PHP don’t recognize row variables

I am having trouble posting a mysql table to my website.

Basically, the php error says that the row variable that I used was undefined. I tried using a code from a youtube tutorial but to no avail.

I also checked my sql query on phpmyadmin and it seems to work just fine.

Here is my test code for your reference:

 <!DOCTYPE html>
<html>
<head>
<title>Table with database</title>
<style>
table {
border-collapse: collapse;
width: 100%;
color: #588c7e;
font-family: monospace;
font-size: 25px;
text-align: left;
}
th {
background-color: #588c7e;
color: white;
}
tr:nth-child(even) {background-color: #f2f2f2}
</style>
</head>
<body>
<table>
<tr>
<th>Name</th>
<th>Transaction</th>
<th>Website</th>
</tr>
<?php
$conn = new mysqli('localhost','sasuke', 'sharinganrox', 'uchiha_db');
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT clientData.Client_ID, clientData.Client_Name, transactionData.Transaction_ID, transactionData.Transaction_Date, websiteProduct.Website_ID, websiteProduct.Website_Name,
            FROM clientData CROSS JOIN transactionData on clientData.Client_ID = transactionData.Client_ID CROSS JOIN websiteProduct on transactionData.Website_ID = websiteProduct.Website_ID WHERE monthname(transaction_date)='January' ORDER BY transaction_date ASC;";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {

// output data of each row
while($row = mysqli_fetch_array($result)) {
echo "<tr><td>" . $row["client_name"]. "</td><td>".$row["transaction_ID"]."</td><td>".$row["transaction_date"]."</td><td>".$row["website_Name"]."</td><td>".$row["website_Price"]."</td></tr>";
}
echo "</table>";
} else { echo "0 results"; echo "number of rows: " . $result->num_rows; }

$conn->close();
?>
</table>
</body>

JavaScript
​x
 
 <!DOCTYPE html><html><head><title>Table with database</title><style>table {border-collapse: collapse;width: 100%;color: #588c7e;font-family: monospace;font-size: 25px;text-align: left;}th {background-color: #588c7e;color: white;}tr:nth-child(even) {background-color: #f2f2f2}</style></head><body><table><tr><th>Name</th><th>Transaction</th><th>Website</th></tr><?php$conn = new mysqli('localhost','sasuke', 'sharinganrox', 'uchiha_db');// Check connectionif ($conn->connect_error) {die("Connection failed: " . $conn->connect_error);}$sql = "SELECT clientData.Client_ID, clientData.Client_Name, transactionData.Transaction_ID, transactionData.Transaction_Date, websiteProduct.Website_ID, websiteProduct.Website_Name,            FROM clientData CROSS JOIN transactionData on clientData.Client_ID = transactionData.Client_ID CROSS JOIN websiteProduct on transactionData.Website_ID = websiteProduct.Website_ID WHERE monthname(transaction_date)='January' ORDER BY transaction_date ASC;";$result = mysqli_query($conn, $sql);if (mysqli_num_rows($result) > 0) {​// output data of each rowwhile($row = mysqli_fetch_array($result)) {echo "<tr><td>" . $row["client_name"]. "</td><td>".$row["transaction_ID"]."</td><td>".$row["transaction_date"]."</td><td>".$row["website_Name"]."</td><td>".$row["website_Price"]."</td></tr>";}echo "</table>";} else { echo "0 results"; echo "number of rows: " . $result->num_rows; }​$conn->close();?></table></body>​

The error says: Notice: Undefined index: client_name in C:wamp64wwwaugusta_webapptestingtable.php on line 41

Notice: Undefined index: transaction_ID in C:wamp64wwwaugusta_webapptestingtable.php on line 41

Notice: Undefined index: transaction_date in C:wamp64wwwaugusta_webapptestingtable.php on line 41

Notice: Undefined index: website_Name in C:wamp64wwwaugusta_webapptestingtable.php on line 41

Notice: Undefined index: website_Price in C:wamp64wwwaugusta_webapptestingtable.php on line 41

Answer

In PHP arrays, string keys are case-sensitive.

The DB column is Client_Name (capital C and N) but you’re asking for the array key client_name (lowercase).

Check the column names, and the corresponding keys that you’re asking for, and you’ll see the mismatch. If you var_dump($row) you should see the problem clearly.

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Answer