I have a problem on my code it said Undefined variable: row in C:\xampp\htdocs\test\transaksi_toko\read_stock.php on line 85

Question

As it said on the title I'm pretty sure I've been defined it, btw I am new to PHP so anybody could help me, and also I'm sorry that some variable is in bahasa but it's not the issue so...

I try every method that stack overflow provides in dealing similar issue but nothing work.

<?php
// Check existence of id parameter before processing further
if(isset($_GET["id"]) && !empty(trim($_GET["id"]))){
    // Include config file
   include "koneksi.php";

    // Prepare a select statement
    $sql ="SELECT tabel_barang.nama_barang, tabel_stock.quantity
        FROM tabel_barang
        LEFT JOIN tabel_stock ON tabel_barang.tabel_barang id=tabel_stock.tabel_stock id
        ORDER BY tabel_barang.nama_barang";

    if($stmt = mysqli_prepare($db, $sql)){
        // Bind variables to the prepared statement as parameters
        mysqli_stmt_bind_param($stmt, "i", $param_id);

        // Set parameters
        $param_id = trim($_GET["id"]);

        // Attempt to execute the prepared statement
        if(mysqli_stmt_execute($stmt)){
            $result = mysqli_stmt_get_result($stmt);

            if(mysqli_num_rows($result) == 1){
                /* Fetch result row as an associative array. Since the result set contains only one row, we don't need to use while loop */
                $row = mysqli_fetch_array($result, MYSQLI_ASSOC);

                // Retrieve individual field value
                $nama_barang = $row["nama_barang"];
                $quantity = $row["quantity"];

            } else{ 
                // URL doesn't contain valid id parameter. Redirect to error page
                header("location: error.php");
                exit();
            }

        } else{
            echo "Terjadi kesalahan, silahkan coba lagi.";
        }


   // Close statement
    mysqli_stmt_close($stmt);

    // Close connection
    mysqli_close($db);

    } 

} else{
    // URL doesn't contain id parameter. Redirect to error page
    header("location: error.php");
    exit();
}
?>
<!DOCTYPE html>
<html>
<head>
    <meta charset="UTF-8">
    <title>Check Data</title>
    <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.css">
    <style type="text/css">
        .wrapper{
            width: 500px;
            margin: 0 auto;
        }
    </style>
</head>
<body>
    <div class="wrapper">
        <div class="container-fluid">
            <div class="row">
                <div class="col-md-12">
                    <div class="page-header">
                        <h1>Check Data</h1>
                    </div>
                    <div class="form-group">
                        <label>Nama Barang</label>
                        <p class="form-control-static"><?php echo $row["nama_barang"]; ?></p>
                    </div>
                    <div class="form-group">
                        <label>Quantity</label>
                        <p class="form-control-static"><?php echo $row["quantity"]; ?></p>
                    </div>
                    <p><a href="stock.php" class="btn btn-primary">Kembali</a></p>
                </div>
            </div>        
        </div>
    </div>
</body>
</html>

I expect that the data will be shown, but it giving me an error report like this

Notice: Undefined variable: row in C:\xampp\htdocs\test\transaksi_toko\read_stock.php on line 85

and also this

Notice: Undefined variable: row in C:\xampp\htdocs\test\transaksi_toko\read_stock.php on line 89

I'm sorry if my information or code doesn't give enough information it's my first time asking in this platform.

Your query has no placeholders, so it fails. Therefor you get to the else block of `if(mysqli_stmt_execute($stmt)){`, and there's nothing stopping the code there - so it continues down to your HTML where you use `$row`- — Qirel, Jul 16 '19 at 08:03
your prepared statement does not have a placeholder yet you attempt to bind a parameter/value. You do not need to use a prepared statement with a simple query such as this where there is no user supplied data as input. — Professor Abronsius, Jul 16 '19 at 08:03

I have a problem on my code it said Undefined variable: row in C:\xampp\htdocs\test\transaksi_toko\read_stock.php on line 85

0 Answers0