[RESOLVED] Help finding coding problem

robkir

Hi,
I've spent forever trying to figure out what's wrong with this little bit of code. I suspect it is dying at

$result = mysql_query($link, $sql_result)or die(mysql_error($link));

but I can't see why.

Sorry to ask such an elementary question--but i'm stuck.

Bob

<?php
//check for required fields from the form

if ((!isset($POST["username"])) || (!isset($POST["password"]))) {
header("Location: userlogin.html");
exit;
}

//connect to server and select database (removed for obvious reasons)
$link = mysql_connect(etc);

//create and issue the query
$sql_result = "SELECT last, first FROM Dad_Profile WHERE
username = '".$POST["username"]."' AND
password = PASSWORD('".$POST["password"]."')";
$result = mysql_query($link, $sql_result)or die(mysql_error($link));

//get the number of rows in the result set; should be 1 if a match
if (mysql_num_rows($result) == 1) {

//if authorized, get the values of f_name l_name
while ($info = mysql_fetch_array($result)) {
	$f_name = stripslashes($info['first']);
	$l_name = stripslashes($info['last']);
}

//set authorization cookie
setcookie("auth", "1", 0, "/", "mywebsite.com", 0);

//create display string
$display_block = "
<p>".$f_name." ".$l_name." is authorized!</p>
<p>Authorized Users' Menu:</p>
<ul>
<li><a href=\"nextpage.php\">next page</a></li>
</ul>";

} else {
//redirect back to login form if not authorized
header("Location: userlogin.html");
exit;

}

echo "<br/> end of script";
?>
<html>
<head>
<title>User Login</title>
</head>
<body>
<?php echo "$display_block"; ?>
</body>
</html>

NogDog

You have your arguments transposed:

$mysql_query($sql_result, $link);

Note that the second "link" argument is optional and only needed if you have multiple database connections open at the same time.

akeane

Not too sure but don't you have to select a database after connecting and before doing an inquiry? i.e. mysql_select_db

:queasy:

robkir

akeane;10902973 wrote:
Not too sure but don't you have to select a database after connecting and before doing an inquiry? i.e. mysql_select_db

:queasy:

Hi,

mysql_select_db was in the original program--I took out the whole "connect to database" section in the code I posted, since it contained my username and log in for the database.

Thanks,

Bob

robkir

NogDog;10902965 wrote:
You have your arguments transposed:
$mysql_query($sql_result, $link);
Note that the second "link" argument is optional and only needed if you have multiple database connections open at the same time.

Thanks for catching the error. I amended the code and it worked like a charm.

Bob