我想获得以下结构：
Day 1 | 2 | 3 | 4 ... Total
stackoverflow 04:00 01:00 05:00
google 05:30 02:00 07:30
但是使用我的代码，我得到：
Day 1 | 2 | 3 | 4 ... Total
stackoverflow 04:00 04:00
google 05:30 05:30
stackoverflow 01:00 01:00
google 02:00 02:00
我正在运行这个查询（在此处给出）：
select
agenc.name,
sum(cast(servic.end_hour as time) - cast(servic.begin_hour as time)) as time_, jobs.description
from
services as servic
join services_jobs AS jobs ON jobs.id = servic.job_id
join agency as agenc ON agenc.id = jobs.agency_id
where
extract(month from servic.service_date) = 9 and
extract(day from servic.service_date) = 16
group by
agenc.name
在 python 中，我执行相同的查询，不同之处在于查询“extract(day from service.service_date)”中的“day column”等于循环（从 1 到 31 -> 天数）
query_A = cr.fetchall()
if query_A:
for values in query_A:
sheet.write(line_, 0, values[0], style_main) # Writes the name
sheet.write(line_, 1, values[2], style_main) # Writes the job
current_value = str(values[1]);
hours_ = current_value.split(':')[0]
minutes_ = current_value.split(':')[1]
total_ = hours_ +':'+ minutes_ # This will get 02:00 - for example
sheet.write(line_, i + 1, total_, style_main) # write the result
total_vertical += timedelta(hours=int(hours_), minutes=int(minutes_)) # here should sum the results along the days
sheet.write(line_, 33, str(total_vertical), style_main) # write it on "total" column
line_ += 1
total_vertical = timedelta(hours=0, minutes=0) # Reset
关于我应该如何继续获得我想要的结果的任何建议？谢谢！