Trying to display values from a SQL table, recieveing error, trying to get property of non-object

Question

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "comp4";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} 


$sql = "SELECT OrderID, id, items FROM orders WHERE id= $user";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$results =   $row["OrderID"]. " " . $row["id"]. $row["items"]. "<br>" ;

}

$loop = implode( " ", $items );

echo $loop;
}
?>

So, I have this code and I'm trying to display OrderID, id and items from the user thats logged in $user=whoever's logged in that's in a different part of my code, however I keep getting an error Notice: Trying to get property of non-object in C:\xampp\htdocs\myorders.php on line 35 After looking around, I'm still not quite sure how to fix this. Any help is appreciated

which line is number 35 and try to learn what does that error mean http://stackoverflow.com/questions/5891911/trying-to-get-property-of-non-object-in — Shaiful Islam, Jan 26 '15 at 14:46
different part ?? you should assign $user in this script only — ɹɐqʞɐ zoɹǝɟ, Jan 26 '15 at 14:48
`$user` error reporting would tell you Undefined variable. http://php.net/manual/en/function.error-reporting.php - if it's in your code, show it. On screen here, that's undefined in my view. As is `$items`; undefined. — Funk Forty Niner, Jan 26 '15 at 14:52

score 0 · Answer 1 · answered Jan 26 '15 at 17:46

I'm guessing your error lies in this line of code:

$sql = "SELECT OrderID, id, items FROM orders WHERE id= $user";

It could be that ID is a reserved keyword of SQL, or that $user is undefined and/or not a numeric value.

Try the following:

$sql = "SELECT `OrderID`, `id`, `items` FROM `orders` WHERE id = $user";

The reason for the error is your query fails. Then you later do the following:

if ($result->num_rows > 0) {

And since the query failed, $result is not an object, and you try to access num_rows of this non-object.

Of course I am only guessing because we need more information :)

Trying to display values from a SQL table, recieveing error, trying to get property of non-object

1 Answers1