Question

表的

列

date                |count|ip
2014-05-14 11:21:00 |2    | 222.222.222.22
2014-05-14 11:22:00 |4    | 232.232.232.23
2014-05-14 11:24:00 |1    | 222.222.222.22

this is working fine for me但是这个解决方案为我提供了系列值null，但我需要值0而不是null

在当前查询中，如果没有填充缺失序列，我就无法得到确切的序列

当前查询：

SELECT sum(date),minute
FROM generate_series('2014-05-14 0:00:00'::timestamp, '2014-05-14 23:59:59', 
              '1 minute') AS minutes(minute)
    LEFT JOIN table ON minute=date_trunc('minute',table.date) 
where switch_ip='232.232.232.23'
GROUP BY minute

当前输出：

2014-05-14 11:22:00 |4

必需输出：

1）如果在ip = 232.232.232.23且日期＆gt; ='2014-05-14 00:00:00'且日期＆lt; ='2014-05-14 23:59:59'

2014-05-14 11:21:00 | 0
2014-05-14 11:22:00 | 4
2014-05-14 11:23:00 | 0

2）如果在ip 222.222.222.22和日期＆gt; ='2014-05-14 00:00:00'和日期＆lt; ='2014-05-14 23:59:59'

2014-05-14 11:21:00 | 2
2014-05-14 11:22:00 | 0
2014-05-14 11:23:00 | 0    
2014-05-14 11:24:00 | 1

3）如果在日期＆gt; ='2014-05-14 00:00:00'和日期＆lt; ='2014-05-14 23:59:59'

2014-05-14 11:21:00 | 2
2014-05-14 11:22:00 | 4
2014-05-14 11:23:00 | 0    
2014-05-14 11:24:00 | 1

Answer 1

通过将ip条件放在where子句中，您将外部联接转换为内部联接。将其移至连接条件

select coalesce(sum("count"), 0), minute
from
    generate_series(
        '2014-05-14 0:00:00'::timestamp,
        '2014-05-14 23:59:59', 
        '1 minute'
    ) minutes(minute)
    left join
    t on
        minutes.minute = date_trunc('minute', t.date)
        and
        ip = '232.232.232.23'
group by minute
order by minute

http://sqlfiddle.com/#!15/31483/5

将时间戳作为逗号分隔的字符串替换

select coalesce(sum("count"), 0), minute

通过

select 
    coalesce(sum("count"), 0), 
    to_char(minute, 'YYYY,MM,DD,HH24,MI') as minute

按分钟分组并填写缺少的地方

1 个答案: