Forum Moderators: coopster
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/xxx/public_html/xxx/xxx/update.php on line 63
Here's the code:
<html>
<head>
<title>UPDATE</title>
</head>
<body>
<?php
$username = "xxx";
$password = "xxx";
$hostname = "localhost";
$dbh = mysql_connect($hostname, $username, $password)
or die("Unable to connect to MySQL");
mysql_select_db("xxx",$dbh);
if ($submit) {
// here if no ID then adding else we're editing
if ($id) {
$sql = "UPDATE employees SET first='$first',last='$last',address='$address',position='$position' WHERE id=$id";
} else {
$sql = "INSERT INTO employees (first,last,address,position) VALUES ('$first','$last','$address','$position')";
}
// run SQL against the DB
$result = mysql_query($sql);
echo "Record updated/edited!<p>";
} elseif ($delete) {
// delete a record
$sql = "DELETE FROM employees WHERE id=$id";
$result = mysql_query($sql);
echo "$sql Record deleted!<p>";
} else {
// this part happens if we don't press submit
if (!$id) {
// print the list if there is not editing
$result = mysql_query("SELECT * FROM employees",$db);
while ($myrow = mysql_fetch_array($result)) {
printf("<a href=\"%s?id=%s\">%s %s</a> \n", $PHP_SELF, $myrow["id"], $myrow["first"], $myrow["last"]);
printf("<a href=\"%s?id=%s&delete=yes\">(DELETE)</a><br>", $PHP_SELF, $myrow["id"]);
}
}
?>
<P>
<a href="<?php echo $PHP_SELF?>">ADD A RECORD</a>
<P>
<form method="post" action="<?php echo $PHP_SELF?>">
<?php
if ($id) {
// editing so select a record
$sql = "SELECT * FROM employees WHERE id=$id";
$result = mysql_query($sql);
$myrow = mysql_fetch_array($result);
$id = $myrow["id"];
$first = $myrow["first"];
$last = $myrow["last"];
$address = $myrow["address"];
$position = $myrow["position"];
// print the id for editing
?>
<input type=hidden name="id" value="<?php echo $id?>">
<?php
}
?>
First name:<input type="Text" name="first" value="<?php echo $first?>"><br>
Last name:<input type="Text" name="last" value="<?php echo $last?>"><br>
Address:<input type="Text" name="address" value="<?php echo $address?>"><br>
Position:<input type="Text" name="position" value="<?php echo $position?>"><br>
<input type="Submit" name="submit" value="Enter information">
</form>
<?php
}
?>
</body>
</html>
I guess many of you at the forum already know this script, i'm trying to adapt it to another (and bigger) database, the error messages are the same, so I went back to the original one.
Any help will be appreciated. Thanks. Danny
Try something like
$db = "****";
mysql_select_db($db, $dbh) or die ('Database error: ' . mysql_error());
