検索の並べ替えの問題

出会い系サイトでプロフィール写真のないメンバーがプロフィール写真のあるメンバーの後に順位を付けるように検索する方法。最後に、検索時にエラーが発生しました: 解析エラー: 構文エラー。予期しない T_STRING、E:PHPnow-1.5.6htdocssourceserviceindexservice.user.php の 147 行目で ')' を期待しています
エラー行は最後の行です: order by avatarflag desc 誰が上記のエラーの場所を確認できるでしょうか?次のように:
public function validSearch( )
$args = array( );
$s_searchtype = XRequest::getargs( "s_searchtype" );
$s_sex = XRequest::getint( " s_sex" );
$s_sage = XRequest::getint( "s_sage" );
$s_eage = XRequest::getint( "s_eage" );
$s_dist1 = XRequest::getint( "s_dist1" );
$s_dist2 = $s_dist3 = XRequest::getint( "s_dist3" );
$s_lovesort = XRequest::getint( "s_lovesort" );
$s_salary = XRequest::getint( "s_ssalary" );
$s_esalary = XRequest::getint( "s_esalary" );
$s_sedu = = XRequest::getint( "s_eedu" )
{
$s_marry = );
else
{
$s_marry = XRequest::getargs( "s_marry" );
$s_havechild = XRequest::getargs( "s_havechild" );
$s_car = XRequest::getint( " s_car" );
$s_avatar = => $s_sage,
"s_eage" => $s_eage,
"s_dist1" => $s_dist1,
"s_dist2" => $s_dist2,
"s_dist3" => sort" => $s_lovesort、
"s_sheight" => $s_height、
"s_eheight" => $s_eheight、
"s_salary" => _sedu" => $s_sedu、
"s_eedu" => $ s_eedu、
"s_marry" => $s_marry、
"s_havechild" => => $s_car、
"s_avatar" => $sql = "";
if ( 0 < $s_sex )
{
$sql .= " AND v.gender='". $s_sex."'";
$countwhere .= " AND ps.gender='"; s_sex."'";
}
if ( 0 < $s_sage && 0 < $s_eage )
{
$year = date( "Y", time( ));
$eageline = $year - $s_sage;
$sql .= " AND p.ageyear >= ".$sageline. " AND p.ageyear $countwhere .= "; .ageyear >= ".$sageline." AND ps.ageyear <= {$eageline}";
}
if ( 0 < $s_dist1 )
{
$sql .= " AND p.provinceid='" .$s_dist1."'";
$countwhere .= " AND ps.provinceid='".$s_dist1."'"
}
if ( 0 < $s_dist2 )
{
$sql .= " AND p.cityid='".$s_dist2."'";
$countwhere .= " AND ps.cityid='".$s_dist2."'";
}
if ( 0 < $s_dist3 )
{
$sql .= " AND p.distid='".$s_dist3."'";
$countwhere .= " AND ps.distid='".$s_dist3."'";
}
if ( 0 < $s_lovesort )
{
$sql .= " AND p.lovesort='".$s_lovesort."'";
$countwhere .= " AND ps.lovesort='".$s_lovesort."'";
}
if ( 0 < $s_sheight && 0 < $s_eheight )
{
$sql .= " AND p.height >= ".$s_sheight." AND p.height <= {$s_eheight}" ;
$countwhere .= " AND ps.height >gt;= ".$s_sheight." AND ps.height <= {$s_eheight}";
}
if ( 0 < $s_salary && 0 < $s_esalary )
{
$sql .= " AND p.salary >= ".$s_ssalary." AND p.salary <= {$s_esalary}" ;
$countwhere .= " AND ps.salary >gt;= ".$s_ssalary." AND ps.salary <= {$s_esalary}";
}
if ( 0 < $s_sedu && 0 < $s_eedu )
{
$sql .= " AND p.education >= ".$s_sedu." AND p.education <= {$s_eedu}" ;
$countwhere .= " AND ps.education >gt;= ".$s_sedu." AND ps.education <= {$s_eedu}";
}
if ( TRUE === XValid::iscomchar( $s_marry ) )
{
$sql .= " AND p.marrystatus IN (".$s_marry.")";
$countwhere .= " AND ps.marry IN (".$s_marry.")";
}
if ( TRUE === XValid::iscomchar( $s_havechild ) )
{
$sql .= " AND p.childrenstatus IN (".$s_havechild.")";
$countwhere .= " AND ps.child IN (".$s_havechild.")";
}
if ( 0 < $s_house )
{
$sql .= " AND p.housing='".$s_house."'";
$countwhere .= " AND ps.house='".$s_house."'";
}
if ( 0 < $s_car )
{
$sql .= " AND p.caring='".$s_car."'";
$countwhere .= " AND ps.car='".$s_car."'";
}
if ( $s_avatar == 1 )
{
$sql .= " AND v.avatar != '' AND v.avatarflag = '1'";
$countwhere .= " AND ps.avatar='1'";
}

return Array(
$sql,
$countwhere,
$args
order by avatarflag desc
);

}

戻り承認(解決案)

代案中検査次の SQL 句は何か。

if ( $s_avatar == 1 ) {
$sql .= " AND v.avatar != '' AND v.avatarflag = '1'";
$countwhere .= " AND ps.avatar='1'";
}

ただこの条件を満たすとき候、才有 avatarflag
order by avatarflag , 是不是要加上别名：v.

できれば完全な SQL 句を開くことをお勧めします。

Web サイトを実行すると、エラーが発生します: 解析エラー: 構文エラー、予期しない T_CONSTANT_ENCAPSED_STRING、E:PHPnow-1.5.6htdocssourceactionindexaction.user.php に T_STRING または T_VARIABLE または '$' が必要です。 51 行目
No. 51 コードの行は unset ("service");
コードは次のとおりです:
private function _unset()
{
unset ("service") }

private function _getListItems( )