Java日期区域格式至简单格式

Question

是否可以在Java（可能ISO 8601）中转换这种格式>哪个是字符串，以日期或时间类型的类别的“ yyyy-mm-dd”为字符串，还是应该使用字符串方法完成？

例子：

you receive this -> "2022-11-11T00:00:00"
you convert to this -> "2022-11-11"

Answer 1

推荐方法：java.time

如果您需要根据值计算任何内容，或者可能会发现一周中的一天，则最好使用java.Time

public static void main(String[] args) {
    // example input in ISO format
    String first = "2022-11-11T00:00:00";
    String second = "2022-11-11T12:00:00+01:00";
    // parse them to suitable objects
    LocalDateTime ldt = LocalDateTime.parse(first);
    OffsetDateTime odt = OffsetDateTime.parse(second);
    // extract the date from the objects (that may have time of day and offset, too)
    LocalDate firstDate = ldt.toLocalDate();
    LocalDate secondDate = odt.toLocalDate();
    // format them as ISO local date, basically the same format as the input has
    String firstToBeForwarded = firstDate.format(DateTimeFormatter.ISO_LOCAL_DATE);
    String secondToBeForwarded = secondDate.format(DateTimeFormatter.ISO_LOCAL_DATE); 
    // print the results (or forward them as desired)
    System.out.println(firstToBeForwarded);
    System.out.println(secondToBeForwarded);
}

： 不建议使用此示例的输出

2022-11-11
2022-11-11

---

，但可能会出现：字符串操纵

如果您只需要

• 提取日期零件（年度和每月的年度，每月），并且
• 您确定它始终是第一个您收到的String s的10个字符

您可以简单地采用前10个字符：

String toBeForwarded = "2022-11-11T00:00:00".substring(0, 10);

此行将存储“ 2022-11-11”在tobeforwarded 。