I keep getting this error: warning: mysqli_query() expects parameter 1 to be mysqli, null given in C:\wamp64\www\fish R us website\details.php on line 12

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I have been working on this for a while and I just can't figure out how to make it work! I keep getting the same error. For reference, I am trying to generate a dynamic link so when I click on a button from my list page it generates content on a second page.

If you want to see the list page code, let me know. Here is the dynamic page:

What I have tried:

<?php
if(isset($_GET['id'])) {
require_once  'login1.php';

$id=null;
$dbc = null;

$conn = new mysqli($hn, $un, $pw, $db); //these three lines connect to the db with the login file
if($conn->connect_error) die($conn->connect_error); //all the lines above here will be in all your codes

	$sql = "SELECT * FROM freshwater WHERE id='$id' ";
	$result = mysqli_query($dbc, $sql) or die("Bad Query: $sql");
	$row = mysqli_fetch_array($result);
	
}	
?>
<html>
<head>
<title>Fish R Us - Saltwater Fish</title>

</head>
<body>
<div class="toptext"><center>
<img src="banner.png" height="300"></img></center>
</div>
<center>
<h2><?php echo $row['name'] ?></h2>
width="150" height="150"
<img src="<?php echo $row['picture'] ?>" width="150" height="150"></img>
<table>
  <tr>
    <th>Name</th>
    <th>Type</th>
    <th>Quantity</th>
	<th>Price</th>
  </tr>
  <tr>
    <td><?php echo $row['name'] ?></td>
    <td><?php echo $row['type'] ?></td>
    <td><?php echo $row['qty'] ?></td>
	<td><?php echo $row['price'] ?></td>
  </tr>
</table><br>
<table class="fish-text">
<th><?php echo $row['description'] ?></th>
</center></table><br>
<button padding: 10px 24px;><a href="products.php">Return to Products List</a></button></body>
<button padding: 10px 24px;><a href="menu.php">Return to Menu</a></button>
</html>

Posted 18-Apr-19 6:36am

Member 14315682

Updated 18-Apr-19 7:25am

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Richard Deeming · Accepted Answer · 2019-04-18T07:26:00

Solution 1

Quote:

$dbc = null;
...
$result = mysqli_query($dbc, $sql) or die("Bad Query: $sql");

You forgot to initialize the $dbc variable. Did you perhaps mean to pass $conn instead?

Posted 18-Apr-19 7:26am

Richard Deeming

Comments

Member 14315682 18-Apr-19 14:35pm

Thank you so much! I switched to Conn and it's not throwing that error anymore! I so appreciate the help. Now my issue is that it's not pulling data from the database but I'm not getting any errors. Hmm...