2013-06-26 50 views
0
$mysqli = new mysqli("localhost", $musername, $mpass, $mdatabase); 
if ($mysqli->connect_errno) { 
    printf("Connect failed: %s\n", $mysqli->connect_error); 
    exit(); 
} 
$sqquer = $mysqli->prepare('SELECT * FROM ? where id=?'); 
$sqquer->bind_param('is', $mtable, $_SESSION['User']['id']); 
if ($sqquer==false) { 
    echo "fail"; 
} 
echo "<br>Test 2"; 
$sqquer->execute(); 
echo "<br>Test 3"; 
$sqquer->store_result(); 

此輸出沒有測試。我錯過了什麼?我試圖改變參數功能,但沒有成功,我相信我錯過了一些微不足道的東西。無錯誤庫MySQLi退出

+0

是你的表名是整數類型 – Afsar

+0

是你實際在任何地方定義$ musername,$ mpass,$ mdatabase? – KyleK

+1

如果將'error_reporting(E_ALL | E_STRICT);'添加到腳本的頂部,會得到什麼結果? – Jon

回答

1

你確定啓用了mysqli嗎?

檢查php.ini文件並取消(即撈出;)此行....

;mysqli.dll 

要...

mysqli.dll (mysqli.so on linux)