Hi,
I will write a function that returns the result of a recordset.
I tried this:
Function getPersonen()
{
openDB(); //function that I implements in another file
$query = "SELECT * FROM Cursisten";
$result = mysql_query($query)
or die("Fout bij uitvoeren query");
$resultrow = mysql_fetch_array( $result );
closeDB(); //function that I implements in another file
mysql_free_result($result);
return $resultrow;
}
I call this function:
$resultaat = getPersonen();
// Printen resultaten in HTML
print "<table>\n";
while ($line = mysql_fetch_array($resultaat, MYSQL_ASSOC)) {
print "\t<tr>\n";
foreach ($line as $col_value) {
print "\t\t<td>$col_value</td>\n";
}
print "\t</tr>\n";
}
print "</table>\n";
In runtime I have the following error
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
result resource in
/home/schoolre/public_html/Hitek/Online/Registratie/showPerson.php on line 8
What is wrong?
Alain
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