Error of undefined variable in my php code

Question

I am new to php and I am trying to get products name from my product table of database and storing that data into $product array. I am getting error

count(): Parameter must be an array or an object that implements Countable in C:\xampp\htdocs\moed\cart.php on line 57

$sqlprod = "SELECT pname FROM products limit 10";
$sqlprice = "SELECT price FROM products limit 10";
$result = $conn->query($sqlprod);
$result1 = $conn->query($sqlprice);

if ($result->num_rows And $result1->num_rows > 0) {
    $products = array();
    $amounts = array();
    while($row = mysql_fetch_assoc($sqlprod)){
        $products[] = $row; 
    }
    while($row1 = mysql_fetch_assoc($sqlprice)){
        // add each row returned into an array
        $products[] = $row1;
    }
} 

// I am getting error of undefined variable products at the for loop below 
if ( !isset($_SESSION["total"]) ) {
    $_SESSION["total"] = 0;
    for ($i=0; $i< count($products); $i++) {
        $_SESSION["qty"][$i] = 0;
        $_SESSION["amounts"][$i] = 0;
    }
}

Well, what exactly is the error? I bet it tells you what you're doing wrong. — Patrick Q, Mar 29 '18 at 17:29
it tells me that $products is not array, this is the exact error count(): Parameter must be an array or an object that implements Countable in C:\xampp\htdocs\moed\cart.php on line 57 — Moeed Azhar, Mar 29 '18 at 17:30
If either of the queries returns no rows, you never set `$products` and `$amounts`. — Barmar, Mar 29 '18 at 17:30
`$row = mysql_fetch_assoc($sqlprod)` should be `$result`. Same with next loop should be `$result1`. — AbraCadaver, Mar 29 '18 at 17:33
I needed them like that to show pname (product name ) and price separately in the cart — Moeed Azhar, Mar 29 '18 at 17:33
you are defining $products inside if statement and using it outside. what happens when the if doesnt pass — JoshKisb, Mar 29 '18 at 17:33
You're also mixing APIs. You can't use the `mysql_` functions with a `mysqli` connection. — Barmar, Mar 29 '18 at 17:34
Shouldn't you be putting the results of the second query into `$amounts` instead of `$products`? — Barmar, Mar 29 '18 at 17:36
There's no need for the `for` loop. Just use `$_SESSION['qty'] = array_fill(0, count($products), 0);` — Barmar, Mar 29 '18 at 17:37
@Barmar you pointed out errors righly can you tell me how to solve this problem pointed out by JoshKisb — Moeed Azhar, Mar 29 '18 at 17:41
@AbraCadaver Really, that's the duplicate you think applies best to this question? He's made a half dozen mistakes, I doubt any of them are explained there.' — Barmar, Mar 29 '18 at 17:42

score 0 · Accepted Answer · answered Mar 29 '18 at 17:44

Your code is totally messed up.

There's no need for multiple queries, just do one query that fetches both columns. You can still save each column in a separate array if you want.

When you're fetching, you need to use $result, not $sqlprod.

Since you're using the mysqli extension, you need to use mysqli_fetch_assoc(), not mysql_fetch_assoc(). But since you're using the OO style, it would be better to use $result->fetch_assoc().

You were putting the prices in $products, not $amounts.

You don't need a loop to create an array with the same elements multiple times, you can use array_fill() for this.

$sqlprod = "SELECT pname, price FROM products limit 10";
$result = $conn->query($sqlprod);

$products = array();
$amounts = array();
while($row = $result->fetch_assoc()){
    $products[] = $row['pname'];
    $amounts[] = $row['price'];
}

if ( !isset($_SESSION["total"]) ) {
    $_SESSION["total"] = 0;
    $_SESSION['qty'] = $_SESSION['amounts'] = array_fill(0, count($products), 0);
}

Error of undefined variable in my php code

1 Answers1