php接收地址欄上的id,該怎麼解決

php接收地址欄上的id

PHP code

檔案命名為test.php,編輯環境為editplus，按ctrl+b，在頁面上彈出

HTML code

Notice: Undefined index: id in D:\server\apache\www\test.php on line 2參數有誤

於是我便在地址欄上輸入

HTML code

http://localhost/test.php[color=#FF0000]?id=2[/color]

可是這次顯示的錯更多

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Warning: mysql_connect() [function.mysql-connect]: php_network_getaddresses: getaddrinfo failed: 不知道這樣的主機。 in D:\server\apache\www\test.php on line 7Warning: mysql_connect() [function.mysql-connect]: [2002] php_network_getaddresses: getaddrinfo failed: 不知道這樣的主機。 (trying to connect via tcp://loaclhost:3306) in D:\server\apache\www\test.php on line 7Warning: mysql_connect() [function.mysql-connect]: php_network_getaddresses: getaddrinfo failed: 不知道這樣的主機。 in D:\server\apache\www\test.php on line 7Warning: mysql_query() expects parameter 2 to be resource, boolean given in D:\server\apache\www\test.php on line 9Warning: mysql_query() expects parameter 2 to be resource, boolean given in D:\server\apache\www\test.php on line 11Warning: mysql_fetch_assoc() expects parameter 1 to be resource, null given in D:\server\apache\www\test.php on line 12鍙傛暟鏈夎

我尋找原因好久，實在沒招了，哪位大大能幫忙分析原因，感激不盡！！！！！

------解決方案--------------------
mysql_connect(), 資料庫連接錯誤

PHP code

$conn = mysql_connect('loaclhost','root','123456') or die("can not connect Mysql Server");mysql_select_db("資料庫名",$conn);//你的資料庫名叫 use msg ？$sql = "select * from denglu where id = $id";$var = mysql_query($sql);
------解決方案--------------------
if(isset($_REQUEST['id'])){
$id = $_REQUEST['id'];
if($id == 0){
exit("The parameter is wrong!");
}
}
echo $id;
@ $conn = mysql_connect('localost', 'root', '123456') or die("Connection Failure:" .mysql_errno());
mysql_select_db('msg', $conn);
mysql_query("set names ''utf8");
$sql =  "select * from denglu where id = $id";
$result = mysql_query($sql);
while($rs = mysql_fetch_assoc($result)){
   var_dump($rs);
}
輸入url: localhost/index.php?id=2即可擷取到id
------解決方案--------------------
$id=isset($_GET["id"])?$_GET["id"]:0;
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