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@NogDogFeb 15.2006 — #Try this to maximize your chance of getting some useful debug info:
MySQL Error messages
Howdy
Trying to make a drop down menu that interacts with my database and some fields on the page.
Trying to figure out these two errors
Warning: mysql_select_db(): supplied argument is not a valid MySQL-Link resource in /home/scott/public_html/drop_down.php on line 3
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/scott/public_html/drop_down.php on line 3
[code=php]
<?php
$username = “xx”;
$password = “xx”;
$hostname = “xx”;
$dbh = mysql_connect($hostname, $username, $password) or die(“Unable to connect to MySQL”);
$selected = mysql_select_db(“xx”,$dbh) or die(“Could not select database”);
$query_seminars = (‘SELECT * FROM `Contacts`’);
$seminars = mysql_query($query_seminars) or die(mysql_error());
$row_seminars = mysql_fetch_assoc($seminars);
$totalRows_seminars = mysql_num_rows($seminars);
$colname_chosen_seminar = “-1”;
if (isset($_POST[‘Attend’])) {
$colname_chosen_seminar = (get_magic_quotes_gpc()) ? $_POST[‘Attend’] : addslashes($_POST[‘Attend’]);
}
mysql_select_db($database_CHF_PHP, $CHF_PHP);
$query_chosen_seminar = sprintf(“SELECT * FROM Permission_To_Camp WHERE id = %s”, $colname_chosen_seminar);
$chosen_seminar = mysql_query($query_chosen_seminar, $CHF_PHP) or die(mysql_error());
$row_chosen_seminar = mysql_fetch_assoc($chosen_seminar);
$totalRows_chosen_seminar = mysql_num_rows($chosen_seminar);
?>
<!DOCTYPE HTML PUBLIC “-//W3C//DTD HTML 4.01 Transitional//EN” “http://www.w3.org/TR/html4/loose.dtd”>
<html>
<head>
<meta http-equiv=”Content-Type” content=”text/html; charset=iso-8859-1″>
<title>Untitled Document</title>
<script type=”text/javascript”>
function chkFrm(el){
if(el.selectedIndex == 0){
alert(“Please choose an option”);
return false
}
else{
el.form.submit();
}
}
</script>
</head>
<body>
<form name=”form1″ method=”post” action=””>
<p>
<select name=”Attend” onchange=”chkFrm(this)”>
<option value=”Choose” <?php if (!(strcmp(“Choose”, $_POST[‘Attend’]))) {echo “SELECTED”;} ?>>Choose Seminar</option>
<?php
do {
?>
<option value=”<?php echo $row_seminars[‘ID’]?>”<?php if (!(strcmp($row_seminars[‘ID’], $_POST[‘Attend’]))) {echo “SELECTED”;} ?>><?php echo $row_seminars[‘Attend’]?></option>
<?php
} while ($row_seminars = mysql_fetch_assoc($seminars));
$rows = mysql_num_rows($seminars);
if($rows > 0) {
mysql_data_seek($seminars, 0);
$row_seminars = mysql_fetch_assoc($seminars);
}
?>
</select>
</p>
<p>Location
<input name=”textfield” type=”text” value=”<?php echo $row_chosen_seminar[‘Location’]; ?>”>
<br>
From
<input name=”textfield” type=”text” value=”<?php echo $row_chosen_seminar[‘Fromto’]; ?>”>
<br>
Leader
<input name=”textfield” type=”text” value=”<?php echo $row_chosen_seminar[‘Leadership’]; ?>”>
</p>
</form>
</body>
</html>
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@NogDogFeb 15.2006 — #Try this to maximize your chance of getting some useful debug info:[code=php]
<?php
$username = "xx";
$password = "xx";
$hostname = "xx";
$dbh = @mysql_connect($hostname, $username, $password) or die("Unable to connect to MySQL");
$selected = @mysql_select_db("xx", $dbh) or die("Could not select database: " . mysql_error());
$query_seminars = ('SELECT * FROM Contacts');
$seminars = @mysql_query($query_seminars) or die("Query failed: $query_seminars - " . mysql_error());
[/code]