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Bird

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Posts posted by Bird

    FOR LOOP not working

    in PHP

    Posted

    1 hour ago, justsomeguy said:

    Why are you using a for loop at all?  It's an array with a single value in it, you don't need to loop.  And if there's nothing in $_SESSION, then it won't loop to set the value at all.

    You're absolutely right. I'm very new at this and thought I had to use a loop for an array. Realized I was already returning the row when I set the $_SESSION['user'].

      

    // Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} else{
	echo 'Connected! ';
$sql = "SELECT * FROM KJE WHERE username = '$user' AND password = '$pass' ";
$result = mysqli_query($conn,$sql);
$check = mysqli_fetch_array($result);
$_SESSION['currency'] = $check['currency'];
$_SESSION['level'] = $check['level'];

    ^much better

    FOR LOOP not working

    in PHP

    Posted · Edited by Bird

    Both of these are using the same DB connection($conn) and both currency and level are in the table, yet when I run my page, it shows currency, but level comes up blank.

      

    // Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} else{
	echo 'Connected! ';
$sql = "SELECT * FROM KJE WHERE username = '$user' AND password = '$pass' ";
$result = mysqli_query($conn,$sql);
$check = mysqli_fetch_array($result);

if(isset($check)){
	$_SESSION['username'] = $user;
	//BELOW WAS WORKING. WHILE TROUBLESHOOTING I CHANGED IT. WHEN I CTRL+Z BACK TO IT'S ORIGINAL IT STOPPED WORKING. I GUESS I'M DOING SOMETHING FUNDAMENTALLY WRONG AND GOT LUCKY?
	//------currency------
	$sql2 = ("SELECT currency FROM KJE WHERE username = '{$_SESSION['username']}' ");
	$result2 = mysqli_query($conn,$sql2);
	$currency = mysqli_fetch_array($result2);
	for($a = 0 ; $a < count($_SESSION['currency']) ; $a++){
		$_SESSION['currency'] = $currency[$a];
	}
	//BELOW DOES NOT WORK
	//------level------
	$sql3 = ("SELECT `level` FROM KJE WHERE username = '{$_SESSION['username']}' ");
	$result2 = mysqli_query($conn,$sql3);
	$level = mysqli_fetch_array($result2);
	for($b = 0 ; $b < count($_SESSION['level']) ; $b++){
		$_SESSION['level'] = $level[$b];
	}
}

     

    $_POST not passing variables

    in PHP

    Posted · Edited by Bird

    Hello!

    I know there are a lot of posts on the internet about this but I can't find one that solves my problem.

    I'm following the PHP Login Page tutorial on w3schools.com and am having some trouble.

    https://www.w3schools.com/howto/howto_css_login_form.asp

     

    It is not passing the 'uname' and 'psw' to action_page.php. Not sure why... Everything looks referenced correctly

     

    I have 2 scripts here:

    1) index.html(This seems to be working as-far as sending the credentials to action_page.php, I see the URL at the top is to action_page.php and includes the input username and password) 

    <body>
<form action="/action_page.php">
  <div class="imgcontainer">
    <img src="kje_logo.png" alt="Avatar" class="avatar">
  </div>

  <div class="container">
    <label for="uname"><b>Username</b></label>
    <input type="text" placeholder="Enter Username" name="uname" required>

    <label for="psw"><b>Password</b></label>
    <input type="password" placeholder="Enter Password" name="psw" required>

    <button type="submit">Login</button>
  </div>
</form>
</body>

    2*) action_page 

    <?php
$dbservername = "localhost";
$dbusername = "****";
$dbpassword = "****";
$dbname = "****";
$username = $_POST['uname'];
$password = $_POST['psw'];

// Create connection
$conn = new mysqli($dbservername, $dbusername, $dbpassword, $dbname);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} else{
	echo 'Connected! ';
$sql = "SELECT * FROM KJE WHERE username = '$username' AND password = '$password' ";//when i change $username and $password to the correct credentials, it says "Login Success"
$result = mysqli_query($conn,$sql);
$check = mysqli_fetch_array($result);
if(isset($check)){
echo 'Login Success.';
}else{
echo 'Login Failure!';
}
}
?>

     

     

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