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Java:使用 java.time 将 object.toString 解析为 LocalDate 时遇到问题

转载 作者:行者123 更新时间:2023-11-30 06:14:12 25 4
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我正在解决一个问题集,并尝试构建一个返回日期的类。对于这个问题,我需要返回星期几,例如给定日期的“星期一”、“星期二”。我在 Stack Overflow 上搜索,并决定尝试使用 java.time:

package Chapter1.Section2;

import edu.princeton.cs.algs4.StdIn;
import edu.princeton.cs.algs4.StdOut;

import java.time.*;
import java.time.format.DateTimeFormatter;
import java.time.format.TextStyle;
import java.util.Locale;

public class exercise_1_2_12 {
private int day;
private int month;
private int year;

public exercise_1_2_12(int month, int day, int year){
if (!isDateValid(month, day, year)){
throw new IllegalArgumentException("Invalid Date!");
}
this.month = month;
this.day = day;
this.year = year;
}

public int day(){
return day;
}

public int month(){
return month;
}

public int year(){
return year;
}

public String toString(){
//return month() + "-" + day() + "-" + year();
return year + "/" + String.format("%02d", month) + "/" +
String.format("%02d", day);
}

public boolean isDateValid(int month, int day, int year) {
int[] dayInEachMonthLeapYr = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int[] dayInEachMonthNonLeapYr = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
boolean valid = true;


if ((year % 400 == 0) || ((year % 4 == 0) && (year % 100 != 0))) {
if (year < 1 || month < 1 || month > 12 || day < 1 || day > dayInEachMonthLeapYr[month - 1]) {
valid = false;
}
} else {
if (year < 1 || month < 1 || month > 12 || day < 1 || day > dayInEachMonthNonLeapYr[month - 1]) {
valid = false;
}
}
return valid;
}

public String dayOfWeekName(){
DateTimeFormatter format = DateTimeFormatter.ofPattern("mm/dd/yyyy");
String input = this.toString();
LocalDate date = LocalDate.parse(input, format);
DayOfWeek dayOfWeek = date.getDayOfWeek();
String result = dayOfWeek.getDisplayName(TextStyle.FULL, Locale.US);
return result;
}

public static void main(String[] args){

StdOut.print("Enter month: ");
int month = StdIn.readInt();
StdOut.print("Enter day: ");
int day = StdIn.readInt();
StdOut.print("Enter year: ");
int year = StdIn.readInt();
exercise_1_2_12 smartDate = new exercise_1_2_12(month,day,year);
StdOut.println(smartDate);
smartDate.dayOfWeekName();
}
}

问题可能出在实例方法dayOfWeekName()中,我尝试解析对象this。转换为String格式,然后将其存储在LocalDate日期中,然后返回工作日的名称。我测试了一下,错误信息如下:

Enter month: 4
Enter day: 1
Enter year: 2018
2018/04/01
Exception in thread "main" java.time.format.DateTimeParseException: Text '2018/04/01' could not be parsed at index 2
at java.time.format.DateTimeFormatter.parseResolved0(DateTimeFormatter.java:1949)
at java.time.format.DateTimeFormatter.parse(DateTimeFormatter.java:1851)
at java.time.LocalDate.parse(LocalDate.java:400)
at Chapter1.Section2.exercise_1_2_12.dayOfWeekName(exercise_1_2_12.java:64)
at Chapter1.Section2.exercise_1_2_12.main(exercise_1_2_12.java:85)

Process finished with exit code 1

我不太确定如何解决这个问题。

最佳答案

看看你的日期模式:

DateTimeFormatter format = DateTimeFormatter.ofPattern("mm/dd/yyyy");

日期:

'2018/04/01'

自然不匹配。

按照该模式输入日期,例如:

'04/01/2018'

或者更改模式:

DateTimeFormatter format = DateTimeFormatter.ofPattern("yyyy/MM/dd");

重要提示:正如 @Ivar 所精确指出的那样在评论中,另请注意 the pattern for months is uppercase M 。当您使用 mm 时,它会匹配分钟。

关于Java:使用 java.time 将 object.toString 解析为 LocalDate 时遇到问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/49592900/

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