De PHPhulp kroeg
Resource id #3
Code (php)
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<?PHP
mysql_connect('', '', '');
mysql_select_db('');
$result = mysql_query("SELECT naam FROM members WHERE id='1'");
echo $result;
?>
mysql_connect('', '', '');
mysql_select_db('');
$result = mysql_query("SELECT naam FROM members WHERE id='1'");
echo $result;
?>
Het geeft een error, namelijk "Resource id #3"
hoe moet ik dit oplossen?
Dat doe je bvb zo
Code (php)
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<?php
while($data = mysql_fetch_assoc($result)){
echo '<pre>'; print_r($data); echo '</pre>';
}
?>
while($data = mysql_fetch_assoc($result)){
echo '<pre>'; print_r($data); echo '</pre>';
}
?>
Voorbeeld:
Code (php)
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<?PHP
mysql_connect('', '', '');
mysql_select_db('');
$query = "SELECT naam FROM members WHERE id='1'";
$result = mysql_query($query) or trigger_error('Error trying: '.$query);
$row = mysql_fetch_assoc($result);
echo $row['naam'];
?>
mysql_connect('', '', '');
mysql_select_db('');
$query = "SELECT naam FROM members WHERE id='1'";
$result = mysql_query($query) or trigger_error('Error trying: '.$query);
$row = mysql_fetch_assoc($result);
echo $row['naam'];
?>
Gewijzigd op 04/01/2012 11:34:00 door TJVB tvb