為什么PHP在sql查詢中顯示解析錯誤

Question

$rows = ( "SELECT * FROM `client` WHERE `clientid` = '".$_SESSION['clientid']."' LIMIT 1" );
$result1 = ( "SELECT `serverid`, `ipid`, `name`, `game`, `status`, `online`, `slots`, `type`, `port` FROM `server` WHERE `clientid` = '".$_SESSION['clientid']."' ORDER BY `serverid`" );
$servers = array( );
while ( $rows1 = ( $result1 ) )
{
    if ( !empty( $rows1['ipid'] ) )
    {
        $rows2 = ( "SELECT `ip` FROM `ip` WHERE `ipid` = '".$rows1['ipid']."' LIMIT 1" );
        $rows1 = ( $rows1, $rows2 );
    }
    ( $servers, $rows1 );
}

所以，我收到此錯誤：

Parse error: syntax error, unexpected ',' in /path/to/index.php on line 10

live 10 是這個：

$rows1 = ( $rows1, $rows2 );

我的朋友給了我這個 php 應用程序，所以我不太熟悉這個問題。

Answer 1

PHP 不支持在括號內聯合兩個表達式：

$rows1 = ( $rows1, $rows2 );

所以，應該是

$rows1 = new Array( $rows1, $rows2 );

或者

$rows1 = $rows1.$rows2;