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$stmt = mysqli_prepare($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($stmt))
echo 'Request Sent';
else
echo 'Something went wrong !';
在上面的代码中,我已经写mysqli_stmt_bind_param($stmt, "ii", $fid, $uid);
我应该转换$fid = (int) $fid
作出改善?我是否需要一个变量的方法,用语句时,INT转换 - PHP
[ToUID的数据类型,数据库中的FromUID的数据类型是否真的有区别吗?
mysqli_stmt_bind_param($stmt, "ii", $fid, $uid);
mysqli_stmt_bind_param($stmt, "ss", $fid, $uid);