1
$stmt = mysqli_prepare($con,"INSERT INTO friend_request (ToUID, FromUID) VALUES (?,?)");
mysqli_stmt_bind_param($stmt, "ii", $fid, $uid);
mysqli_stmt_execute($stmt);
if (mysqli_stmt_affected_rows($stmt))
echo 'Request Sent';
else
echo 'Something went wrong !';
在上面的代碼中,我已經寫mysqli_stmt_bind_param($stmt, "ii", $fid, $uid);
我應該轉換$fid = (int) $fid
作出改善?我是否需要一個變量的方法,用語句時,INT轉換 - PHP
[ToUID的數據類型,數據庫中的FromUID的數據類型是否真的有區別嗎?
mysqli_stmt_bind_param($stmt, "ii", $fid, $uid);
mysqli_stmt_bind_param($stmt, "ss", $fid, $uid);