ajaxphp传递和接收变量实现思路及代码_php实例

$.ajax({ 
url: 'query.php', 
data: {id:10}, 
datatype: json 
success: function(results) { 
if (results.msg == 'success') { 
for (var i in data) { 
$('#content').append( 
'id = ' + results.data[i].id + ', description = ' + results.data[i].description + ', msrp = ' + results.data[i].msrp 
); 
} 
} else { 
$('#content').append(results.msg); 
} 
} 
}); 
 
And your php.... 
 代码如下:

if (isset($_GET['id'])) { 
$sql = "SELECT id, description, msrp FROM tbl WHERE id = '{$_GET['id']}'"; 
$return = array(); 
if ($result = mysql_query($sql)) { 
if (mysql_num_rows($result)) { 
$return['msg'] = 'success'; 
while ($row = mysql_fetch_assoc($result)) { 
$return['data'][] = $row; 
} 
} else { 
$return['msg'] = 'No results found'; 
} else { 
$return['msg'] = 'Query failed'; 
} 
header("Content-type: application/json"); 
echo json_encode($result); 
}