[英]How to fetch substring from java string starting with the range of \n+ to \n and \n- to \n in below given string?
Input string is as below: 输入字符串如下:
@@ -106,12 +106,12 @@ end loop\n loop map dummm56\n \tdummy data/path/u/u_op/kl-sc45\n end loop\n-\n-loop map {$df=56,$kl=20564300,$testId=\"jk: message1 message2:48667697\",$kl3=true,$kl=true, $kl=[2],$kl=$kl1, $kl1=true, $kl1=[1],$kl14=$kl15,$kl16=$kl14}\n+##### message2:48667697\n+loop map {val56}\n \tl1 l2/l3/i3/l7_l90/l90-SC21_l90/l90-l90_l90_l90\n end loop\n-\n-loop map {val56}\n+#####kl message45:48667697\n+loop map {val34}\n \ttestcases data/testcases/path1/[ath5+path6/UC20015-SC21/UC20015-SC21\n end loop\n
In above string, I want to fetch all the sub-strings starting from \\n+ and ending with \\n . 在上面的字符串中,我想获取从\\ n +开始并以\\ n结尾的所有子字符串。 Also, starting from \\n- and ending with \\n using Java regular expression.
此外,从\\ n开始并使用Java正则表达式以\\ n结尾。
Expected Output is as below: 预期产出如下:
blank // here its blank as first \n- to next \n nothing is there.
loop map {$df=56,$kl=20564300,$testId=\"jk: message1 message2:48667697\",$kl3=true,$kl=true, $kl=[2],$kl=$kl1, $kl1=true, $kl1=[1],$kl14=$kl15,$kl16=$kl14} // as second \n- to next \n
##### message2:48667697 //third \n+ to next \n
loop map {val56}\n \tl1 l2/l3/i3/l7_l90/l90-SC21_l90/l90-l90_l90_l90 //fourth \n+ to next \n
blank // \n- to next \n
loop map {val56} // \n- to next \n
#####kl message45:48667697 // as \n+ to \n
loop map {val34} // as \n+ to \n
Actually I wanted to make two different sets, one for \\n+ to \\n and another for \\n- to \\n. 实际上我想制作两个不同的集合,一个用于\\ n +到\\ n,另一个用于\\ n-到\\ n。 As I wanted to use this for later purposes.
因为我想将它用于以后的目的。
below is the java code I have tried with: 下面是我试过的java代码:
String str="Pasted above string making sure no additional string literals should come.";
Pattern p0 = Pattern.compile("^(\\\\n[+|-])(.*?)(\\\\n.*)?$");
Matcher m = p0.matcher(str);
while (m.find()) {
System.out.printf( m.group(0));// here I am expecting my output to get printed.
}
Can anyone help me out with the same. 任何人都可以帮助我同样的。 Any help would be highly appreciated.
任何帮助将受到高度赞赏。 Thanks.
谢谢。
This is the regex that might help you 这是可以帮助你的正则表达式
^(\\\\\\\\n[+|-])(.*)$
^ - Start of line
\\\\\\\\n - escaping \\
^(\\\\\\\\n[+|-]) - Starting point of line followed by \\n with either + or -
(.*?) anything can follow after that (? for non-greedy search if \\n follows after that)
(\\\\\\\\n.*)? - Might be that \\n follows (4th line in text)
$ - End of line. Assuming that after end of line what follows is \\n-new line
From the output that you have provided, it seems that there are other conditions also that you have forgot to mention in your question, like why ###problem statement1
is not in the output because it starts with \\n+
and ends with new line. 从您提供的输出中,似乎还有其他条件也在您的问题中忘记提及,例如为什么
###problem statement1
不在输出中,因为它以\\n+
开头并以新行结束。 Below is the code snippet for the same 以下是相同的代码段
public static void main(String[] args) throws Exception {
Pattern pattern = Pattern.compile("^(\\\\n[+|-])(.*?)(\\\\n.*)?$");
Matcher matcher = null;
BufferedReader br = new BufferedReader(new FileReader(
"filePath"));
String line = null;
while ((line = br.readLine()) != null) {
matcher = pattern.matcher(line);
while (matcher.find()) {
System.out.println(matcher.group(2));
}
}
}
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