mysqli查询。有没有办法显示错误？即使我的查询工作正常

Question

我试图将数据库中的一些数据显示到html表中。我的其他查询工作正常，我能够检索一些其他表并将其显示到我的网站，虽然这个特定的查询甚至没有，如果有一个函数mysqli_error（）。我尝试直接从我的数据库（phpmyadmin）执行查询，它检索它就好了。

这是我的代码。

注意：$ p_id变量有一些数据是＆＃39; 3＆＃39;。

PHP

$p_sql = "SELECT p.package_name, p.package_price, p.package_details, p.package_categories, p.package_id FROM event_table as e inner join package as p on e.package_id = p.package_id where e.package_id = '$p_id' group by e.package_id";
$data_p = mysqli_query($conn, $p_sql) ;
$result = mysqli_fetch_assoc($data_p);

$p_amount = $result['package_price'];



  while ($record_p = mysqli_fetch_array($data_p)) {


echo "<table border = 1>";
echo "<tr>";
echo "<th>" . "Package Name" . "</th>";
echo "<th>" . "Package Price" . "</th>";
echo "<th>" . "Package Details" . "</th>";
echo "<th>" . "Package Categories" . "</th>";

echo "</tr>";
echo "<tr>"; 
echo "<td>" . "<br />" .  $record_p['package_name'] . "<br />" . "</td>" ;
echo "<td>" . "<br />" .  $record_p['package_price'] . "<br />" . "</td>" ;
echo "<td>" . "<br />" .  $record_p['package_details'] . "<br />" . "</td>" ;
echo "<td>" . "<br />" .  $record_p['package_categories'] . "<br />" . "</td>" ;

echo "<tr/>";





}
echo "</table>";

if(!$data_p){
        echo("Error description: " . mysqli_error($conn));
    }