我有一個對話表和一個受邀表。會話表將所有消息與conversationId一起存儲(不需要了解更多信息)。被邀請的表格有兩個名爲conversationId和profileId的列。我想選擇與邀請列表相匹配的conversationId。我如何在MYSQL和PHP中做到這一點?從邀請中選擇對話ID
到目前爲止,我已經在PHP中做到了這一點:
$sql = "
SELECT invited.conversationId, count(invited.conversationId) AS cof
FROM conversation_invited AS invited
WHERE invited.profileId = '$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
";
,我要創建的SQL語句要匹配邀請準確。例如,如果邀請數組由['27','34','36']組成,則比我想要找到恰好匹配受邀請數組的對話(Id)。也沒少,也沒有更多的邀請
看在上帝的份上,轉義你在SQL查詢中使用的變量! http://php.net/manual/en/function.mysql-real-escape-string.php – 2011-03-12 18:33:46
謝謝布魯諾雷斯 – einstein 2011-03-12 18:39:50