Kez323
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Everything posted by Kez323
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Hi! I'm getting this PHP syntax error on line 87 of my file. ERROR: PHP Parse error: syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE), expecting identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING) The function works fine, if I take the function out of that file and put it into a separate file, it works fine. function get_UserId($username){ include('config.php'); // Check connection if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); } $sql = "SELECT id FROM users WHERE username='$username'"; $result = $conn->query($sql); $row = $result->fetch_assoc(); $id = $row['id']; $conn->close(); return $id; } Thanks
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Is there any other ways of transferring data instead of $_GET[] or $_POST[] ?
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Hi, I need help getting this thing to work, I'm not sure on how to $_GET[] data after a fragment in the URL. I've seen it used on websites before, for example: example.com/#home?q=test A working example is: http://craftland.org/#map&p=aether This is the code I'm using for the pages: http://pastebin.com/vAHppEyr Any help would be great! Thanks, Kez.
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Thanks trq. .josh, This is a forums. Im here to ask questions, you don't have to be a grumpy fuck. Doesn't matter what its for.
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Hey, i want the '***' to change depending on the length of the persons password. any way to echo a variable like this?
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this expire system wont work.. whats wrong with it? When you login: (login page) <?php session_start(); $_SESSION['start'] = time(); $_SESSION['expire'] = $_SESSION['start'] + (3600); ?> The page you goto after you login: (logged in page) <?php session_start(); $now = time(); if($now > $_SESSION['expire']) { session_destroy(); echo "session expired!"; } else { echo $myusername; } ?> When im at the logged in page, and i refresh it just says "session expired!"
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Still dont echo all from database
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Hey, so to display an image from a database I was using $id = addslashes($_REQUEST['id']); $image = mysql_query("SELECT * FROM store_image WHERE id=$id"); $image = mysql_fetch_assoc($image); $image = $image['image']; header("Content-type: image/jpeg"); echo $image; That worked fine to return an image via matching the id. But I want to get ALL images from the table.. I tried this: $image = mysql_query("SELECT * FROM store_image"); while($image = mysql_fetch_array($image)){ $image = $image['image']; header("Content-type: image/jpeg"); echo $image; } But did not work.. what am I doing wrong?
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hey, this should echo all of the results from the query, right? in the database there is more than 1 fields to echo, but it only echos 1, whats wrong with it? $getid = mysql_query("SELECT * FROM $tbl_name WHERE userid='$id'", $connect); $idresults= mysql_fetch_array($getid); while($idresults= mysql_fetch_array($getid)){ echo $idresults['frienduserid']; }
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Oh! Thanks
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Hey, Im new to PHP and am trying to make a simple form where it puts the form data into a table called "users" This is the form: <p><form method="post" action="register.php"> <table border="0" align="center"> <tr> <td>Username</td><td><input type="text" name="username" size="15" > </tr> <br /> <tr> <td>Password</td><td><input name="password" type="password" size="15"></td> </tr> <br /> <td><input type="submit" value="Sign Up"/></td><td></td> </table> This is the php code: <?php $dbhost = ''; $dbname = ''; $dbuser = ''; $dbpass = ''; mysql_connect($dbhost,$dbuser,$dbpass); mysql_select_db($dbname); $order = "INSERT INTO users (username, password) VALUES ('$username', '$password')"; $result = mysql_query($order); if($result){ echo("<br>It Worked!"); } else{ echo("<br>It Failed!"); } ?> But when i upload it to the server and try it, it says there was a entry but its blank..... When i upload it to a server and try it, it automaticly seems to post the data to the database before i even enter a username/password! when i type a username/password in and post it again it says there was another entry but its blank..... any help? - Thanks!