[英]Return Yes or No PHP Query
I have the following query: 我有以下查询:
$result = query("SELECT IdUser, followingID FROM following WHERE IdUser = '%d' AND followingID = '%d'", $id, $followingID);
I was wondering how I would return if the result is YES or NO based off the 0 and 1 count for if the relationship exists? 我想知道如果基于存在关系的0和1计数,结果是是还是否,我将如何返回?
Here is one way: 这是一种方法:
select (case when exists (SELECT IdUser, followingID
FROM following
WHERE IdUser = '%d' AND followingID = '%d'
)
then 'YES' else 'NO'
end) as YesOrNo
, $id, $followingID
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