PHP/MYSQL Problem Passing Data

squidman

Here is the problem?

I have a link on one page that passes a $num variable to this file (which it does fine). Then this file 1st of all needs a password to get to main area. Having trouble passing $num variable thru the password part. Because when it does the query, all I get is Resource ID #5

<?
// Defines password
$password = "change";
// Checks Password
if ($_POST['txtPassword'] != $password) {
?>
<html>
<head>
<title>Comment Login</title>
</head>
<body bgcolor=#3C507D>
<center>
<form name="form" method="post" action="<? echo $_SERVER['PHP_SELF']; ?>">
	<input type="hidden" name="num" value="<? echo "$num"; ?>">
    <input type="password" title="Enter your password" name="txtPassword"><br><br>
	<input type="submit" name="Submit" value="Login">
</form>
</center>
<?
} else {
include ("conf.php");
mysql_connect ("localhost", "$sqlname", "$sqlpw") or die ("Could not connect");	# connect mySQL
mysql_select_db("$sqldb");			# select database

$query = "SELECT reply FROM eg_book WHERE (num = '$num')"; 
$result = mysql_query ($query);
echo "$result";
?>
<title>Please Comment</title>
<body bgcolor=#3C507D>
<center>
<font color="#FFFFFF">Administrators Comment</font><br>
<form name="form1" method="post" action="<? echo $_SERVER['PHP_SELF']; ?>">
<input type="hidden" name="num" value="<? echo "$num"; ?>">
<textarea rows="10" name="reply" cols="60" style="font-family: Verdana; font-size: 10px"><? echo "$result"; ?></textarea><br><br>
<input type="submit" name="Submit" value="Add Comment">
</form>
</center>
</body>
</html>
<?
mysql_query("UPDATE eg_book SET reply = '$reply' WHERE (num = '$num')") or die("MYSQL Returned: " . mysql_error());
}
?>

My mind is so jumbled right now, and I can't think straight, can anyone help out this poor sap, LOL

Thanks

rgermain

Cause you are echoing out the result. What you need to do is write something similar to this:

<?
while($row = mysql_fetch_array($result))
{
echo $row['fieldName'];
echo "<BR>";
echo $row['fieldName2'];
}
?>

For each result that is select by your query it will display what ever you have in those fields.

squidman

This is what works after 6 hours of turmoil, I think my brain is fixed now.

// Defines password
$password = "change";
// Checks Password
if ($_POST['txtPassword'] != $password) {

include ("conf.php");
mysql_connect ("localhost", "$sqlname", "$sqlpw") or die ("Could not connect");	# connect mySQL
mysql_select_db("$sqldb");			# select database

$query = "SELECT * FROM eg_book WHERE (num = '$num')"; 
$result = mysql_query ($query);
while($row = mysql_fetch_array($result)) { 
$num = $row["num"];
$reply = $row["reply"];
$reply = stripslashes($reply);
}
?>
<html>
<head>
<title>Comment Login</title>
</head>
<body bgcolor=#3C507D>
<center>
<form name="form" method="post" action="<? echo $_SERVER['PHP_SELF']; ?>">
	<input type="hidden" name="num" value="<? echo "$num"; ?>">
	<font color=FFFFFF>Administrators Comment</font><br>
	<textarea rows="10" name="reply" cols="60" style="font-family: Verdana; font-size: 10px"><? echo "$reply"; ?></textarea><br><br>
    <font color=FFFFFF>Password: </font><input type="password" title="Enter your password" name="txtPassword"><br><br>
	&nbsp;&nbsp;
	<input type="submit" name="Submit" value="Add Comment">
</form>
</center>
<?
} else {
include ("conf.php");
mysql_connect ("localhost", "$sqlname", "$sqlpw") or die ("Could not connect");	# connect mySQL
mysql_select_db("$sqldb");			# select database

$query = "UPDATE eg_book SET reply = '$reply' WHERE (num = '$num')"; 
$result = mysql_query ($query);
?>
<title>Close Window</title>
<body bgcolor=#3C507D>
<center>
<input type="button" value="Close Window" name="close" onClick="window.close()">
<?
}