Undefined variable: con in /dashboard/index.php on line 101

Question

I am trying to show in a while loop the status (Active or Suspended) of a list of users.

0 = Active 1 = Suspended

<?php 
    $result = mysqli_query($con,"SELECT status FROM `e_customer`");
        while($row = mysqli_fetch_array($result)) {
        if ($row['status'] == "1") {
         echo "Suspended";
        } else {
          echo "Active";
        }
    } 
    ?>

The above gives me:

Notice: Undefined variable: con in /dashboard/index.php on line 101

Warning: mysqli_query() expects parameter 1 to be mysqli, null given in /dashboard/index.php on line 101

Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, null given in /dashboard/index.php on line 102

Answer 1

Be sure to answer the questions in the comments. Lot of smart people trying to help you.

I don't use mysqli. If I did I might try something like this...

<?php
if ($stmt = mysqli_prepare($con, "SELECT status FROM e_customer WHERE account = ?")) {

    /* bind parameters for markers */
    mysqli_stmt_bind_param($stmt, "s", $username);

    mysqli_stmt_execute($stmt);

    /* bind result variables */
    mysqli_stmt_bind_result($stmt, $status);

    mysqli_stmt_fetch($stmt);

    //is 0 really active and 1 suspended because than seems pretty backwards?
    if (!$status) { //per your comment then `!$status`. cringing...
        echo 'Active';
    } else {
        echo 'Suspended';
    }

    mysqli_stmt_close($stmt);
}

When expecting one row there is no need to loop. Not sure where username comes from - try and used parametrized (see PDO) statements.

Answer 2

First i would try your query directly in the database to make sure that it works, then try the following:

<?php 
    $result = mysqli_query($con,"SELECT status FROM `e_customer` WHERE account='$username'");
    $row = mysqli_fetch_assoc($result);
    echo ($row['status'] == "1") ? "Suspended" : "Active";
?>

If nothing still then try doing a var_dump of the $con variable and also the $row variable. see if you are even connecting and/or getting a result back.

...and also the $username variable.

Answer 3

Now that we have the real PHP error, we can deal with it.

Make sure $con is initialized before your query. You might have forgotten an include or something. For debugging purpose add it above the query. Let me know if this solves your problem.

NOTE: mysqli_connect() needs 4 param

Host: 127.0.0.1 or localhost or the IP/DNS of the remote server

User: The username to connect to the DB

Password: The password to connect to the DB

DB Name: The name of the database

$con = mysqli_connect("127.0.0.1", "my_user", "my_password", "my_db");

So the complete code would look like this:

<?php 
error_reporting( E_ALL );
ini_set('display_errors', 1);
$con = mysqli_connect("127.0.0.1", "my_user", "my_password", "my_db");

$result = mysqli_query($con,"SELECT status FROM `e_customer`") or die(mysqli_error($con));
    while($row = mysqli_fetch_array($result)) {
    if ($row['status'] == "1") {
     echo $row['account']." is Suspended</br>";
    } else {
      echo $row['account']." is Active</br>";
    }
} 
?>