sql을 묶어서 하나의 리스트로 만들려면
본문
최신글뽑아오는 lib소스인데요.
빨간부분에 표시한 코드를 한번에 묶어서
0~3 $i 로 출력을 하고싶거든요.
function latest_mark($skin_dir='', $code='')
{
global $g5;
$latest_skin_path = G5_SKIN_PATH.'/latest/'.$skin_dir;
$latest_skin_url = G5_SKIN_URL.'/latest/'.$skin_dir;
$list = array();
$sql = " select * from g5_mark where mark2 !='' order by CAST(mark2 AS UNSIGNED) DESC limit 0, 1 ";
$sql = " select * from g5_mark where mark6 !='' order by CAST(mark6 AS UNSIGNED) DESC limit 0, 1 ";
$sql = " select * from g5_mark where mark10 !='' order by CAST(mark10 AS UNSIGNED) DESC limit 0, 1 ";
$sql = " select * from g5_mark where mark14 !='' order by CAST(mark14 AS UNSIGNED) DESC limit 0, 1 ";
$result = sql_query($sql);
for ($i=0; $row = sql_fetch_array($result); $i++)
$list[$i] = $row;
$list[$i]['id'] = $row['wr_id'];
$list[$i]['href'] = G5_BBS_URL.'/board.php?bo_table=mysys&wr_id='.$row['wr_id'];
$list[$i]['mb_id'] = $row['mb_id'];
$list[$i]['nick'] = get_sideview($row['mb_id']);
ob_start();
include $latest_skin_path.'/latest.skin.php';
$content = ob_get_contents();
ob_end_clean();
return $content;
}
답변 1
구조를 잘 알고 있다면 JOIN등을 써서 가져오는게 일반적이나...
위 질문사항에서 제가 가장 쉽게 해드릴수 있는 건 union 을 쓰라는 답변밖에 못드리겠습니다.
$sql = "
select * from g5_mark where mark2 !='' order by CAST(mark2 AS UNSIGNED) DESC limit 0, 1
union
select * from g5_mark where mark6 !='' order by CAST(mark6 AS UNSIGNED) DESC limit 0, 1
union
select * from g5_mark where mark10 !='' order by CAST(mark10 AS UNSIGNED) DESC limit 0, 1
union
select * from g5_mark where mark14 !='' order by CAST(mark14 AS UNSIGNED) DESC limit 0, 1
";
