Question

我写了这个选择

SELECT
  t1.id t1_id, t1.title t1_title, t1.subtitle t1_subtitle, t1.content t1_content,
  t3.picture t3_picture

FROM
  webcms_cms_content t1

INNER JOIN webcms_mod_galleries_pictures t3
  ON t1.gallery_id = t3.gallery_id

WHERE t1.structure_id =3
AND t1.status=1
AND t1.flag=1
ORDER BY t1.newsdatum DESC

和这个小小的php片段：

$i=0;
while($db->next_record()) {
    $con[$db->get_content('t1_id')]['id']           = $db->get_content('t1_id');
    $con[$db->get_content('t1_id')]['title']        = $db->get_content('t1_title');
    $con[$db->get_content('t1_id')]['link']         = $db->get_content('t1_subtitle');
    $con[$db->get_content('t1_id')]['content']      = $db->get_content('t1_content');
    $con[$db->get_content('t1_id')]['pics'][$i]     = $db->get_content('t3_picture');
    $i++;
}

返回一个这样的数组：

array
    8 => 
        array
            'id' => string '8' (length=1)
            'title' => string 'Studios (2 Personen)' (length=20)
            'link' => string '350.- bis 550.- ' (length=16)
            'content' => string 'Gem&uuml;tlich eichenm&ouml;biliertes Studio bestehend aus einem Wohnraum mit zwei Schrankbetten, Fernseher mit Radio, Telefon, Wireless, K&uuml;che mit Backofen, Wasserkocher und Kaffeemaschine, Bad/Dusche, WC und auf der S&uuml;dseite mit Terasse und Matterhornblick.' (length=269)
            'pics' => 
                array
                    0 => string 'p186rgot3ohd612pljp8m8h12d74' (length=28)
                    1 => string 'p186rgot3o1cso2nj4h81e47ils2' (length=28)
                    2 => string 'p186rgl7f84r9h671jkr1vctip61' (length=28)
                    3 => string 'p186rgot3o6fjk7c1j1nc1p1ukj1' (length=28)
                    4 => string 'p186rhu0bjuq9tfe1ca13ll1qte1' (length=28)

Soo现在可以了，但我还想要更多:)不知道这是否可行（这是我写的第一个更大的SQL查询）。

有一个＆＃34;元素＆＃34;在t3.picture

1.1所以我希望pics数组只是空的

我需要订购t1.newsdatum DESC和ORDER BY t3.rang ASC

这只能在一个查询中实现吗？有多个查询没问题，但我希望只有一个。

创建了sqlfiddle

提前感谢您的时间和建议！

Answer 1

你可以尝试“订购t1.newsdatum DESC，t3.rang ASC”吗？

Answer 2

您可以使用LEFT OUTER JOIN获取另一个表中没有匹配行的行： -

SELECT t1.id AS t1_id, 
        t1.title AS t1_title, 
        t1.subtitle AS t1_subtitle, 
        t1.content AS t1_content,
        t3.picture AS t3_picture
FROM webcms_cms_content t1
LEFT OUTER JOIN webcms_mod_galleries_pictures t3
ON t1.gallery_id = t3.gallery_id
WHERE t1.structure_id = 3
AND t1.status = 1
AND t1.flag = 1
ORDER BY t1.newsdatum DESC, t3.rang ASC

然后一些php填充你的数组： -

$prev_id = 0;
while($db->next_record()) 
{
    if ($prev_id != $db->get_content('t1_id'))
    {
        $con[$db->get_content('t1_id')]['id']       = $db->get_content('t1_id');
        $con[$db->get_content('t1_id')]['title']    = $db->get_content('t1_title');
        $con[$db->get_content('t1_id')]['link']     = $db->get_content('t1_subtitle');
        $con[$db->get_content('t1_id')]['content']  = $db->get_content('t1_content');
        $con[$db->get_content('t1_id')]['pics']     = array();
        $prev_id = $db->get_content('t1_id');
    }
    $con[$db->get_content('t1_id')]['pics'][]     = $db->get_content('t3_picture');
}

mysql选择加入2个不同的ORDER BY

2 个答案: