Question

我试图从mysql数据库中获取数据...但我只需要数组指定的结果...例如我有一个数据库，其中填充了成员数据：用户名，电子邮件等。另一张桌子上我按照会员的方式存储了他们的联系人列表......所以我正在寻找一种方式来展示他们的联系人名单...我不知道最好的方式是什么...欢迎任何提示......例如：

$result = mysql_query("select contact_id from member_contact_list");

$contacts = array();

while ($row = mysql_fetch_array($result)){  
   $c_array[] = $row[ 'username' ];
}

$c_array = implode( ',', $existing_users );

$sql = "SELECT * FROM members WHERE id="c_array[]"";

有点难以解释......我希望有人明白我的意思。

Answer 1

IN会帮助您

$sql = "SELECT * FROM members WHERE id IN (".$c_array.")";

请注意$c_array是由member id内爆,组成的字符串。

Answer 2

使用In

$sql = "SELECT * FROM members WHERE id in (".$c_array.")";

Answer 3

使用JOIN，您可以获得给定成员的联系人列表：

SELECT * FROM members AS m
LEFT JOIN
contacts AS c ON (m.user_id = c.user_id)
WHERE m.user_id = 1;

JOINS比子查询更有效

Answer 4

奇怪的程序。

$result = mysql_query("select contact_id from member_contact_list");

$contacts = array(); //**not used anywhere else in the given code**

while ($row = mysql_fetch_array($result)){  
   $c_array[] = $row[ 'username' ]; //**was that supposed to be $contacts?**
   //** also was this the right version: $contacts = $row['id']; ??? **//
}

$c_array = implode( ',', $existing_users ); 
//** where was $existing_users defined? Was that supposed to be $contacts, too?**

$sql = "SELECT * FROM members WHERE id="c_array[]"";
//** correct version is indeed with the clause "WHERE id IN (".$c_array.")" **;

//** last but not least, the $c_array seems to be the collection of user names, not user ids. **//

从mysql获取数组特定数据

4 个答案: