Question

这是我的数据及其模式：

// _23.02_ANTALYA____________FRANKFURT___________DE_7461_18:20-21:00________________
public static final String FLIGHT_DEFAULT_PATTERN = "\\s+\\d{2}.\\d{2}\\s[A-Z]+\\s+[A-Z]+\\s+[A-Z\\s]{3}[\\d\\s]{5}\\d{2}:\\d{2}-\\d{2}:\\d{2}\\s+";

下划线是空间角色。现在我需要一个将每个正则表达式术语划分为数据的类。例如

\\s+ = " "
\\d{2} = "23"
. = "."
\\d{2} = "02"
\\s = " "
[A-Z]+ = "ANTALYA"

等......必须按模式排序。

我该怎么办？或者是否有图书馆？

Answer 1

正如@devnull所提到的，你应该使用capturing groups：

(\s+)(\d{2})(.)(\d{2})(\s)([A-Z]+)(\s+)([A-Z]+)(\s+)([A-Z\s]{3})([\d\s]{5})(\d{2}:\d{2})(-)(\d{2}:\d{2})(\s+)

在Regex101上查看此正则表达式的完整说明。

然后，您将使用以下内容匹配文本并提取单个值：

String text = " 23.02 ANTALYA            FRANKFURT            DE 7461 18:20-21:00                 ";
Pattern pattern = Pattern.compile("(\\s+)(\\d{2})(.)(\\d{2})(\\s)([A-Z]+)(\\s+)([A-Z]+)(\\s+)([A-Z\\s]{3})([\\d\\s]{5})(\\d{2}:\\d{2})(-)(\\d{2}:\\d{2})(\\s+)");
Matcher matcher = pattern.matcher(text);
if (matcher.find()) {
    for (int i = 1; i < matcher.groupCount(); i++) {
        System.out.println(matcher.group(i));
    }
}

为了更容易提取特定字段，您可以（在Java 7及更高版本中）使用命名捕获组：

(?<LeadSpace>\s+)(?<Day>\d{2})(.)(?<Month>\d{2})...

然后您可以使用以下内容来获取每个命名组：

...
if (matcher.find()) {
    System.out.println(matcher.group("LeadSpace"));
    System.out.println(matcher.group("Day"));
    System.out.println(matcher.group("Month"));
    ...
}

Answer 2

我发现了一种不同的方式。我用手分开了碎片。

// _24.02_MAURITIUS_________HAMBURG________________via:FRA_DE/LH____08:30-20:05_____
public static final List<String> FLIGHT_VIA_PATTERN = Arrays.asList( "\\s+", "\\d{2}", "\\.", "\\d{2}", "\\s+", "[A-Z]+", "\\s+", "[A-Z]+", "\\s+", "via:", "[A-Z\\s]{4}", "[A-Z]{2,3}", "/",
        "[A-Z]{2,3}", "\\s+", "\\d{2}", ":", "\\d{2}", "\\-", "\\d{2}", ":", "\\d{2}", "\\s+" );

在此之后我使用了一个循环，一切都很好。这个问题可以结束。

正则表达式解析器一步一步Java

2 个答案: