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Hydrian

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    Order by newest

    in PHP Coding Help

    Posted

    Im running a testing forum iv made from scratch, i want the posts to be order by newest, but there going as, order by oldest. How come this happens.

     

    <?php

session_start();
error_reporting(-1);ini_set('display_errors', 1);
    $myServer = "mysql8.000webhost.com";
    $myUser = "a1761169_chat";
    $myPass = "site123";
    $myDB = "a1761169_chat";
    //Connection to the database
    $dbhandle = mysql_connect($myServer, $myUser, $myPass)
    or die("Couldn't connect to SQL Server on $myServer");
    //Select a database to work with
    $selected = mysql_select_db($myDB, $dbhandle)
    or die("Couldn't open database $myDB");
$where = "";	
	if(isset($_GET['id'])){
	$id = $_GET['id'];
	$where = " WHERE id = $id";
}    



//Declare the SQL statement that will query the database
    $query = "SELECT * FROM `topics` $where ORDER BY topic_date";
    //Execute the SQL query and return records
    $result = mysql_query($query)
    or die('A error occured: ' . mysql_error());
?>
<html>
<head>
</head>

<body>
	<h2>
		My First Forum
	</h2>
		<br>
		<div style="background-color:#CCCCCC; color:#; border-radius:5px;" align=''>
			     <a href='index.php'><b>Board Index<b></a>     <a href='create_topic.php'><b>New Topic<b></a>
		</div>
		<br>
		<?php

		while ( $record = mysql_fetch_array($result) )
		{	

		echo '<div id="content" >' . '<div style="background-color:#CCCCCC; color:#; border-radius:5px;">' . '<a href="index.php">'. $record["topic_username"] .'</a>' . '     |     ' . '<a href="index.php?id=' . $record["id"] . ' ">'. $record["topic_name"] .'</a>' . '</div> ' . "<br>" . '<div style="word-wrap:break-word;div-layout:fixed; background-color:#CCCCCC; color:#; border-radius:5px;" width="500px" border="1">';
		if(isset($_GET['id'])){echo $record['topic_date']; echo "<br><br>"; echo $record['topic_input']; }
		else { echo ""; }
		echo '</div></div>';

		}
		 echo "<br><br>";
		while ( $record = mysql_fetch_array($result) )
		{	

		echo '<div id="content" >' . '<div style="background-color:#CCCCCC; color:#; border-radius:5px;">' . '<a href="index.php">'. $record["reply_username"] .'</a>' . '     |     ' . '<a href="index.php?id=' . $record["id"] . ' ">'. $record["reply_name"] .'</a>' . '</div> ' . "<br>" . '<div style="word-wrap:break-word;div-layout:fixed; background-color:#CCCCCC; color:#; border-radius:5px;" width="500px" border="1">';
		if(isset($_GET['id'])){echo $record['reply_date']; echo "<br><br>"; echo $record['reply_input']; }
		else { echo ""; }
		echo '</div></div>';

		}
		 echo "<br><br>";
		?>
</body>
</html>

     

    Thats the page i want the topics to go order by descending

    Warning on page load

    in PHP Coding Help

    Posted

    When i load my page. i get

     

    Warning: mysql_fetch_array(): 9 is not a valid MySQL result resource in /home/a1761169/public_html/a40g/index.php on line 46

     

    Here is my script

     

    <?php

session_start();
error_reporting(-1);ini_set('display_errors', 1);
    $myServer = "";
    $myUser = "";
    $myPass = "";
    $myDB = "";
    //Connection to the database
    $dbhandle = mysql_connect($myServer, $myUser, $myPass)
    or die("Couldn't connect to SQL Server on $myServer");
    //Select a database to work with
    $selected = mysql_select_db($myDB, $dbhandle)
    or die("Couldn't open database $myDB");
$where = "";	
	if(isset($_GET['id'])){
	$id = $_GET['id'];
	$where = " WHERE id = $id";
}    



//Declare the SQL statement that will query the database
    $query = "SELECT * FROM `topics`$where";
    //Execute the SQL query and return records
    $result = mysql_query($query)
    or die('A error occured: ' . mysql_error());
    //Show result
    //Free result set memory
    mysql_free_result($result);
    //Close the connection 
    mysql_close($dbhandle);	
?>
<html>
<head>
</head>

<body>
	<br>
	<div style="background-color:#CCCCCC; color:#; border-radius:5px;" align=''>
		     <a href='create_topic.php'><b>New Topic<b></a>
	</div>
	<br><br>
	<?php

	while ( $record = mysql_fetch_array($result) )
	{	

	echo '<div id="content" >' . '<div style="background-color:#CCCCCC; color:#; border-radius:5px;">' . '<a href="index.php?id=' . $record["id"] . ' ">'. $record["topic_username"] .'</a>' . '       ' . '<a href="index.php?id=' . $record["id"] . ' ">'. $record["topic_name"] .'</a>' . '</div> ' . "<br>" . '<div style="word-wrap:break-word;div-layout:fixed; background-color:#CCCCCC; color:#; border-radius:5px;" width="500px" border="1">';
	if(isset($_GET['id'])){echo $record['topic_date']; echo "<br><br>"; echo $record['topic_input']; }
	else { echo ""; }
	echo '</div></div>';

	}

	?>
</body>
</html>

     

    What have i done wrong?

    Php Application

    in PHP Coding Help

    Posted

    Im working on the login stuff

     

    I get an unexpected '=' on line 9

     

    <?php

session_start();
include_once('connect.php');

if (isset($_POST['username'])) {
$username = $_POST['username'];
$password = $_POST['password'];
$sql = " SELECT * FROM members WHERE username='".$username."'" AND password='".$password." LIMIT 1';
$res = mysql_query($sql) or die(mysql_error());
if (mysql_num_rows($res) == 1) {
	$row = mysql_fetch_assoc($res);
	$_SESSION['uid'] = $row['id'];
	$_SESSION['username'] = $row['username'];
	header ("Location: index.php");
	exit();
} else {
	echo "Invalid login information. Please return to previous page.";
	exit();
}
}

?>

    Php Application

    in PHP Coding Help

    Posted

    I get  

    mysql_fetch_assoc():

    supplied argument is not a valid MySQL result resource. when i load the page

     

    <?php

session_start();
include 'config.php';

mysql_select_db("my_db");

$result = mysql_query("SELECT * FROM `categorys`, SELECT * FROM `topics` ");	

while ($row = mysql_fetch_assoc($result)) {
	echo $row["cat_name"];
}


?>

    Php Application

    in PHP Coding Help

    Posted

    I get Resource id #10 when i load my page

     

     

    Full Script

    <?php

session_start();
include 'config.php';

mysql_select_db("my_db");

$result = mysql_query(" SELECT * FROM `categorys`, SELECT * FROM `topics` ");	


?>
<?php include 'header.php'; ?>
	<div id="nar">
		 
	</div>
        <div id="content">
	<?php print $result; ?>
	</div>
	<div id="nar">
		 
	</div>
<?php include 'footer.php'; ?>

    Application for a site

    in Application Design

    Posted

    I get  Resource id #10  when i do  

    <?php print $result; ?>

    when i load my page.

     

    Here is my page code

     

    <?php

session_start();
include 'config.php';

mysql_select_db("my_db");

$result = mysql_query(" SELECT * FROM `categorys` ");
$result = mysql_query(" SELECT * FROM `topics` ");

$num = mysql_numrows($result);


?>
<?php include 'header.php'; ?>
	<div id="nar">
		 
	</div>
        <div id="content">
	<?php print $result; ?>
	</div>
	<div id="nar">
		 
	</div>
<?php include 'footer.php'; ?>

     

     

    Application for a site

    in Application Design

    Posted

    I want to know if you can make a application that allows a user to make a post(topic) and allow other users to reply to that post. Is there any way to do this. If i can do that, i want to try and create a forum with it. And style it how i want it. And then maybe after beta testing upload it to my site. Please i want to know if this is possible, it would be good if this can happen.

     

    Cheers Hydroo

    File Hosting Site

    in PHP Coding Help

    Posted

    For the file names just have a user having to register before uploading a file. Then set a var that randomises a url and set another function making sure that each url isnt the same. if a post is say like hasent been visited for 5months delete it

    unexpcpected {

    in PHP Coding Help

    Posted

    Iv got a script. Now it it tells me there is a unexpected { at the end of line 9 which is i dont know why it tells me its unexpected. Please help, im a learning phper

     

    while (($row = mysql_fetch_assoc($result)) !== false{
	$users[] = $row;
}

     

     

    Full Script

     

    <?php

// fetchs all of the users from the table
function fetch_users(){
$result = mysql_query('SELECT `user_id` AS `id`, `user_username` AS `username` FROM `users` ');

$users = array();

while (($row = mysql_fetch_assoc($result)) !== false{
	$users[] = $row;
}

return $users;
}


?>

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