phpmysqlhtmldatabaseprepared-statement

How to output MySQL table in PHP


I'm testing this code

$query = "SELECT * from `items` where category_id=?";
$stmt = $mysqli->query($query);
$stmt->bind_param("s", $category_id);
$stmt->execute();
$result = $stmt->get_result();
while ($row = $stmt->fetch_assoc()) {
    printf ("%s (%s)\n", $row['name'], $row['code']);
}

Is there any easier way to output whole table to HTML markup?


Solution

  • The following code snippet can match your requirements..but i'm not sure whether this is easier way or not..Hope this will help.

    For the php,

    $pricequery="SELECT price FROM technoxchange";
    $result=mysql_query($pricequery);
    
    while($row= mysql_fetch_array($result)){
       $prices [] = $row['price'];
    }
    
    echo json_encode( array( 'prices' => $prices ) ); 
    

    For the js,

    var p;
    $.get("getTechnoXchange.php", function(data){
        p = data.prices;
    });
    
    $('#priceUnicus').html( p[0] ); 
    $('#priceHire').html( p[1] ); 
    $('#priceMonsterArena').html( p[2] );