将符合 ISO 8601 的字符串转换为 java.util.Date [英] Converting ISO 8601-compliant String to java.util.Date

问题描述

我正在尝试将 ISO 8601 格式的字符串转换为 java.util.日期.

I am trying to convert an ISO 8601 formatted String to a java.util.Date.

我发现模式 yyyy-MM-dd'T'HH:mm:ssZ 如果与区域设置(比较示例)一起使用，则符合 ISO8601.

I found the pattern yyyy-MM-dd'T'HH:mm:ssZ to be ISO8601-compliant if used with a Locale (compare sample).

但是，使用 java.text.SimpleDateFormat，我无法转换正确格式的字符串 2010-01-01T12:00:00+01:00.我必须先将它转换为 2010-01-01T12:00:00+0100，没有冒号.

However, using the java.text.SimpleDateFormat, I cannot convert the correctly formatted String 2010-01-01T12:00:00+01:00. I have to convert it first to 2010-01-01T12:00:00+0100, without the colon.

所以，目前的解决方案是

So, the current solution is

SimpleDateFormat ISO8601DATEFORMAT = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ssZ", Locale.GERMANY);
String date = "2010-01-01T12:00:00+01:00".replaceAll("\+0([0-9]){1}\:00", "+0$100");
System.out.println(ISO8601DATEFORMAT.parse(date));

这显然不是那么好.我是否遗漏了什么或有更好的解决方案?

which obviously isn't that nice. Am I missing something or is there a better solution?

回答

感谢 JuanZe 的评论，我发现了 Joda-Time 魔法，它也是 此处描述.

Thanks to JuanZe's comment, I found the Joda-Time magic, it is also described here.

所以，解决方案是

DateTimeFormatter parser2 = ISODateTimeFormat.dateTimeNoMillis();
String jtdate = "2010-01-01T12:00:00+01:00";
System.out.println(parser2.parseDateTime(jtdate));

或者更简单地说，通过构造函数使用默认解析器:

Or more simply, use the default parser via the constructor:

DateTime dt = new DateTime( "2010-01-01T12:00:00+01:00" ) ;

对我来说，这很好.

推荐答案

不幸的是，时区格式可用于 SimpleDateFormat(Java 6 及更早版本)不是 符合 ISO 8601.SimpleDateFormat 理解时区字符串如GMT+01:00"或+0100"，后者根据 RFC # 822.

Unfortunately, the time zone formats available to SimpleDateFormat (Java 6 and earlier) are not ISO 8601 compliant. SimpleDateFormat understands time zone strings like "GMT+01:00" or "+0100", the latter according to RFC # 822.

即使 Java 7 根据 ISO 8601 添加了对时区描​​述符的支持，SimpleDateFormat 仍然无法正确解析完整的日期字符串，因为它不支持可选部分.

Even if Java 7 added support for time zone descriptors according to ISO 8601, SimpleDateFormat is still not able to properly parse a complete date string, as it has no support for optional parts.

使用正则表达式重新格式化您的输入字符串当然是一种可能性，但替换规则并不像您的问题那么简单:

Reformatting your input string using regexp is certainly one possibility, but the replacement rules are not as simple as in your question:

• 某些时区并非完全偏离 UTC，因此字符串不一定以:00".
• ISO8601 只允许时区中包含小时数，因此+01"相当于+01:00"
• ISO8601 允许使用Z"来表示 UTC，而不是+00:00".

• Some time zones are not full hours off UTC, so the string does not necessarily end with ":00".
• ISO8601 allows only the number of hours to be included in the time zone, so "+01" is equivalent to "+01:00"
• ISO8601 allows the usage of "Z" to indicate UTC instead of "+00:00".

更简单的解决方案可能是在 JAXB 中使用数据类型转换器，因为 JAXB 必须能够根据 XML 模式规范解析 ISO8601 日期字符串.javax.xml.bind.DatatypeConverter.parseDateTime("2010-01-01T12:00:00Z") 会给你一个 Calendar 对象，你可以简单地使用 getTime()在它上面，如果你需要一个 Date 对象.

The easier solution is possibly to use the data type converter in JAXB, since JAXB must be able to parse ISO8601 date string according to the XML Schema specification. javax.xml.bind.DatatypeConverter.parseDateTime("2010-01-01T12:00:00Z") will give you a Calendar object and you can simply use getTime() on it, if you need a Date object.

您也可以使用 Joda-Time，但我不知道为什么你应该为此烦恼.

You could probably use Joda-Time as well, but I don't know why you should bother with that.

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