java 正则表达式从字符串中获取未知长度的前两个单词
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Regex to get first two words of unknown length from a string
提问by TookTheRook
Say I have a string with various words of unknown length. I plan to split the string by using a regular expression. Something like:
假设我有一个字符串，其中包含各种长度未知的单词。我计划使用正则表达式拆分字符串。就像是：
String resString = origString.split(".*//s.*//s")[0];
What would be the regular expression to get the first two words? I was thinking .*//s.*//s, so all characters, followed by a space, then all characters, followed by another space. But using that gives me the exact same string I had before. Am I going about this the wrong way?
获取前两个单词的正则表达式是什么？我在想.*//s.*//s，所以所有字符，后跟一个空格，然后是所有字符，然后是另一个空格。但是使用它给了我与以前完全相同的字符串。我会以错误的方式解决这个问题吗？
Any help would be appreciated!
任何帮助，将不胜感激！
回答by icyrock.com
If you have only spaces between words, split by \\s+. When you split, the array would be the words themselves. First two would be in arr[0]and arr[1]if you do:
如果单词之间只有空格，请按\\s+. 拆分时，数组将是单词本身。前两个将在arr[0]，arr[1]如果你这样做：
String[] arr = origString.split("\s+");
回答by SuperJulietta
With regular expressions you can do something like this:
使用正则表达式，您可以执行以下操作：
public static ArrayList<String> split2(String line, int n){
line+=" ";
Pattern pattern = Pattern.compile("\w*\s");
Matcher matcher = pattern.matcher(line);
ArrayList<String> list = new ArrayList<String>();
int i = 0;
while (matcher.find()){
if(i!=n)
list.add(matcher.group());
else
break;
i++;
}
return list;
}
if you want the first n words, or simply this:
如果你想要前 n 个单词，或者只是这个：
public static String split3(String line){
line+=" ";
Pattern pattern = Pattern.compile("\w*\s\w*\s");
Matcher matcher = pattern.matcher(line);
matcher.find();
return matcher.group();
}
if you want only the first and second words.
如果你只想要第一个和第二个词。
回答by Bubba
If you want to split it on exactly the space character:
如果您想将其完全拆分为空格字符：
String[] parts = args[i].split(" ");
If you want to split it on any whitespace character (space, tab, newline, cr):
如果要将其拆分为任何空白字符（空格、制表符、换行符、cr）：
String[] parts = args[i].split("\s");
To treat multiple adjacent spaces as one separator:
将多个相邻空格视为一个分隔符：
String[] parts = args[i].split(" +");
Same for whitespace:
空格相同：
String[] parts = args[i].split("\s+");
The first two words would be parts[0]and parts[1]
前两个词是parts[0]andparts[1]
回答by buru
Assuming your "words" consist of alphanumeric characters, the following regex will match the first 2 words:
假设您的“单词”由字母数字字符组成，以下正则表达式将匹配前 2 个单词：
\w+\s+\w+
回答by Ashok Reddy M
Use below method.
使用下面的方法。
public static String firstLettersOfWord(String word) {
StringBuilder result = new StringBuilder();
String[] myName = word.split("\s+");
for (String s : myName) {
result.append(s.charAt(0));
}
return result.toString();
}}
