I am currently programming a system to tell users bans from a database. However, It is only returning one result.
How would I change this to echo more than one result?
<?php
include("check.php");
?>
<?php
mysql_connect("localhost", "root", "[PASSWORD]") or die(mysql_error());
mysql_select_db("bans") or die(mysql_error());
$query= mysql_query("SELECT * FROM bans WHERE name = '".$_SESSION['username']."' ")or die(mysql_error());
$arr = mysql_fetch_array($query);
$num = mysql_numrows($query); //this will count the rows (if exists)
?>
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Home</title>
<link rel="stylesheet" href="style.css" type="text/css" />
</head>
<body>
<h1 class="hello">Hello, <em><?php echo $login_user;?>!</em></h1>
<a href="logout.php" style="font-size:18px">Logout?</a>
</body>
</html>
<html>
<?php if($num > 0){ ?>
<h1>Your Punishments</h1>
<p>Type: <?php echo $arr['type']; ?></p>
<p>Reason: <?php echo $arr['reason']; ?></p>
<?php }else{ ?>
User not found.
<?php } ?>
</html></div>
check this code, this is demo code from
https://www.w3schools.com/php/php_mysql_select.asp
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, firstname, lastname FROM MyGuests";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
Msql library is deprecated and removed as of PHP 7.0.0. php.net says:
mysql_connect — Open a connection to a MySQL Server
Warning: This extension was deprecated in PHP 5.5.0, and it was removed in PHP 7.0.0. Instead, the MySQLi or PDO_MySQL extension should be used.
mysqli library. Is very easy to make the changes in your code: replace mysql_ with mysqli_ overall. Almost 98% enough.$bans). This way, you can close your db connection immediately after fetching. All steps further you'll make only on the array, without needing to worry about the database anymore.<body> in your page.<body>. Use <span>, <div>, etc to encapsulate it.<?php
// Fetch the user name, always assign default, even if NULL again.
// Like so, you can move it later in a function.
$userName = isset($_SESSION['username']) ? $_SESSION['username'] : NULL;
// Create an array to hold the query results.
$bans = array();
// Connect to db.
$connection = mysqli_connect('localhost', 'root', '[PASSWORD]', '[DB]');
if (!$connection) {
die('Connect error: ' . mysqli_connect_errno() . ' >> ' . mysqli_connect_error());
}
// Run query on db.
$sql = "SELECT * FROM bans WHERE name = '" . $userName . "'";
$queryBans = mysqli_query($connection, $sql);
// Loop through results and add each record to $bans array.
// I recommend you to fetch what you need from db,
// save in arrays and close db connection immediately.
while ($row = mysqli_fetch_array($queryBans, MYSQLI_ASSOC)) {
$bans[] = $row;
}
// Close connection.
mysqli_close($connection);
// Count bans.
$numBans = count($bans);
?>
<!DOCTYPE HTML>
<html>
<head>
<meta charset="utf-8">
<title>Home</title>
<link rel="stylesheet" href="style.css" type="text/css" />
</head>
<body>
<h1 class="hello">Hello, <em><?php echo $login_user; ?>!</em></h1>
<a href="logout.php" style="font-size:18px">Logout?</a>
<?php
if ($numBans > 0) {
// Loop through $bans and display.
foreach ($bans as $ban) {
?>
<h1>Your Punishments</h1>
<p>Type: <?php echo $ban['type']; ?></p>
<p>Reason: <?php echo $ban['reason']; ?></p>
<?php
}
} else {
?>
<span>User not found.</span>
<?php
}
?>
</body>
</html>