php - mssql_rows_affected 返回错误的行数

Question

下面的代码是一个 PHP 脚本，它通过单击链接来激活用户的帐户。 执行更新表的 SQL 查询。更新很好，但最后我会显示一条消息，无论激活是否成功。但 mssql_rows_affected 返回可能 0 因此，即使更新成功，我也会收到错误的消息。无法理解出了什么问题。提前致谢。

$account = $_GET['account'];
$code = $_GET['code'];

$account = antiinjection($account);
$code = antiinjection($code);

connectdb($CONFIG['dbdbname'], $CONFIG['dbaddress'], $CONFIG['dbuser'], $CONFIG['dbpass']);
mssql_query (sprintf(UPDATE_PAY_STAT, $account, $code));
mssql_query("COMMIT");
$rowsaffected = mssql_rows_affected();
$message =  "
                <div align=center>The account {$account} has been activated! Have fun!</div>
            ";

if( $rowsaffected != 0 ){
echo "$message";
}
else
{
    $message =  "
                <div align=center>The account {$account} has not been activated due to an error. Please contact the Account Support.</div>
                ";
}

echopage('header', 'Activate');
echo($message);
echopage('footer', '');

Answer 1

因为我无法让 mssql_affected_rows() 工作，所以我这样做了。我添加了 1 个 SQL 查询来检查 x 帐户名称是否具有 pay_stat ='1'。感谢您的回答!

   $link=mssql_query (sprintf(SELECT_COUNT_PS, $account));
$row = mssql_fetch_row($link);

if( $row[0] == '1' ){
$message =  "
                <div align=center>The account {$account} has been activated! Have fun!</div>
            ";
}
else
{
    $message =  "
                <div align=center>The account {$account} has not been activated due to an error. Please contact the Account Support.</div>
                ";
}