使用正则表达式捕获任意多个组

Question

我正在尝试使用 Java 中的正则表达式从以下字符串中捕获数据：

SettingName = "Value1",0x2,3,"Value4 contains spaces", "Value5 has a space before the string that is ignored"

字符串前面可能有任意数量的空格，可以忽略。它也可能包含比此处列出的更多或更少的值，这只是一个示例。

我的目的是捕捉这些群体：

第 1 组：设置名称

第 2 组："Value1"（带引号）

第 3 组：0x2

第 4 组：3

第 5 组："Value4 contains spaces"

第 6 组："Value5 has a space before the string that is ignored"

我尝试使用的正则表达式是：

\s*([\w\/.-]+)\s*=(?:\s*(\"?[^\",]*\"?)(?:,|\s*$))+
\s* -> Consume an arbitrary number of whitespace
   ( -> Start a capturing group (group 1)
    [\w\/.-] -> Get a letter of the SettingName, which may be contain alphanumberic, /, ., and -
            + -> Get the previous token one or more times (so group 1 is not blank)
             ) -> End the capturing group
              \s* -> Consume an arbitrary amount of whitespace
                 = -> Consume the equals sign
                  (?: -> Start an uncaptured group
                     \s* -> Consume an arbitrary amount of whitespace
                        ( -> Start a captured group
                         \"? -> Consume a quote, if it exists
                            [^\",] -> Consume any nonquote, noncomma character
                                  \"? -> Consume the end quote, if it exists
                                     ) -> End the captured group
                                      (?: -> start a uncaptured group
                                         ,|\s*$ -> capture either a comma or end of line (string?) character
                                               ) -> end the uncaptured group
                                                ) -> end the outer uncaptured group
                                                 + -> match the outer uncaptured group 1 or more times

我正在使用此代码：

private static final String regex = "\s*([\w\/.-]+)\s*=(?:\s*(\"?[^\",]*\"?)(?:,|\s*$))+";
private static final Pattern settingPat = Pattern.compile(regex);
...
public String text;
public Matcher m;
...
public void someMethod(String lineContents)
{
    m = settingPat.matcher(text);
    if(!m.matches())
        ... (do other stuff)
    else
    {
        name = m.group(1);     // should be "SettingName"
        value[0] = m.group(2); // should be "\"Value1\""
        value[1] = m.group(3); // should be "0x2"
        ...
    }
}

使用这段代码，它匹配到字符串，但似乎我只捕获了最后一组。 Java and/or 正则表达式是否支持使用 + 修饰符重复任意捕获组？

Answer 1

您只有 2 个捕获组，因此您不能在结果中获得超过 2 个组。您将不得不运行一个循环来匹配所有重复项

您可以在 while 循环中使用此正则表达式来获取所有匹配项：

(?:([\w/.-]+)\h*=|(?!^)\G,)\h*((\"?)[^\",]*)

\G 断言位置在前一个匹配的末尾或第一个匹配的字符串的开头，因为我们使用 (?!^) 我们强制 \G 只匹配 上一场比赛结束时的位置

RegEx Demo

CODE DEMO

代码：

final String regex = "(?:([\w/.-]+)\h*=|(?!^)\G,)\h*((\"?)[^\",]*\3)";
final String string = "SettingName = \"Value1\",0x2,3,\"Value4 contains spaces\", \"Value5 has a space before the string that is ignored\"";
    
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
    
while (matcher.find()) {
    if (matcher.group(1) != null)
        System.out.println(matcher.group(1));
    System.out.println("\t=> " + matcher.group(2));
}

使用正则表达式捕获任意多个组

Capturing Arbitrary Multiple Groups with Regex

java

regex

parsing

capturing-group