mysql/php更新語法，傳遞php變量

Question

我有一個mysql表，列：name，active（active is int）。我想將圖片上傳到服務器，當它們已經在服務器上時，我想讓它們處於活動狀態，以便將它們用作新的標誌。 （當活動= 1，圖片是新的標誌）我不明白爲什麼我的「$查詢」不起作用。其他一切正常。

$name = $_FILES["file"]["name"]; 
if (file_exists("/home/a1829256/public_html/admin/logo/" .$name)){   //if file already on the server 
     echo "file is already on the server"; 
     echo ". Making it the new logo"; 
     $deselectlogo = "UPDATE newlogo SET active=0 WHERE active>0";    //deselect old 
     $deselectlogoResult = mysql_query($deselectlogo); 
     $query = ("UPDATE newlogo SET active=1 WHERE name=" .$name);   // make the new logo active 
     $queryResult = mysql_query($query); 
     if (!$queryResult){echo "error";} 
     if ($queryResult){echo "success";} 

Answer 1

$query = "UPDATE newlogo SET active=1 WHERE name='".$name."'";   // make the new logo active