はどのように私は1つのクエリでこれらの2つの選択クエリを使用します

Question

$rget_comments = mysql_query("SELECT DISTINCT comment.id, comment.comment_name, comment.comment_body FROM rs_comments AS comment WHERE comment.comment_parent ='0' ORDER BY comment.id ASC"); 
while($get_comment = mysql_fetch_array($rget_comments)) 
{ 

    echo '+ <a href="comment.php?id=' . $get_comment['id'] . '">' . $get_comment['comment_name'] . '</a><br />'; 
    echo $get_comment['body'] . '<br />'; 

    $rget_comments = mysql_query("SELECT DISTINCT 
    (sub_comment.id) AS sub_id, (sub_comment.comment_name) AS sub_comment_name, (sub_comment.comment_body) AS sub_comment_body 
    FROM rs_comments AS sub_comment WHERE sub_comment ON sub_comment.comment_parent = '1'"); 
    while($get_comment = mysql_fetch_array($rget_comments)) 
    { 

     echo '- <a href="comment.php?id=' . $get_comment['sub_id'] . '">' . $get_comment['sub_comment_name'] . '</a><br />'; 
     echo $get_comment['sub_body'] . '<br />'; 
    } 

} 

私のテーブル情報：

$rget_comments = mysql_query(" 
    SELECT DISTINCT 
    comment.id, comment.comment_name, comment.comment_body, (sub_comment.id) AS sub_id, (sub_comment.comment_name) AS sub_comment_name, (sub_comment.comment_body) AS sub_comment_body 
    FROM rs_comments AS comment 
    LEFT OUTER JOIN rs_comments AS sub_comment ON sub_comment.comment_parent = comment.id 
    WHERE comment.comment_parent ='0' 
    GROUP BY comment.id 
    ORDER BY comment.id ASC"); 
    while($get_comment = mysql_fetch_array($rget_comments)) 
    { 
     echo '+ <a href="comment.php?id=' . $get_comment['id'] . '">' . $get_comment['comment_name'] . '</a><br />'; 
     echo $get_comment['body'] . '<br />'; 

     echo '- <a href="comment.php?id=' . $get_comment['sub_id'] . '">' . $get_comment['sub_comment_name'] . '</a><br />'; 
     echo $get_comment['sub_body'] . '<br />'; 
    } 

結果：私は、クエリを得たが、最初の行だけが返されます

CREATE TABLE IF NOT EXISTS `rs_comments` (
    `id` int(10) unsigned NOT NULL AUTO_INCREMENT, 
    `comment_type` varchar(255) NOT NULL, 
    `comment_post_ID` int(11) NOT NULL, 
    `comment_parent` int(11) NOT NULL, 
    `comment_name` varchar(128) CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL DEFAULT '', 
    `comment_body` text CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL, 
    PRIMARY KEY (`id`) 
) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=5 ; 

INSERT INTO `rs_comments` (`id`, `comment_type`, `comment_post_ID`, `comment_parent`, `comment_name`,`comment_body`) VALUES 
(1, 'post', 1, 0, 'test comment', 'test comment body'), 
(2, 'post', 1, 0, 'test comment 2', 'test comment 2 body'), 
(3, 'post', 1, 1, 'sub comment 1', 'sub comment 1 body'), 
(4, 'post', 1, 1, 'sub comment 2', 'sub comment 2 body'); 

：

+test comment 
test comment 1 body 
-sub comment 1 
sub comment 1 body 
+test comment 2 
test comment 2 body 

予想結果：

+test comment 
test comment 1 body 
-sub comment 1 
sub comment 1 body 
-sub comment 2 
sub comment 2 body 
+test comment 2 
test comment 2 body 

Answer 1

GROUP BY comment.idがこの問題を引き起こす。 GROUP BYは、同じidと異なるsub_idを持つ2つのレコードを1つにグループ化します。このようにそれを削除します。

SQL：

SELECT DISTINCT 
    comment.id, comment.comment_name, comment.comment_body, (sub_comment.id) AS sub_id, (sub_comment.comment_name) AS sub_comment_name, (sub_comment.comment_body) AS sub_comment_body 
    FROM rs_comments AS comment 
     LEFT OUTER JOIN rs_comments AS sub_comment ON sub_comment.comment_parent = comment.id 
    WHERE comment.comment_parent ='0' 
    ORDER BY comment.id ASC 

PHP：

$flag = array(); 
while($get_comment = mysql_fetch_array($rget_comments)) { 
    if(!isset($flag[$get_comment['id']])) { 
     echo '+ <a href="comment.php?id=' . $get_comment['id'] . '">' . $get_comment['comment_name'] . '</a><br />'; 
     echo $get_comment['body'] . '<br />'; 
     $flag[$get_comment['id']] = true; 
    } 
    if($get_comment['sub_id']) { 
     echo '- <a href="comment.php?id=' . $get_comment['sub_id'] . '">' . $get_comment['sub_comment_name'] . '</a><br />'; 
     echo $get_comment['sub_body'] . '<br />'; 
    } 
}