Support to show the schedule for a requested day AFTER holidays

functions/les_rework.py

@@ -1,5 +1,5 @@
 from datetime import datetime, timedelta
-from timeFormatters import dateTimeNow, weekdayToInt
+from functions.timeFormatters import dateTimeNow, weekdayToInt, forward_to_weekday, skip_weekends
 from typing import Optional
 
 
@@ -28,31 +28,3 @@ def find_target_date(arg: Optional[str]) -> datetime:
     day = skip_weekends(day)
 
     return day
-
-
-def skip_weekends(day: datetime) -> datetime:
-    """
-    Increment the current date if it's not a weekday
-    """
-    weekday = day.weekday()
-
-    # Friday is weekday 4
-    if weekday > 4:
-        return day + timedelta(days=(7 - weekday))
-
-    return day
-
-
-def forward_to_weekday(day: datetime, weekday: int) -> datetime:
-    """
-    Increment a date until the weekday is the same as the one provided
-    Finds the "next" [weekday]
-    """
-    current = day.weekday()
-
-    # This avoids negative numbers below, and shows
-    # next week in case the days are the same
-    if weekday >= current:
-        weekday += 7
-
-    return day + timedelta(days=(weekday - current))

functions/timeFormatters.py

@@ -138,13 +138,15 @@ def getPlural(amount, unit):
     return dic[unit.lower()]["s" if amount == 1 else "p"]
 
 
-def weekdayToInt(day) -> int:
+def weekdayToInt(day: str) -> int:
     days = {"maandag": 0, "dinsdag": 1, "woensdag": 2, "donderdag": 3, "vrijdag": 4, "zaterdag": 5, "zondag": 6}
 
-    if day.lower() not in days:
-        return -1
+    # Allow abbreviations
+    for d, i in days.items():
+        if d.startswith(day):
+            return i
 
-    return days[day.lower()]
+    return -1
 
 
 def intToWeekday(day):
@@ -164,3 +166,31 @@ def fromArray(data: List[int]) -> datetime:
     year = str(data[2])
 
     return fromString(f"{day}/{month}/{year}")
+
+
+def skip_weekends(day: datetime) -> datetime:
+    """
+    Increment the current date if it's not a weekday
+    """
+    weekday = day.weekday()
+
+    # Friday is weekday 4
+    if weekday > 4:
+        return day + datetime.timedelta(days=(7 - weekday))
+
+    return day
+
+
+def forward_to_weekday(day: datetime, weekday: int) -> datetime:
+    """
+    Increment a date until the weekday is the same as the one provided
+    Finds the "next" [weekday]
+    """
+    current = day.weekday()
+
+    # This avoids negative numbers below, and shows
+    # next week in case the days are the same
+    if weekday <= current:
+        weekday += 7
+
+    return day + datetime.timedelta(days=(weekday - current))