[英]How can I capture a multiline pattern using a regular expressions in java?
I have a text file that I need to parse using regular expressions. 我有一个文本文件,我需要使用正则表达式解析。 The text that I need to capture is in multiline groups like this: 我需要捕获的文本是多行组,如下所示:
truck
zDoug
Doug's house
(123) 456-7890
Edoug@doug.com
30
61234.56
8/10/2003
vehicle
eRob
Rob's house
(987) 654-3210
Frob@rob.com
For this example I need to capture truck followed by the next seven lines.In other words, in this "block" I have 8 groups. 在这个例子中,我需要捕获卡车,然后是接下来的七行。换句话说,在这个“块”中我有8个组。 This is what I've tried but it will not capture the next line: 这是我尝试过但它不会捕获下一行:
(truck)\n(\w).
NOTE: I'm using the program RegExr to test my regex before I port it to Java. 注意:在将其移植到Java之前,我正在使用程序RegExr来测试我的正则表达式。
(?m)^truck(?:(?:\r\n|[\r\n]).+$)*
This assumes the whole text has been read into a single string (ie, you're not reading a file line-by-line), but it doesn't assume the line separator is always \\n
, as your code does. 这假设整个文本已被读入单个字符串(即,您不是逐行读取文件),但它并不认为行分隔符始终是\\n
,正如您的代码所做的那样。 At the minimum you should allow for \\r\\n
and \\r
as well, which is what (?:\\r\\n|[\\r\\n])
does. 至少你应该允许\\r\\n
和\\r
,这是(?:\\r\\n|[\\r\\n])
作用。 But it still matches only one separator, so the match stops before the double line separator at the end of the block. 但它仍然只匹配一个分隔符,因此匹配在块结尾处的双线分隔符之前停止。
Once you've matched a block of data, you can split it on the line separators to get the individual lines. 匹配数据块后,可以将其拆分为行分隔符以获取各行。 Here's an example: 这是一个例子:
Pattern p0 = Pattern.compile("(?m)^truck(?:(?:\r\n|[\r\n]).+$)*");
Matcher m = p0.matcher(data);
while (m.find())
{
String fullMatch = m.group();
int n = 0;
for (String s : fullMatch.split("\r\n|[\r\n]"))
{
System.out.printf("line %d: %s%n", n++, s);
}
}
output: 输出:
line 0: truck line 1: zDoug line 2: Doug's house line 3: (123) 456-7890 line 4: Edoug@doug.com line 5: 30 line 6: 61234.56 line 7: 8/10/2003
I'm also assuming each line of data contains at least one character, and that the blank lines between data block are really empty--ie, no spaces, TABs, or other invisible characters. 我还假设每行数据至少包含一个字符,并且数据块之间的空行实际上是空的 - 即没有空格,TAB或其他不可见字符。
(BTW: To test that regex in RegExr, remove the (?m)
and check the multiline
box instead. RegExr is powered by ActionScript, so the rules are a little different. For a Java -powered regex tester, check out RegexPlanet .) (顺便说一句:要在RegExr中测试该正则表达式,请删除(?m)
并检查multiline
框.RegExr由ActionScript提供支持,因此规则略有不同。对于Java驱动的正则表达式测试程序,请查看RegexPlanet 。)
这种模式应该有效((.*|\\n)*)
我认为,为了跨越多行,你的Pattern应该在DOTALL模式下编译,就像
Pattern p = Pattern.compile("truck\\n(.*\\n){7}", Pattern.DOTALL);
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