如何用Python获取每月各周起止日期并避免生成下月周数据

需求 #

获取指定月份每周（周一至周六）的起止日期，若该周的周一在上月，则以上月的周一作为周起始日期，结束日期为当月日期。2023年2月的期望输出如下：

{
  "week_1": { "start_date": "2023-01-30", "end_date": "2023-02-04" },
  "week_2": { "start_date": "2023-02-06", "end_date": "2023-02-11" },
  "week_3": { "start_date": "2023-02-13", "end_date": "2023-02-18" },
  "week_4": { "start_date": "2023-02-20", "end_date": "2023-02-25" },
  "week_5": { "start_date": "2023-02-27", "end_date": "2023-03-04" }
}

现有实现代码 #

import datetime
import json


def get_weeks(year, month):
    # get the first day of the month
    first_day = datetime.date(year, month, 1)
    
    # get the date of the previous Monday
    prev_monday = first_day - datetime.timedelta(days=first_day.weekday())

    # get the date of the first Saturday
    first_saturday = prev_monday + datetime.timedelta(days=5)
    weeks = {}
    start_date = prev_monday
    for i in range(1, 6):
        # get the end date for the current week
        end_date = start_date + datetime.timedelta(days=5)
        # check if the end date is in the next month
        if end_date.month != month:
            # merge the current week with the next week
            next_monday = end_date + datetime.timedelta(days=1)
            next_week = {
                "start_date": next_monday.strftime("%Y-%m-%d"),
                "end_date": (next_monday + datetime.timedelta(days=6)).strftime(
                    "%Y-%m-%d"
                ),
            }
            weeks[f"week_{i}"] = {
                "start_date": start_date.strftime("%Y-%m-%d"),
                "end_date": end_date.strftime("%Y-%m-%d"),
            }
            weeks[f"week_{i+1}"] = next_week
            start_date = next_monday + datetime.timedelta(days=7)
        else:
            # add the current week to the dictionary
            weeks[f"week_{i}"] = {
                "start_date": start_date.strftime("%Y-%m-%d"),
                "end_date": end_date.strftime("%Y-%m-%d"),
            }
            start_date = end_date + datetime.timedelta(days=2)
    return weeks


week_dates = get_weeks(2023, 1)
print(json.dumps(week_dates, indent=4))

当前问题 #

当前代码运行后会额外生成完全属于下月的周数据，例如调用get_weeks(2023,1)时，输出中包含了完全属于3月的week_6。需要修改代码避免生成这类无关周数据。

问题根源在于原代码固定循环5次，且跨月处理逻辑会错误生成额外周。修改思路为：

• 动态判断循环终止条件，不再固定循环次数
• 以目标月份的最后一天为边界，仅保留与目标月份有交集的周

修改后的代码：

import datetime
import json


def get_weeks(year, month):
    first_day = datetime.date(year, month, 1)
    # 计算目标月份的最后一天
    if month == 12:
        last_day = datetime.date(year + 1, 1, 1) - datetime.timedelta(days=1)
    else:
        last_day = datetime.date(year, month + 1, 1) - datetime.timedelta(days=1)
    
    # 找到第一个需要处理的周一（可能在上月）
    prev_monday = first_day - datetime.timedelta(days=first_day.weekday())
    weeks = {}
    week_num = 1
    start_date = prev_monday
    
    while True:
        end_date = start_date + datetime.timedelta(days=5)
        # 若起始日期已超过目标月份最后一天，直接终止循环
        if start_date > last_day:
            break
        # 添加当前周数据
        weeks[f"week_{week_num}"] = {
            "start_date": start_date.strftime("%Y-%m-%d"),
            "end_date": end_date.strftime("%Y-%m-%d")
        }
        week_num += 1
        # 跳转到下一周的周一
        start_date = end_date + datetime.timedelta(days=2)
        # 提前判断下一周是否完全超出目标月份范围
        if start_date > last_day:
            break
    return weeks


# 测试2023年2月
week_dates = get_weeks(2023, 2)
print(json.dumps(week_dates, indent=4))

代码说明 #

1. 新增目标月份last_day计算，作为日期范围边界
2. 将固定循环改为while循环，动态判断是否继续生成周数据
3. 每次循环后检查下一周起始日期是否超出边界，避免生成完全属于下月的周
4. 移除原代码中错误的跨月周合并逻辑，该逻辑会导致额外周生成

测试2023年2月的输出与期望完全一致：

{
    "week_1": {
        "start_date": "2023-01-30",
        "end_date": "2023-02-04"
    },
    "week_2": {
        "start_date": "2023-02-06",
        "end_date": "2023-02-11"
    },
    "week_3": {
        "start_date": "2023-02-13",
        "end_date": "2023-02-18"
    },
    "week_4": {
        "start_date": "2023-02-20",
        "end_date": "2023-02-25"
    },
    "week_5": {
        "start_date": "2023-02-27",
        "end_date": "2023-03-04"
    }
}

内容的提问来源于stack exchange，提问作者user21372338