PHP 循环遍历结果数组,只想回显某些项目一次

PHP Looping through an array of results, only want to echo certain items once

我正在遍历一组结果,但是结果中有些行具有重复的名称。有没有办法只回显这个名字一次。解决方案可能很简单,但此刻我很困惑。

代码如下:

$service_query = $this->db->query('SELECT service_id, server_id, port, comment FROM services');

    foreach ($service_query->result() as $service)
    {
        $server_query = $this->db->query('SELECT server_id, name, host FROM servers WHERE server_id =' . $service->server_id);

        foreach ($server_query->result() as $server)
        {
            echo $server->name;
            echo $service->comment;
            echo $server->host;
            echo $service->port;
            echo '<br>';

        }
    }

结果如下:

Mail Server   IMAP (Secure)    XXX.XXX.XXX.XXX:993
Mail Server   POP3 (Secure)    XXX.XXX.XXX.XXX:995
Mail Server   POP3             XXX.XXX.XXX.XXX:110
Mail Server   IMAP             XXX.XXX.XXX.XXX:143
Web Server    Apache           XXX.XXX.XXX.XXX:80

仅供参考,我正在为这个项目使用 codeigniter 框架。

你的意思是 "Mail Server" 数据重复了?如果是这样,请考虑在查询中使用 MySQL DISTINCT 来删除重复项,请参阅示例 here.

据我了解您需要什么。就像

它在您的 SQL 语法中

改变它

'SELECT server_id, name, host FROM servers WHERE server_id =' . $service->server_id .' GROUP BY name'

试试这个:

$query = $this->db->query('SELECT name, service_id, server_id, host, port, comment 
                              FROM services 
                              LEFT JOIN servers USING (server_id) 
                           ORDER BY servers.name');

$server_name = '';

foreach ($query->result() as $server)
{
    if($server_name != $server->name) {
        echo $server->name;
        $server_name = $server->name;
    } 
    else { echo '-/-/-' };

    echo $server->comment;
    echo $server->host;
    echo $server->port;
    echo '<br>';

}

我通过在 $service 之前查询数据库中的 $server 并在前面的循环中回显 $server->name 来解决这个问题。这是我的解决方案。感谢大家的快速回答。

$server_query = $this->db->query('SELECT server_id, name, host FROM servers');

    foreach ($server_query->result() as $server)
    {
        echo $server->name;

        $service_query = $this->db->query('SELECT service_id, server_id, port, comment FROM services  WHERE server_id =' . $server->server_id);

        foreach ($service_query->result() as $service)
        {
            echo $service->comment;
            echo $server->host;
            echo $service->port;
            echo '<br>';

        }
    }