PHP MySQL的 - 列 'ID' 在IN/ALL/ANY查詢是不明確的

Question

目前，我有以下查詢工作：

$sql = $db->Query("SELECT a.id FROM surf a LEFT JOIN users b ON b.id = a.user LEFT JOIN surfed c ON c.user = '".$data['id']."' AND c.site = a.id WHERE a.active = '0' AND (b.coins >= a.cpc AND a.cpc >= '2') AND (c.site IS NULL AND a.user !='".$data['id']."')".$dbt_value." ORDER BY a.cpc DESC LIMIT 1"); 

但我想補充的後續和聲明它：

AND `cpc`<='10' AND (`id` NOT IN (SELECT `site` FROM `surfed_site`)) 

所以我做它看起來像：

$sql = $db->Query("SELECT a.id FROM surf a LEFT JOIN users b ON b.id = a.user LEFT JOIN surfed c ON c.user = '".$data['id']."' AND c.site = a.id WHERE a.active = '0' AND (b.coins >= a.cpc AND a.cpc >= '2') AND `cpc`<='10' AND (`id` NOT IN (SELECT `site` FROM `surfed_site`)) AND (c.site IS NULL AND a.user !='".$data['id']."')".$dbt_value." ORDER BY a.cpc DESC LIMIT 1"); 

但後來我收到此消息：

IN/ALL/ANY子查詢中的列'id'不明確

我的查詢出了什麼問題？

Answer 1

嘗試改變

AND `cpc`<='10' AND (`id` NOT IN (SELECT `site` FROM `surfed_site`)) 

到

AND `cpc`<='10' AND (`a.id` NOT IN (SELECT `site` FROM `surfed_site`)) 

你需要a.id，因爲它不知道id列在表中引用其中。

Answer 2

該錯誤的原因是因爲有多個表包含列ID。爲了執行查詢時，必須提供在列ID所在，以便服務器能夠識別哪個表有關的表格，如

AND `cpc`<='10' AND (a.`id` NOT IN (SELECT `site` FROM `surfed_site`)) 

如果你的意思，你是比較ID從表surf

Answer 3

您必須指定所需的id。 a.id,b.id等

Answer 4

另外我假設$數據包含轉義數據（單引號）？如果不是這是第二個問題。