如何通过jquery $.ajax函数从php脚本中理解和检索多个值

instead of using $.ajax you can also use this...
var id=$("#id").val();
 $.getJSON('script.php', {id:id}, function(json) {
                        var id=json.id;
                        var name=json.name;
                        var email=json.email;
        });
            }

in your php scrip..    
<?php 
    mysql_connect($host,$dbuser,$dbpass);
    mysql_select_db($db_name);
    $id=$_GET['id'];
    $query="select * from tbl_name where id='$id'";
    $rs=mysql_query($query);
    $row=mysql_fetch_assoc($rs);
    $id=$row['id'];
    $name=$row['name'];
    $email=$row['email'];
    $data=array("id"=>$id,"name"=>$name,"email"=>$email);
    echo json_encode($data);
    ?>

最好的方法是将其作为 json 对象发送回 javascript。所以对于您的用户示例

 // assuming post containing a username, using pseudo code for db access
$row = $db->fetchAssoc("select * from users where username = '".dbEscape($_POST["username"]."'");
header("Content Type: application/json");
echo json_encode($row);

ajax 代码如下所示

$.ajax({
 type: "POST",
 url: "your_page.php",
 data: { param1: "val1", param2: "val2" }
 }).done(function( data) {
        // do your callback operation
});

从your_page.php中获取值，如下所示

$_POST["param1"] , $_POST["param2"]

如果您想了解更多信息，请单击此处

我认为这将解决您的问题。

<script type="text/javascript">
$.ajax({ // ajax call starts
          url: 'serverside.php', // JQuery loads serverside.php
          data: 'button=' + $(this).val(), // Send value of the clicked button
          dataType: 'json', // Choosing a JSON datatype
          success: function(data) // Variable data contains the data we get from serverside
          {
          }
      });

</script>
dataType: json..so jason will return multiple values. from you php script 
echo json_encode($array_of_val);

$.ajax({
 type: "POST",
 dataType:"html",
 url: "ajax_page.php",
 data:"params1="$params1"&params2="+$params2,
 }).done(function(value) {
        // write your callback operation
});

您可以使用 POST 方法获取ajax_page.php上的数据，例如

$_POST['参数1'];和$_POST['参数2'];

$query= select statement
$items=array();
while ($row = mysql_fetch_object($rs)) {
  array_push($items, $row);
}
$result["rows"] = $items;
echo json_encode($result);