Question

我将如何在php中处理这个Ajax。 我想要做的是将数据发送到process.php如果模式= loadlinks它会做一个MySQL查询

function PresentLinks(div_id){ 

     $("#loading-status").fadeIn(900,0); 
     $("#loading-status").html("<img src='img/bigLoader.gif' />");   

$.ajax({ 

      type: "POST", 
      url: "process.php", 
      data: "mode=loadlinks", 

      success: function(msg){ 

       $("#loading-status").fadeOut(900,0); 
       $("#"+div_id).html(msg); 


      } 

     });} 

我想过程

if($_POST['mode'] == loadlinks){ // this is what i want to ask 
$query = "SELECT * FROM site ORDER BY link_id DESC"; 
$result = MYSQL_QUERY($query) or die (mysql_error()); 
while($data = mysql_fetch_row($result)){ 
echo ("$data[1]"); 
}} 
else { 
} 

Answer 1

你需要引用字符串值

if($_POST['mode'] == 'loadlinks'){..... 

Answer 2

您需要在PHP中引用字符串。否则，它们将被假定为常量。您还应该使用PDO。

if($_POST['mode'] == 'loadlinks'){ 
    $pdo = new PDO('mysql:host=HOST;dbname=DATABASE'), 'username', 'password'); 
    $stmt = $pdo->execute('SELECT * FROM site ORDER BY link_id DESC'); 

    $sites = $stmt->fetchAll(); 
    foreach($sites as $site) { 
     echo "<div>" . $site['name'] . "</div>"; // Or whatever info you want to output 
    } 
} 

对于性能，您应该指定表列名称来检索而不是使用*。