PHP错误：“mysqli_num_rows（）期望参数1为mysqli_result，null为”

Question

There isnt an actual error, but it's still displaying an error message when a user who is not the buyer or seller goes to the page. Is it possible to suppress the error message?

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  //transaction id
$transactionid = $_GET['id'];

//Retrieve info about transaction
$query = "SELECT ads.*, feedback.*, transactions.* FROM (ads INNER JOIN transactions ON ads.id=transactions.ad_id) INNER JOIN feedback ON transactions.id=feedback.transaction_id WHERE transaction_id = '$transactionid'";
$data = mysqli_query($dbc, $query);
$row = mysqli_fetch_array($data);
$seller = $row['seller'];
$buyer = $row['buyer'];

//check if user is buyer or seller
if ($_SESSION['user_id'] == $seller) {
    $query = "SELECT * FROM feedback WHERE transaction_id = '$transactionid' AND seller_comment IS NULL";
    $data1 = mysqli_query($dbc, $query);
} else if ($_SESSION['user_id'] == $buyer) {
    $query = "SELECT * FROM feedback WHERE transaction_id = '$transactionid' AND buyer_comment IS NULL";
    $data1 = mysqli_query($dbc, $query);
}

//if user is buyer/seller echo form to them to submit feedback
if (mysqli_num_rows($data1) == 1) {
    echo '<p><form method="post" action="feedback.php?id=' . $transactionid . '&action=submitfeedback">
    <textarea id="feedback" name="feedback" rows="10" cols="30"></textarea><br/>
<input type="submit" value="Submit" name="submit" /></form></p>';
} else {
    echo '<p>feedback already given</p>';
}

Answer 1

You may want to resolve the warning, not just cover it up. You should be able to resolve this by changing the if statement to:

if(!empty($data1) && mysql_num_rows($data1) == 1) {

The problem is trying to pass NULL to the mysql function.