TYPE, last_name users.last_name%TYPE) ASBEGIN
IF(SELECT gift_lists.users_id, COUNT (gift_lists.gift_lists_id) as number_giftlists < 5 FROM gift_lists LEFT JOIN users ON gift_lists.users_id=
users.u
Gift可以通过一个表关联拥有多个gift_images。我正在尝试返回至少有一个gift_images关联的具有特定隐私级别的有限#礼物。本质上,我想返回: gift条目和它的第一个关联的gift_image (gift_image应该按照它拥有的位置值排序,位置1是第一个)。没有关联gift_image的礼物应该被忽略。SELECT gifts.* FROM gifts LEFT JOIN gift_images ON gifts.id = <em
function gift_giver() $people = array ("$heroname", "$friendname", "$wizardname", "Captain Rumbeard", "$frogname");
$gifts = array("a magic compass", "the gift of no fear", "all seeing powers"