2014-05-06 105 views
0

我在嘗試上傳視頻並將鏈接保存到數據庫。我寫了代碼,但我得到這個錯誤。有人知道我做錯了什麼嗎?Php視頻上傳錯誤

$name = $_FILES['file']['name']; 
$temp = $_FILES['file']['tmp_name']; 

move_uploaded_file($temp,"uploaded/".$name); 
$file = "http://127.0.0.1/webapp/video%20upload%20and%20playback/uploaded/$name"; 

mysql_query(" 
    INSERT INTO oefeningen (`Id`, `Fname`, `Lname`, `namevideo`, `video`, `expl`) 
    VALUES (NULL, '$fname', '$lname', ,'$namevideo', '$file','$expl') 
") or die (mysql_error()); 

thx!

我得到的錯誤是:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''test.mov', 'http://127.0.0.1/webapp/video%20upload%20and%20playback/uploade' at line 2 
+0

你的代碼的短段是易受[SQL注入]( https://www.owasp.org/index.php/PHP_Security_Cheat_Sheet#Database_Cheat_Sheet)和[遠程執行](https://www.owasp.org/index.php/PHP_Security_Cheat_Sheet#File_uploads)代碼。所以你應該真正閱讀安全。 – Yoshi

回答

0
$lname', ,'$namevideo', 

你有一個額外的逗號所以它應該是

$lname', '$namevideo', 
+0

Thx這是有史以來最快的答案! – user3356007

+0

選擇正確的上述答案並投票。謝謝.. –