Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/accounts_e/emte/public_html/php/dataout.php3 on line 13
Tak wyglada kod strony (sluzy do pokazywania wpisow do ksiegi gosci)
Kod
<html>
<head>
<title>ksiega</title>
</head>
<body bgcolor="#ffffff">
<table border="1" width="100%" style="border-collapse:collapse" bordercolor="#BE7A27">
<?php
$db = mysql_connect("localhost", "emte", "*****");
mysql_select_db("ksiega", $db);
$sql = "select * from ksiega";
$result = mysql_query($sql);
while($row=mysql_fetch_array($result))
{
print("<tr><td width=\"20%\" bgcolor=\"#D1A978\">");
printf("<b>%s</b>\n",
$row["wpisimie"]);
printf("<a href=\"mailto:%s\">%s</a>\n",
$row["wpismail"]);
printf("<a href=\"%s\" target=\"_blank\">%s</a>\n",
$row["wpissite"]);
printf("%s</td>\n",
$row["data"]);
printf("<td width=\"80%\">%s</td></tr>\n",
$row["tekst"]);
}
?>
</table>
</body>
</html>
<head>
<title>ksiega</title>
</head>
<body bgcolor="#ffffff">
<table border="1" width="100%" style="border-collapse:collapse" bordercolor="#BE7A27">
<?php
$db = mysql_connect("localhost", "emte", "*****");
mysql_select_db("ksiega", $db);
$sql = "select * from ksiega";
$result = mysql_query($sql);
while($row=mysql_fetch_array($result))
{
print("<tr><td width=\"20%\" bgcolor=\"#D1A978\">");
printf("<b>%s</b>\n",
$row["wpisimie"]);
printf("<a href=\"mailto:%s\">%s</a>\n",
$row["wpismail"]);
printf("<a href=\"%s\" target=\"_blank\">%s</a>\n",
$row["wpissite"]);
printf("%s</td>\n",
$row["data"]);
printf("<td width=\"80%\">%s</td></tr>\n",
$row["tekst"]);
}
?>
</table>
</body>
</html>
Co ja zrobilem zle?? Pomozcie
