Java 8 LocalDateTime 正在解析无效的日期
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/32823368/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Java 8 LocalDateTime is parsing invalid date
提问by Zeeshan
I wanted to validate date in client side so I wrote the following code. But instead of getting an exception I am getting a proper date object for 31st of February date string, which is clearly an invalid date.
我想在客户端验证日期，所以我编写了以下代码。但是，我没有得到异常，而是得到了 2 月 31 日日期字符串的正确日期对象，这显然是一个无效日期。
public class Test {
public static void main(String[] args) {
String dateFormat = "HH:mm:ss MM/dd/yyyy";
String dateString = "11:30:59 02/31/2015";
DateTimeFormatter dateTimeFormatter = DateTimeFormatter.ofPattern(dateFormat, Locale.US);
try {
LocalDateTime date = LocalDateTime.parse(dateString, dateTimeFormatter);
System.out.println(date);
} catch (Exception e) {
// Throw invalid date message
}
}
}
Output: 2015-02-28T11:30:59
输出：2015-02-28T11:30:59
Does anyone know why LocalDateTimeis parsing this date instead of throwing an exception.
有谁知道为什么LocalDateTime解析这个日期而不是抛出异常。
采纳答案by Jon Skeet
You just need a strict ResolverStyle.
你只需要一个严格的ResolverStyle.
Parsing a text string occurs in two phases. Phase 1 is a basic text parse according to the fields added to the builder. Phase 2 resolves the parsed field-value pairs into date and/or time objects. This style is used to control how phase 2, resolving, happens.
解析文本字符串分两个阶段进行。阶段 1 是根据添加到构建器的字段进行基本文本解析。阶段 2 将解析的字段值对解析为日期和/或时间对象。此样式用于控制阶段 2（解决）如何发生。
Sample code - where withResolverStyle(ResolverStyle.STRICT)is the important change, along with the use of uuuurather than yyyy(where uuuuis "year" and "yyyy" is "year of era", and therefore ambiguous):
示例代码 -withResolverStyle(ResolverStyle.STRICT)重要的变化在哪里，以及使用uuuu而不是yyyy（哪里uuuu是“年份”，“yyyy”是“年代”，因此不明确）：
import java.time.*;
import java.time.format.*;
import java.util.*;
public class Test {
public static void main(String[] args) {
String dateFormat = "HH:mm:ss MM/dd/uuuu";
String dateString = "11:30:59 02/31/2015";
DateTimeFormatter dateTimeFormatter = DateTimeFormatter
.ofPattern(dateFormat, Locale.US)
.withResolverStyle(ResolverStyle.STRICT);
try {
LocalDateTime date = LocalDateTime.parse(dateString, dateTimeFormatter);
System.out.println(date);
} catch (DateTimeParseException e) {
// Throw invalid date message
System.out.println("Exception was thrown");
}
}
}
回答by jste89
LocalDateTime.parse will only throw an error if the String passed in contains invalid characters, a number of days exceeding 31 or a month exceeding 12.
LocalDateTime.parse 只会在传入的 String 包含无效字符、天数超过 31 或一个月超过 12 时抛出错误。
For example, if you modified your code as such:
例如，如果您这样修改代码：
String dateString = "11:30:59 0zz2/31/2015";
an exception would be thrown caused by the invalid 'zz' characters within your given date. As to why it's 'rounding-down' the date so to speak, that I don't know.
给定日期内的无效“zz”字符会引发异常。至于为什么它是“四舍五入”的日期可以这么说，我不知道。
Source: https://docs.oracle.com/javase/8/docs/api/java/time/LocalDateTime.html#parse-java.lang.CharSequence-
来源：https: //docs.oracle.com/javase/8/docs/api/java/time/LocalDateTime.html#parse-java.lang.CharSequence-
回答by SatyaTNV
try {
SimpleDateFormat df = new java.text.SimpleDateFormat("HH:mm:ss MM/dd/yyyy");
df.setLenient(false);
System.out.println(df.parse("11:30:59 02/29/2015"));
} catch (java.text.ParseException e) {
System.out.println(e);
}
I found one solution to recognize date as a valid date with DateFormat.setLenient(boolean). If you try to parse any invalid date it will throws parse exception.
我找到了一种使用DateFormat.setLenient(boolean)将日期识别为有效日期的解决方案。如果您尝试解析任何无效日期，它将引发解析异常。
Edit:
编辑：
Java 8, but this will raise exception if a month is not between 1and 12, if a day is more than 32. Exactly not working. But for month its working.
Java 8，但如果一个月不在1和之间12，如果一天超过，这将引发异常32。完全不工作。但一个月它的工作。
try {
TemporalAccessor ta = DateTimeFormatter.ofPattern("HH:mm:ss MM/dd/yyyy").parse("11:30:59 02/32/2015");
} catch (Exception e) {
System.out.println(e);
}
Output:
输出：
java.time.format.DateTimeParseException: Text '11:30:59 02/32/2015' could not be
parsed: Invalid value for DayOfMonth (valid values 1 - 28/31): 32
回答by Palamino
The Java 8 DateTimeFormatter uses yyyy to mean YEAR_OF_ERA, and uuuu to mean YEAR. You need to modify your pattern string as follows:
Java 8 DateTimeFormatter 使用 yyyy 表示 YEAR_OF_ERA，使用 uuuu 表示 YEAR。您需要按如下方式修改您的模式字符串：
String dateFormat = "HH:mm:ss MM/dd/uuuu";
The DateTimeFormatter defaults to using the SMART resolver style, but you want it to use the STRICT resolver style. Modify your dateTimeFormatter initialization code as follows:
DateTimeFormatter 默认使用 SMART 解析器样式，但您希望它使用 STRICT 解析器样式。修改您的 dateTimeFormatter 初始化代码如下：
DateTimeFormatter dateTimeFormatter = DateTimeFormatter.ofPattern(dateFormat, Locale.US)
.withResolverStyle(ResolverStyle.STRICT);
回答by Edwin Buck
It is not rounding down. February has never had 31 days, and it is impossible to use a validating date / time object to represent a day that doesn't exist.
它不是四舍五入。二月从来没有 31 天，不可能使用验证日期/时间对象来表示不存在的一天。
As a result, it takes the invalid input and gives you the best approximation to the correct date (the last date of February that year).
因此，它采用无效输入并为您提供正确日期（当年 2 月的最后一个日期）的最佳近似值。
SimpleDateFormatinherits from DateFormatwhich has a setLenient(boolean value)method on it. I would expect that if you called setLenient(true)prior to parsing, it would probably complain more, as detailed in the javadocs.
SimpleDateFormat继承自DateFormat其上有一个setLenient(boolean value)方法。我希望如果您setLenient(true)在解析之前调用，它可能会抱怨更多，如 javadocs 中所述。
