Python: Print structure time in local time

Last update on December 21 2024 07:27:55 (UTC/GMT +8 hours)

Write a Python program that takes a given number of seconds and passes since the epoch as an argument. Print structure time in local time.

Sample Solution:

Python Code:

# Import the time module
import time

# Convert the given timestamp (414538698) into a time structure and store it in the variable 'result'
result = time.localtime(414538698)

# Print a message indicating the result
print("\nResult:", result)

# Print a message indicating the year component of the time structure stored in 'result'
print("\nYear:", result.tm_year)

Output:

Result: time.struct_time(tm_year=1983, tm_mon=2, tm_mday=19, tm_hour=21, tm_min=38, tm_sec=18, tm_wday=5, tm_yday=50, tm_isdst=0)

Year: 1983

Explanation:

In the exercise above,

• The code imports the "time" module, which provides functions for working with time.
• It uses the "time.localtime()" function to convert the given timestamp (414538698) into a time structure representing the local time.
• It stores the resulting time structure in the variable named 'result'.
• It prints a message indicating the result, which is the time structure stored in 'result'.
• It prints a message indicating the year component ('tm_year') of the time structure stored in 'result'.

Flowchart:

Python Code Editor:

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