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slotegraafd

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Posts posted by slotegraafd

    Why will only disable users work in my code?

    in PHP Coding Help

    Posted

    11 hours ago, requinix said:

    What's the code for your HTML form? Specifically for the radio buttons.

    ope sorry 

    if ($keys[$recordCount] == "logo" && $totalRows[$column] != "") {
          echo '<td class="py-2"><img src="' . $totalRows[$column] . '" alt="Client Logo" class="logo-thumbnail" /></td>';
        
        //Check whether or not the client is active or inactive
        } else if ($keys[$recordCount] == "enabled") {
          $userId = $data[$record]['id'];
          $enabled = $data[$record]['status'];
          
          echo '
            <td class="py-2">'. 
              '<form method="POST" action="salesperson.php">
                <div>
                  <input type="radio" name="active[' . $userId . ']" value="Active" '; 
                  if($enabled == "a") { echo 'checked'; };
                   echo '/>
                  <label for="' . $userId . '-Active">Active</label>
                </div> 
                <div>
                  <input type="radio" name="active[' . $userId . ']" value="Inactive" '; 
                  if($enabled == "d") { echo 'checked'; };
                  echo '/>
                  <label for="' . $userId . '-Inactive">Inactive</label>
                </div> 
                <input type="submit" name="submitType" value="Update" />
              </form> 
              Current Status: '. $totalRows[$column] . 
            "</td>";

    (its created as a function)

    Why will only disable users work in my code?

    in PHP Coding Help

    Posted

    So I am trying to create a table where users can be disabled and enabled with the press of a radio button. So for the most part it works but for some reason it only allows me to disable a user it wont let me enable them and I can't seem to figure out why.

    I've checked that all the names were correct with the database and I have also tried both single ' quotes and double " quotes on the prepared statements to see if that makes a difference... it did not

    These are the functions to disable and enable users 

    function inactive($userId){
$conn = db_connect();

$disable_user = pg_prepare($conn, 'disable', "UPDATE users SET Enabled = false, Status = 'd' WHERE Id = $userId");

$result = pg_execute($conn, 'disable', array());
}

//Create a function to activate a salesperson
function active($userId){
$conn = db_connect();

$enable_user = pg_prepare($conn, 'enable', "UPDATE users SET Enabled = true, Status = 'a' WHERE Id = $userId");

$result = pg_execute($conn, 'enable', array());
}
    $id = (array_keys($_POST['active'])[0]);
      $status = $_POST['active'][$id];

      if($status == "t"){
        active($id);
      } 
      else{
        inactive($id);
      }

    And this is where it gets put into action ^^

     

    In the database I have a boolean for enabled which is either true or false and then I have a varchar for status which is either 'd' or 'a'

     

    Any help would be greatly appreciated

     

    Thanks in advance

    How to resolve Trying to access array offset on value of type null error?

    in PHP Coding Help

    Posted

    37 minutes ago, kicken said:

    @slotegraafd then perhaps you have an additional problem or did not make the correct changes.  Provide the updated code after you've made the changes.

    I copied your function from the original post and tested it with the sample data and changes and it works just fine for me.

     

    Is this how you did it. Idk what im missing 

    
  $lastRecord = min($firstRecord + ROWS_PER_PAGE, $rows);
  
  for($record = $firstRecord; $record < $lastRecord; $record++){
    $row = $data[$firstRecord + ROWS_PER_PAGE];
    echo '<tr>';

    for($recordCount = 0; $recordCount < count($keys); $recordCount++){
      $column = $keys[$recordCount];
      echo '<td class="py-2">' . $row[$column] . '</td>';
    }

    echo '</tr>';
  }

     

    How to resolve Trying to access array offset on value of type null error?

    in PHP Coding Help

    Posted

    48 minutes ago, kicken said:

    Your data array only contains 2 rows, but your function is written to expect a minimum of ROWS_PER_PAGE rows.   Once the loop passes the end of your data array, 

    
$row = $data[$record];

    Will result in $row being NULL because the index it's trying to access doesn't exist.

    What you need to do is change your function to only loop until the end of the array if there are not enough rows available.  So you want the end of your loop to be either 

    
$firstRecord + ROWS_PER_PAGE

    or 

    
$rows

    which ever is smallest.  This can be easily determined by using the min() function.

    So change your for loop to 

    
  $lastRecord = min($firstRecord + ROWS_PER_PAGE, $rows);
  for($record = $firstRecord; $record < $lastRecord; $record++){

     

    Additionally, the function currently generates invalid HTML due to various errors in the HTML strings.  You'll want to address those as well.

    Okay so that didnt in fact get rid of the errors I was getting but now it doesnt show any of the data

    How to resolve Trying to access array offset on value of type null error?

    in PHP Coding Help

    Posted

    17 minutes ago, Barand said:

    Not sure how to say it with any more clarity.

    Your definition of the function has 4 parameters, namely $fields, $data, $rows and $page

    When you call the function you provide only the "$fields" parameter. There are no values provided for the other three.

    image.png.80da842f1d02a7121f5252d371a735e9.png 

    display_table(
     array(
      "id" => "Id",
      "emailaddress" => "Email",
      "firstname" => "First Name",
      "lastname" => "Last Name",
      "salesperson" => "Salesperson",
      "phonenumber" => "Phone Number",
      "extension" => "Extension",
      "type" => "Type"
    ),
    client_select_all(),
    client_count(),
    1
  );

    I forgot to add the last part of this, I have those two as other functions to for the data and rows and such

     

    How to resolve Trying to access array offset on value of type null error?

    in PHP Coding Help

    Posted

    Hi,

    WHAT:

    So I cannot for the life of me figure out why I'm getting this error. I'm trying to create a function that will display the user information from the database in a table but I have like a ton of these errors on the page and I don't know whats wrong.

    STEPS TO RESOLVE:

    So I've gone over my code a bunch of types to make sure that all the variables and what not were spelled correctly and as far as I can tell they are.

    I've also googled this issue to see if I can find a solution but none of them are very helpful.

    I honestly don't know whats wrong so kinda hard to find ways to resolve an issue, I don't even really know what this error means.

    THE CODE:

    This is where I put the function into action 

     <?php
   display_table(
     array(
      "id" => "Id",
      "emailaddress" => "Email",
      "firstname" => "First Name",
      "lastname" => "Last Name",
      "salesperson" => "Salesperson",
      "phonenumber" => "Phone Number",
      "extension" => "Extension",
      "type" => "Type"
    )
  );
   ?>
    //This is the function
<?php
function display_table($fields, $data, $rows, $page){
  if(isset($_GET['page'])){
    $page = $_GET['page'];
  } else {
    $page = 1;
  }
  $firstRecord = ($page - 1) * ROWS_PER_PAGE;
  $pageNumbers = ceil($rows / ROWS_PER_PAGE);

  echo '<div class="table-responsive w-75 mx-auto py-3">
  <table class="table table-dark table-bordered table-sm">
  <thead>
  <tr>';

  foreach($fields as $key){
    echo '<th class="py-2">' . $key . '</th>';
  }

  echo '</tr>
  </thead>
  </tbody>';

  $keys = array_keys($fields);

  for($record = $firstRecord; $record < $firstRecord + ROWS_PER_PAGE; $record++){
    $row = $data[$record];
    echo '<tr>';

    for($recordCount = 0; $recordCount < count($keys); $recordCount++){
      $column = $keys[$recordCount];
      echo '<td class="py-2">' . $row[$column] . '</td>';
    }

    echo '</tr>';
  }

  echo '</tbody>
  </table';

  for($pages = 1; $pages <= $pageNumbers; $pages++){
    echo '<a class="btn btn-dark mx-1" href=?page=' . $pages . '</a>';
  }
}
?>

    Any help/advice would be really appreciated

    How do you use arrays to display a form??

    in PHP Coding Help

    Posted

    6 hours ago, cyberRobot said:

    What does your current code look like? 

    // Form Fields Arrays: Field Title, Field Name, Field Type

$display_form = array(
  array(
    "type" => "text",
    "name" => "first_name",
    "value" => "",
    "label" => "First Name"
  ),
  array(
    "type" => "text",
    "name" => "last_name",
    "value" => "",
    "label" => "Last Name"
  ),
  array(
    "type" => "email",
    "name" => "email",
    "value" => "",
    "label" => "Email"
  ),
  array(
    "type" => "number",
    "name" => "extension",
    "value" => "",
    "label" => "Extension"
  ),
);

foreach ($display_form as $key => $value)
    {
    echo "<label for='{$value[1]}'>{$value[3]}</label>\n<input id='{$value[1]}' name='{$value[1]} type=\"{$value[0]}'  value='{$value[2]}' ><br><br>\n\n";
    }

?>
<input type="submit" name="submit" value="Submit" class="button">
  </fieldset>
</form>

     

    How do you use arrays to display a form??

    in PHP Coding Help

    Posted

    7 hours ago, cyberRobot said:

    It sounds like you are trying to use an array key that doesn't exist. Are you using an associative array, like your first post above shows. Or are you using a numeric array, like benanamen's code uses? Your choice will affect how you use the keys within the loop that outputs the form fields.

    Oh so I’m using the array that I have in my code there. (I’m really bad with arrays if you couldn’t tell) I thought that each one in the array corresponded to a number no matter what type of arrays 

    How do you use arrays to display a form??

    in PHP Coding Help

    Posted · Edited by slotegraafd

    On 10/10/2020 at 4:59 PM, benanamen said: 
    

<!DOCTYPE HTML>

<html>

<head>
    <title>Untitled</title>
<style type="text/css">
/* <![CDATA[ */

body{
	background-color:#e4dab8;
}

form fieldset{
	background-color:#fff9e7;

	border-width:2px;
	border-style:solid;
	border-color:#7c5b47;

	font-family:Verdana, Arial, Helvetica, sans-serif;
	font-size:12px;

	margin:20px 0px 20px 0px;
	width:350px;
	position:relative;
	display:block;
	padding: 0px 10px 10px 10px;
}

form fieldset legend{
	background-color:#7c5b47;

	border-width:1px;
	border-style:solid;
	border-color:#FFCC99;

	color:#ffcc99;
	font-weight:bold;
	font-variant:small-caps;
	font-size:110%;

	padding:2px 5px;
	margin:0px 0px 10px 0px;
	position:relative;
	top: -12px;

}

form fieldset legend img{
	padding:0px 5px 0px 5px;
}

label{
	font-size:80%;

	display:block;
	float:left;
	width:100px;
	text-align:right;
	margin:6px 5px 0px 0px;
}

.button{
	background-color:#7c5b47;

	border-width:1px;
	border-style:solid;
	border-color:#FFCC99;

	font-weight:bold;
	font-family:Verdana, Arial, Helvetica, sans-serif;
}

/* ]]> */
</style>
</head>

<body>

<form>
  <fieldset>
  <legend>My Array Form Generator </legend>
<?php
if ($_SERVER['REQUEST_METHOD'] == 'POST'){
    
}

// Form Fields Arrays: Field Title, Field Name, Field Type
$formfields = array(
    array(
        "First Name",
        "name_first",
        "text"
    ),
    array(
        "Last Name",
        "name_last",
        "text"
    ),
    array(
        "Username",
        "username",
        "text"
    ),
    array(
        "Password",
        "password",
        "password"
    ),
);

foreach ($formfields as $key => $value)
    {
    echo "<label for='{$value[1]}'>{$value[0]}</label>\n<input id='{$value[1]}' name='{$value[1]} type=\"{$value[2]}' value='' ><br><br>\n\n";
    }

?>
<input type="submit" name="submit" value="Submit" class="button">
  </fieldset>
</form>
</body>

</html>

     

    @benanamen Ah I see, Notice: Undefined offset: 2 in C:\XAMPP\htdocs\Webd3201\salesperson.php on line 83 though do you know what this error means?

    How do you use arrays to display a form??

    in PHP Coding Help

    Posted

    Hey,

    So I'm doing a school assignment and my professor wants us to use an array to display forms and I have no idea how to do that.

     

    He wants us to create an array of arrays like this: 

    display_form(
	array(
		"type" => "text",
		"name" => "first_name",
		"value" => "",
		"label" => "First Name"
	),

	array(
		"type" => "text",
		"name" => "last_name",
		"value" => "",
		"label" => "Last Name"
	),

	array(
		"type" => "email",
		"name" => "email",
		"value" => "",
		"label" => "Email"
	),
	array(
		"type" => "number",
		"name" => "extension",
		"value" => "",
		"label" => "Extension"
	)
);

    But I dont understand how you use that array to actually display the form. Does anyone know how?

    Why Won't my login page actually login?

    in PHP Coding Help

    Posted

    10 hours ago, requinix said:

    It should probably come as no surprise to learn that the answer is "to check if the input is the right combination".

    You have a query that looks for matching username and password. What will happen if there is a match? What will happen if there is not a match? How can you get the code to tell which is happening?

    I know the answer is to check if the input is the right combination but I'm asking how you do that

    Why Won't my login page actually login?

    in PHP Coding Help

    Posted

    22 minutes ago, Strider64 said:

    Throw that in file 13 and look for a safe secure login using PDO (My Suggestion) or mysqli. I did a Google search and found this https://levelup.gitconnected.com/how-to-build-a-secure-login-page-in-php-954f51d08701 and I am sure there are many others out there. 

    I'm required to use pgadmin, this is an assignment for school, all I am focusing on right now is trying to get it to login

    Why Won't my login page actually login?

    in PHP Coding Help

    Posted

    Hello, I am once again desperately asking for your help,

    I am working on a simple login page and I am having trouble actually getting it  to login. I display error messages for if the user doesn't enter anything but I can't seem to get it to work for if the credentials are wrong. It logs the user in whether the information is right or not and i dont even know what to do now

     

    This is the code any suggestions would be greatly appreciated 

    <?php
    /*
    Name: Deanna Slotegraaf 
    Course Code: WEBD3201
    Date: 2020-09-22
    */

    $file = "sign-in.php";
    $date = "2020-09-22";
    $title = "WEBD3201 Login Page";
    $description = "This page was created for WEBD3201 as a login page for a real estate website";
    $banner = "Login Page";
    require 'header.php';

    $error = "";

    if($_SERVER["REQUEST_METHOD"] == "GET")
    {
        $username = "";
        $password = "";
        $lastaccess = "";        
        $error = "";
        $result = "";
        $validUser = "";
    }
    else if($_SERVER["REQUEST_METHOD"] == "POST")
    {        
    	$conn;
        $username = trim($_POST['username']); //Remove trailing white space
        $password = trim($_POST['password']); //Remove trailing white space

    if (!isset($username) || $username == "") {
      $error .= "<br/>Username is required";
    }
    if (!isset($password) || $password == ""){
      $error .= "<br/>Password is required";
    }
    if ($error == "") {
      $password = md5($password);
      $query = "SELECT * FROM users WHERE EmailAddress='$username' AND Password='$password'";
      $results = pg_query($conn, $query);
      //$_SESSION['username'] = $username;
        //$_SESSION['success'] = "You are now logged in";
        header('location: dashboard.php');
      }else {
        $error .= "Username and/or Password is incorrect";
      }

}

?>

<div class = "form-signin">
    <?php
             echo "<h2 style='color:red; font-size:20px'>".$error."</h2>";
    ?>
<form action = "<?php echo $_SERVER['PHP_SELF'];  ?>" method="post">

      <label for="uname"><b>Login ID</b></label>
      <input type="text" name="username" value="<?php echo $username; ?>"/>
      <br/>
      <label for="psw"><b>Password</b></label>
      <input type="password" name="password" value="<?php echo $password; ?>"/>
      <br/>
        <button type="submit" name="login_user">Login</button>
        <button type="reset">Reset</button></div>
</form>
</div>

<?php require "footer.php"; ?>

     

    Why won't this connect to my database?

    in PostgreSQL

    Posted

    HI, Sorry the question title is really bad. But I'm currently making a simple php login page for school(well it should be simple anyway) and I'm having trouble getting the user to actually login. So Everything else on my page works fine. I have the login form and then I have error messages that displays an error if the user didnt enter any information or if the information is incorrect. So when I try to login with information from the database that should let me login and take me to the dashboard page it still says it is incorrect. Now at the top of my page I am getting an error that says: syntax error at or near "'lastAccess'" LINE 1: UPDATE users SET 'lastAccess' = $1 WHERE 'id' = $2 ^ in C:\XAMPP\htdocs\Webd3201\includes\db.php on line 13. I don't understand why I a,m getting that error. I don't see any syntax errors in my database.
      

    CREATE EXTENSION IF NOT EXISTS pgcrypto WITH SCHEMA public;

-- DROP'ping tables clear out any existing data
DROP TABLE IF EXISTS users;

-- CREATE the table
CREATE TABLE users(
userId VARCHAR(50) PRIMARY KEY NOT NULL,
email VARCHAR(255) UNIQUE,
password VARCHAR(255) NOT NULL,
firstName VARCHAR(128),
lastName VARCHAR(128),
enrolDate TIMESTAMP,
lastAccess TIMESTAMP,
enable BOOLEAN,
type VARCHAR(2) NOT NULL
);

ALTER TABLE users OWNER TO postgres;

INSERT INTO users (userId, email, password, firstName, lastName, enrolDate, lastAccess, enable, type) VALUES (
'DoeJ', 'jdoe@dcmail.ca', crypt('password', gen_salt('bf')), --NOTE: bf stands for blowfish
'John', 'Doe', '2020-09-14 19:10:25', '2020-09-15 11:11:11', true, 'a');

INSERT INTO users (userId, email, password, firstName, lastName, enrolDate, lastAccess, enable, type) VALUES (
'slotegraafd', 'deanna.slotegraaf@dcmail.ca', crypt('deanna1999', gen_salt('bf')),
'Deanna', 'Slotegraaf', '2020-09-15 23:18:27', '2020-09-16 13:40:02', true, 'a');

INSERT INTO users (userId, email, password, firstName, lastName, enrolDate, lastAccess, enable, type) VALUES (
'smithj', 'smithj@dcmail.ca', crypt('smith', gen_salt('bf')),
'Jane', 'Smith', '2020-08-22 09:14:56', '2020-09-01 20:34:14', true, 'a');

    This is the database code ^^^^^^^

      

    <?php
function db_connect()
{
	$conn = pg_connect("host=" .DB_HOST. " port=" .DB_PORT. " dbname=" .DB_NAME. " user=" .DB_USER. " password=" .DB_PASSWORD);
	return $conn;
} 

$conn = db_connect(); 

//Retrieve the user
$stmt1 = pg_prepare($conn, "user_select", 'SELECT * FROM users WHERE userId = $1 AND password = $2');
//Update the login time
$stmt2 = pg_prepare($conn, 'user_update_login_time', "UPDATE users SET 'lastAccess' = $1 WHERE 'id' = $2");
$stmt3 = pg_prepare($conn, 'user_authenticate', "SELECT 'userId', 'password', 'email', 'lastAccess', 'type' FROM users WHERE 'id' = $1 AND 'password' = $2");
$stmt4 = pg_prepare($conn, 'get_info', "SELECT userId, firstName, lastName FROM users WHERE userId = $1;");
?>

    This is the file thats giving the error^^^^ 

    <?php
    /*
    Name: Deanna Slotegraaf
    Course Code: WEBD3201 
    Date: 2020-09-22
    */

    $file = "login.php";
    $date = "2020-09-22";
    $title = "Login Page";
    $coursecode = "WEBD3201";
    $description = "This is a login page created for a real estate website for WEBD3201";
    $banner = "Login Page";
    $heading = "DS REAL ESTATE";
    require 'header.php';

    $error = ""; //Displays the error message to the screen

    //Get method to get  the info from the server
    if($_SERVER["REQUEST_METHOD"] == "GET")
    {
        $username = "";  //Holds the username
        $password = "";  //Holds the password
        $lastaccess = ""; //Displays the last access time       
        $error = "";     //Displays the error message on the screen
        $result = "";   //Gets the result of the query
        $validUser = ""; //Determines if the user is valid

    }
    //Posts to the server
    else if($_SERVER["REQUEST_METHOD"] == "POST")
    {
      //Connects to the database from db.php
        $conn;
        $username = trim($_POST["username"]); //removes trailing white space for username input box
        $password = trim($_POST["password"]);  //Removes trailing white space for password input box

        //Determines if the username input box is blank
        if(!isset($username) || $username == "")
        {
          //Display error message
            $error .= "<br/>You must enter a username!";

        }
        //Determines if the password input box is blank
        if(!isset($password) || $password == "")
        {
          //Display error message
            $error .= "<br/>You must enter a password!";
        }

        //If the error message is blank
        if ($error == "")
        {
          //Hash the password
            $password = hash("md5", $password);
            //Gets the user from the database
            $result = pg_execute($conn, "user_select", array($username, $password));
           
           //IF the query results were successful
            if(pg_num_rows($result))
            {
                
                $user = pg_fetch_assoc($result, 0);
                $_SESSION['user'] = $user;

                //Update the login time
                $result = pg_execute($conn, "user_update_login_time", array(date("Y-m-d"), $login));
                
                //If there is a user on the session
                if($_SESSION['user'])
                {
                    
                    header("Location: ./dashboard.php");
                    ob_flush();
                }
            }

            else
            {
              //Validates if the username and password are correct
              $sql = "SELECT userId FROM users
              WHERE userId = '".$username."'";
              $result = pg_query($conn, $sql);
              $validUser = pg_num_rows($result);
              if($validUser > 0)
              {
                //Display error message
                  $error = "Incorrect Password. Please try again!";
              }
              else
              {
                //Displays error message
                  $error = "Incorrect Username / Password. Please try again!";

              }
            }
        }
    }

?>

<div class = "form-signin">
    <?php
             echo "<h2 style='color:red; font-size: 20px'>".$error."</h2>";
    ?>
    <br/>
<form action = "<?php echo $_SERVER['PHP_SELF'];  ?>" method="post">
      <label for="username"><b>Username:</b></label>
      <input type="text" name="username" value="<?php echo $username; ?>"/>
      <br/>
      <label for="password"><b>Password:</b></label>
      <input type="password" name="password" value="<?php echo $password; ?>"/>
      <br/>
      <button type="submit">Login</button>
      <button type="reset">Reset</button>
</form>
</div>

<?php require "footer.php"; ?>

    And this is the login page(don't know if this one will be neededf.

     

    Anyway any help would be amazing!

    Thanks in advance!

    How to save data in MYSQL database

    in MySQL Help

    Posted

    On 5/22/2020 at 4:30 AM, gizmola said:

    Which is why I linked you to a tutorial that shows you how to fix your mysqli code and use prepared statements.  Did you bother to look at that?  Did you make an attempt to refactor your code? 

    lol yes I did I already fixed it,

    • Like 1

    How to save data in MYSQL database

    in MySQL Help

    Posted

    On 5/14/2020 at 4:16 PM, jodunno said:

    Hi again,

    I have 20+ years experience in programming. Do whatever you want but consider the following:

    When i first tried xampp with phymyadmin i couldn't login to my site either. I had to switch to the console. I rebuilt my database through the console and i logged in just fine. I think that phpmyadmin is difficult to use somehow. I filled in all of the data but it wasn' working. I find that the console makes it easier to build a database. Try to verify that your database is working using the console. If you need help with that, then let us know. Honestly, i have no problems since i switched to the console.

    Good luck.

     

    Oh well, I am currently using MYSQL im not even using phpmyadmin and i wouldnt know how to use the pdo you suggested im more familiar with mysql

     

    How to save data in MYSQL database

    in MySQL Help

    Posted

    46 minutes ago, gizmola said:

    Good to know that you have determined that you only have one row in there.  That is progress, if you are sure.   

    That code does not insure you only have one row for a given username.  That should be enforced at the database level by adding a unique index to the users.username table.  You can do that with phpMyAdmin/workbench etc. or learn to write the SQL statement that does it.  The tools simply write the DDL for you and execute it, but SQL DDL is not difficult to read or understand.

    Here's the actual statement: 

    
CREATE UNIQUE INDEX users_username_idx ON users (username)

    If this can not be run due to a duplicate index error, then you would be covered.  If however you get an error because there are already multiple duplicate usernames in the table, it will not be possible to create the index, and you would need to clean that up before you can create the index.

    Back to your statement: it only checks if there is one and only one row in the database.  If there is an error in your current code (and you have stated that there is) it's possible that more than one identical row exists.  In that case, your code will generate an incorrect result.  Even though there is actually > 1 user rows with that username and pw, your code treats that situation as if there are none.  

    Didn't you say you werent going to help me?

    How to save data in MYSQL database

    in MySQL Help

    Posted

    29 minutes ago, gizmola said:

    Here's one obvious potential issue with your code.

      

    
$query = "SELECT * FROM users WHERE username='$username' AND password='$password'";
			$results = mysqli_query($db, $query);

			if (mysqli_num_rows($results) == 1) {

    This is fine so long as there is one and only one row in the db with that username and password.  We don't know if that's actually true in your case.  Typically you might have a unique constraint/index on username to prevent this, but if you don't have that, you might have unknowingly entered 2+ rows with the same username and password.

    This test is going to fail in that case, at the moment you have 2 rows.  

    Obviously you don't want this to be possible, but without knowing how your user table is set up, we can't be sure.  

    It only has one row of data I entered a test one just to well test. cuz its only supposed to be checking for one because I don't want users with duplicate usernames

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