merging strings. Dont need in your query. Should be like this
LIKE '%Do you an interest in Green IT%'
mysql_query($Requete,$db) or die(mysql_error());
to print the mysql erro - safarov 2012-04-04 18:35
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/a4728588/public_html/charts.php on line 3
which points to Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/a4728588/public_html/charts.php on line 3
Junaid Hussain 2012-04-04 18:49