1Z0-829 Practice Question: Handling Date, Time, Text, Numeric and Boolean Values
Which of the following correctly uses DateTimeFormatter to parse the date "2023-01-15" into a LocalDate?
⚠ Common exam trap
It's easy for candidates to confuse lowercase `mm` (minute) with uppercase `MM` (month) in date/time patterns, leading them to choose option B, or they forget that `MonthDay.parse()` requires a specific format without a year, causing them to select option C.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
LocalDate.parse("2023-01-15", DateTimeFormatter.ofPattern("yyyy-MM-dd"))
Options A and D are both correct. Option A uses DateTimeFormatter.ofPattern("yyyy-MM-dd") with the correct pattern, which parses the date successfully. Option D uses the predefined constant DateTimeFormatter.ISO_LOCAL_DATE, which also correctly parses the input "2023-01-15". Option B is wrong because lowercase "mm" represents minutes, not months. Option C is wrong because MonthDay.parse() expects a format like "--01-15" without a year.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
LocalDate.parse("2023-01-15", DateTimeFormatter.ofPattern("yyyy-MM-dd"))
Why this is correct
Correct. The pattern 'yyyy-MM-dd' matches the input format exactly.
- ✗
LocalDate.parse("2023-01-15", DateTimeFormatter.ofPattern("yyyy-mm-dd"))
Why it's wrong here
Incorrect. The pattern uses lowercase 'mm' for minutes instead of uppercase 'MM' for month, causing a parsing error.
- ✗
MonthDay.parse("2023-01-15")
Why it's wrong here
Incorrect. MonthDay.parse() expects a format like '--01-15' without a year, so it fails on a full date string.
- ✓
LocalDate.parse("2023-01-15", DateTimeFormatter.ISO_LOCAL_DATE)
Why this is correct
Correct. DateTimeFormatter.ISO_LOCAL_DATE is a predefined format that matches ISO-8601 date format (yyyy-MM-dd), so it parses the input correctly.
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