问题描述
我正在创建一个预订系统，我不想让用户在开始时间 11:00 和结束时间 09:00(过去)(我使用 24 小时制)进行预订.我有两个组合框，里面装满了作为开始和结束时间的字符串 (09:00,10:00,11:00,12:00,13:00....)
I am creating a booking system and I don't want to allow users to book with starting time 11:00 and end time 09:00 (past)(I am using 24hour clock). I have two combo boxes filled with Strings that act as start and end time (09:00,10:00,11:00,12:00,13:00....)
我有这个代码:
String start = (String) startTime.getSelectedItem();
String end = (String) endTime.getSelectedItem();
try {
if(new SimpleDateFormat("hh:mm").parse(start).before(new SimpleDateFormat("hh:mm").parse(end))){
System.out.println("test1");// future date - good
}else{
System.out.println("fail2");// old date - bad
}
} catch (ParseException ex) {
System.out.println("error");
}
这非常有效，除非我选择开始/结束时间为 12:00.程序输出与它应该输出的相反，我不确定为什么.
This works perfectly except when I pick start/end time to be 12:00. Program outputs opposite of what it is supposed to output and I am unsure why.
如果我选择开始时间 14:00 和结束时间 12:00，程序将输出失败 2(良好的输出)，
If I pick start time 14:00 and end time 12:00 the program will output fail2(good output),
如果我选择开始时间 09:00 和结束时间 12:00 程序将输出 fail2(应该是 test1)，
If I pick start time 09:00 and end time 12:00 the program will output fail2(should be test1),
如果我选择开始时间12:00和结束时间10:00，程序将输出test1(应该是fail2)，
if I pick start time 12:00 and end time 10:00 the program will output test1(should be fail2),
如果我选择开始时间 12:00 和结束时间 15:00 程序将输出 test1(good output)
if I pick start time 12:00 and end time 15:00 the program will output test1(good output)
这种类型的问题只有在我选择 12:00 时才会出现..
This type of problem only occurs when I pick 12:00..
推荐答案
public static void checkTimes(String start, String end) {
try {
if (LocalTime.parse(start).isBefore(LocalTime.parse(end))) {
System.out.println("test1");// future date - good
} else {
System.out.println("fail2");// old date - bad
}
} catch (DateTimeParseException dtpe) {
System.out.println("error");
}
}
让我们试试吧:
checkTimes("14:00", "12:00");
checkTimes("09:00", "12:00");
checkTimes("12:00", "10:00");
checkTimes("12:00", "15:00");
打印:
fail2
test1
fail2
test1
我相信这与您的意图一致.请注意， LocalTime 无需显式格式化程序即可解析您的字符串.此外，如果你相信你的组合框只包含有效的时间字符串，你可以省略 try-catch 构造，因为 DateTimeParseException 是一个未经检查的异常.
I believe this agrees with what you had intended. Note that LocalTime parses your strings without the need for an explicit formatter. Furthermore, if you trust that your combobox only contains valid time strings, you can leave out the try-catch construct since DateTimeParseException is an unchecked exception.
如果startTime 和endTime 是JComboBox，我相信你甚至可以将LocalTime 对象填入其中.那么当用户从每个中选择一个时，您就不需要解析了.您的 JComboBox 将调用 LocalTime.toString()，它将返回一个类似 09:00 的字符串，反过来组合框将显示和让用户选择.
If startTime and endTime are JComboBox, I believe you can even fill LocalTime objects into them. Then you don’t need to parse when the user selects one from each. Your JComboBox will call LocalTime.toString(), which will return a string like 09:00, which in turn the combo box will display and let the user select.
LocalTime[] times = { LocalTime.of(9, 0), LocalTime.of(10, 0), LocalTime.of(11, 0),
LocalTime.of(12, 0), LocalTime.of(13, 0), LocalTime.of(14, 0) };
JComboBox<LocalTime> combo = new JComboBox<>(times);
展开:
我正在使用 java.time 中的 LocalTime，现代 Java 日期和时间 API.java.time 通常比旧的和过时的日期和时间类(如 SimpleDateFormat、Date 等)更好用.
I am using LocalTime from java.time, the modern Java date and time API. java.time is generally much nicer to work with than the old and outdated date and time classes like SimpleDateFormat, Date and more.
链接: Oracle 教程:日期时间解释如何使用 java.time.
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