Question

请问我的代码有问题。我打算在我的数据库中选择所有细节并循环创建一个表。但我一直收到这个错误：

警告：mysql_fetch_assoc（）要求参数1为资源，第10行的C：\ xampp \ htdocs \ donation_details.php中给出布尔值没有找到捐赠记录

请帮忙。我会感激任何帮助。感谢

$connect = mysql_connect("localhost","hifee","kalimat","hifee");
if(!$connect) die('Unable to fetch data');
else{
$donation = array();
$query = "SELECT * FROM tx_transaction";
echo $query;
$result = mysql_query($query);
while($row = mysql_fetch_assoc($result)){
    $donation[] = $row;
}
$total_records = count($donation);
if($total_records > 0){
    echo '<table>';
    echo    '<tr>Transaction ID</td>
                <td>Amount donated(=N=)</td>
                <td>Donation date</td>
                <td>Donation status</td>
                <td>Response code</td>
                <td>Meno</td>
                <td>Response details</td>
                <td>Donor\'s email</td>
            </tr>';
    foreach($donation as $value){

        echo    '<tr />
                    <td>'.$value['tx_id'].'</td>
                    <td>'.number_format($value['tx_amount'],2).'</td>
                    <td>'.$value['tx_date'].'</td>
                    <td>'.$value['tx_status'].'</td>
                    <td>'.$value['tx_response_code'].'</td>
                    <td>'.$value['tx_memo'].'</td>
                    <td>'.$value['tx_response_details'].'</td>
                    <td>'.$value['tx_buyer_email'].'</td>
                </tr>';
    }

    echo '</table>';
}else echo 'No donation records found';
}

没有使用mysql_fetch_assoc选择的详细信息

0 个答案: