Question

我有两张桌子

帐户表

id INT, username TEXT, password TEXT, proxy_id INT, enabled BOOLEAN

代理表

id INT, proxy_ip TEXT, proxy_port INT

我会有一个循环，一次通过一个帐户。我希望将最少使用的代理分配给table.proxy_id。

例如，如果我们在代理表中有2个代理，在Account表中有5个帐户

1 10.0.0.1  4000
2 10.0.0.1  4001

我们的帐户

1 david    password 2    enabled
2 mark     password 1    enabled
3 jessica  password 1    enabled
4 ashley   password NULL enabled
5 allan    password NULL enabled
6 james    password 2    disabled

我的程序将在Java中循环，遍历所有已启用的帐户，它会将已启用帐户中使用最少的代理分配给该帐户。在上面的例子中，David，Mark和Jessica已经有了一个代理集。因此循环将通过Ashley，并且需要将具有id 2的代理分配给Ashley，因为它是最少使用的。对于Allan代理1或2可以分配，因为它在任何情况下都是最少使用的。詹姆斯应该被忽略，因为他的帐户没有启用。

我希望我的问题清楚。我认为这需要在两个查询中完成？

Answer 1

这是一个想法。您需要展开代理ID以获取分配所需的列表。以下是这样做的基本方法，但并不难。为每个id分配额外的100“行”，并从当前计数开始枚举它们。然后订购这些作为作业。

第一部分：

select p.id, count(a.id) as cnt,
       generate_series(1, 100) + count(a.id) as proxy_seqnum
from proxy p left join
     accounts a
     on p.id = a.proxy_id and a.enabled 
group by p.id;

现在，计算整体顺序排序：

select ap.*, row_number() over (order by proxy_seqnum) as seqnum
from (select p.id, count(a.id) as cnt,
             generate_series(1, 100) + count(a.id) as proxy_seqnum
      from proxy p left join
           accounts a
           on p.id = a.proxy_id and a.enabled 
      group by p.id
     ) ap;

我们可以使用此顺序排序来匹配帐户的顺序排序。这将按照“罕见”的顺序给出代理ID，从而产生更平衡的最终结果。

为此，我们需要计算帐户的seqnum：

update accounts a
    set proxy_id = p.id;
    from (select a.*, row_number() over (order by id) as seqnum
          from accounts
         ) aa join
         (select ap.*, row_number() over (order by proxy_seqnum) as seqnum
          from (select p.id, count(a.id) as cnt,
                       generate_series(1, 100) + count(a.id) as proxy_seqnum
                from proxy p left join
                     accounts a
                     on p.id = a.proxy_id and a.enabled 
                group by p.id
               ) ap
           ) p
           on a.seqnum = p.seqnum
    where a.enabled and a.proxy_id is null and
          aa.id = a.id;

Answer 2

我已经使用Java和SQL。但它使阅读更容易。我测试了它并且它可以工作

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</body>

以下是您的SQL方法

  public void setLeastUsedProxy() throws Exception {
    Database db = new Database();

    ArrayList<Integer> allProxies = db.getAllProxies();
    ArrayList<Integer> allAssignedProxies = db.getAssignedProxies();

    ArrayList<Integer> unusedProxies = allProxies;
    unusedProxies.removeAll(allAssignedProxies);

    // assign the unused proxy
    if (unusedProxies.size() > 0) {

    } else {
      Integer leastUsedProxy = db.getLeastUsedProxy();
      System.out.println(leastUsedProxy);
    }

  }

在另一个表中获取最少使用的id

2 个答案: