використовуючи mysqli, а не mysql, доступ до DB - mysql, phpmyadmin

    please find below a script in php that that fetch data from my database company, the database is in phpMyadmin

Я розумію, що існують безпечні способи отримання даних. Моє питання Чи можу я поліпшити код нижче за допомогою mysqli для вибірки замість mysql_query.

    <?php
get results from database. Can I improve the code below using mysqli to fetch instead of mysql_query in the following line?.

$result = mysql_query("SELECT * FROM customer")
or die(mysql_error());
tables begins here with the customer id and customer name
echo "<table >";
echo "<tr>
<th>ID</th>   the customer id
<th>Name</th> the customer name
</tr>";

// loop through results of database query, displaying them in the table
the table below display the columns in html using mysql fetch array. all data is displayed.

while($row = mysql_fetch_array( $result )) {

// echo out the contents of each row into a table
my purpose is embed the php in html in another way using msqli
echo "<tr>";
echo "<td>" . $row["CustomerID"]."</td>";
echo "<td>" . $row["Name"]. "</td>";
echo "</tr>";
}
echo "</table>";
?>
my purpose is embed the php in html in another way using msqli

Відповіді:

Просто замініть всі mysql з mysqli

      $result = mysqli_query("SELECT * FROM customer");

while($row = mysqli_fetch_array( $result )) {

Якщо ви хочете бути простим

Або зробіть це так ...

  /*Your connection info goes here...*/
$mysqli = new mysqli("localhost", "my_user", "my_password", "world");

if (mysqli_connect_errno()) {
printf("Connect failed: %sn", mysqli_connect_error());
exit();
}

$query = "SELECT * FROM customer";

if ($result = $mysqli->query($query)) {

echo "<table ><tr><th>ID</th><th>Name</th></tr>";

/* fetch results by row */
while ($row = $result->fetch_row()) {
echo "<tr><td>" . $row["CustomerID"]."</td><td>" . $row["Name"]. "</td></tr>";
}

/* free the result set */
$result->close();
}

/* close the connection */
$mysqli->close();