Please help asap
| From: | Jeff Moncrieff | Date: | Fri, 04 Jan 2002 18:27:10 +0000 |
| Subject: | Please help asap | ||
| Groups: | php.db | ||
| Request: | Send a blank email to php-db+get-15403@lists.php.net to get a copy of this message | ||
Hello
I am trying make a script fatch my data form my mysql database. I use
this script but when execute it is says
Warning: printf(): too few arguments in
/home/httpd/html/larken/database.php on line 47
any hints
Thanks Jeff
<html>
<body>
<?php
$db = mysql_connect("****", "root");
mysql_select_db("larken",$db);
$result = mysql_query("SELECT * FROM custgeninfo",$db);
echo "<table border=1>\n";
echo "<tr><td>Name of contact</td><td>Name of
company</td><td>
Address</td><td>Postal Code</td><td>Format Mail</td><td>Email
Address<td>model</td><td>Comment</td><td>telephone</td><td>Notes</td><td>County</td><td>
ID </td></tr>\n";
while ($myrow = mysql_fetch_row($result)) {
// 1 2 3 4
5 6 7
printf("<tr><td>%s%s</td><td>%s%s</td><td>%s%s</td><td>%s%s%s</td><td>%s%s</td><td>%s</td><td>%s%s%s</td><td>%s</td><td>%s%s</td><td>%s</td><td>%s%s%s%s%s</td></tr>\n",$myrow[1],$myrow[2],$myrow[3],$myrow[4],$myrow[5],$myrow[6],$myrow[7],$myrow[8],$myrow[9],$myrow[10]);
}
echo "</table>"
?>
</body>
</html>