Question

我有一个会话表和一张受邀表。会话表存储了所有消息以及conversationId（不需要知道更多）。受邀表有两列名为conversationId和profileId。我想选择与邀请列表匹配的conversationId。我如何在MYSQL和PHP中做到这一点？

到目前为止，我已经在PHP中完成了这个：

 $sql = "
    SELECT invited.conversationId, count(invited.conversationId) AS cof
    FROM conversation_invited AS invited 
    WHERE   invited.profileId = '$profileId'//This is myself
    ";
    $i = 0;
    foreach($invited as $v){//$invited is an array of invited's profileIds
            $sql .= "
    AND     invited.profileId = '$v' ";
        $i++;
    }

    $sql .= "
    GROUP BY invited.conversationId
    ORDER BY count(invited.conversationId) ASC
    LIMIT 1
";

我想要创建的sql语句想要完全匹配邀请。例如，如果被邀请的数组由['27'，'34'，'36']组成，那么我想要找到与被邀请数组完全匹配的对话（Id）。邀请也不少，也不多。

Answer 1

试试这个

$invites = join(',', array_merge($invited, (array)$profileId))
$sql = "
    SELECT invited.conversationId, count(invited.conversationId) AS cof
    FROM conversation_invited AS invited 
    WHERE invited.profileId IN($invites)
    GROUP BY invited.conversationId
    ORDER BY count(invited.conversationId) ASC
    LIMIT 1
";

Answer 2

function add_quotes(&$item) {
    $item = "'$item'";
} 
array_walk($invited, 'add_quotes'); 
$t = implode(',', $invited);
$sql = "
    SELECT invited.conversationId, count(invited.conversationId) AS cof
    FROM conversation_invited AS invited 
    WHERE invited.profileId IN ('$profileId', $t)
    GROUP BY invited.conversationId
    ORDER BY count(invited.conversationId) ASC
    LIMIT 1
";

请注意，上述内容将更改$invited中的值。

从邀请中选择conversationId

2 个答案: