PHP MySQL的 - 列 'ID' 在IN/ALL/ANY查询是不明确的

Question

目前，我有以下查询工作：

$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"); 

但我想补充的后续和声明它：

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

所以我做它看起来像：

$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"); 

但后来我收到此消息：

IN/ALL/ANY子查询中的列'id'不明确

我的查询出了什么问题？

Answer 1

尝试改变

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，因为它不知道id列在表中引用其中。

Answer 2

该错误的原因是因为有多个表包含列ID。为了执行查询时，必须提供在列ID所在，以便服务器能够识别哪个表有关的表格，如

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

如果你的意思，你是比较ID从表surf

Answer 3

您必须指定所需的id。 a.id,b.id等

Answer 4

另外我假设$数据包含转义数据（单引号）？如果不是这是第二个问题。