PHP中的“无法通过引用传递参数2”错误

Question

当我运行此代码以更新我的 likeOne 列并将其设为空 ("")....

    $sql11 = $conn->prepare("UPDATE UserData SET likedOne=? WHERE username=?");
    $sql11->bind_param('ss',"",$Username);
    $Username = "netsgets";
    $sql11->execute();

我收到这个错误....

1 Fatal error: Cannot pass parameter 2 by reference in /xxx/xxx/xxx/test.php on line 36. 

线路是....

$sql11->bind_param('ss',"",$Username);

怎么了？

Answer 1

你需要使用一个变量，bind_param只接受变量而不是直接取值。

   $likedone ="";
   $sql11->bind_param('ss',$likedone,$Username);