![](/img/trans.png)
[英]PHP/MySQL - Select RANDOM value from column where another column equals value
[英]Select where a column equals any value from another query
我有一个查询来查找名为entryEntryPairs的表中的关系:
if (!$result = $mysqli->query(
"SELECT entryIdA AS relatedId FROM entryEntryPairs WHERE entryIdB = ".$inRow["id"].
" UNION ".
"SELECT entryIdB AS relatedId FROM entryEntryPairs WHERE entryIdA = ".$inRow["id"]
)) return "Something went wrong";
但是,我真正想要的是使用所有relatedId
作为对名为entry的表的更大查询的一部分。 形式的东西:
SELECT title FROM entry WHERE id =
id =
右侧的缺失部分将是所有relatedIds
。 这可以在一个更大的查询中完成,还是我必须迭代插入id =
?右边的每个relatedId
的第一个查询的结果?
您可以尝试使用IN
"SELECT title FROM entry WHERE id IN (SELECT entryIdA AS relatedId FROM entryEntryPairs WHERE entryIdB = ".$inRow["id"].
" UNION ".
"SELECT entryIdB AS relatedId FROM entryEntryPairs WHERE entryIdA = ".$inRow["id"].")"
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.