搜索一个字符串并返回特定文本android

Question

嗨如何从下面的字符串中返回Grapes，我想搜索一个字符串并在四个字符后返回字符串中间的文本，并丢弃文本的其余部分。

String grapes = "2 x Grapes @Walmart"; 

Answer 1

感谢您帮助我的家伙下面的代码工作

String grapes = "2 x Grapes @Walmart"; 
String[] split = grapes.split("\\s+"); 
String fsplit = split[2]; 

Answer 2

我的建议是不会为此使用正则表达式。但是为了以防万一，你发现没有倒过来，用这个：

(\w+\s){3} 

，你会在第一时间拿到反向引用的第三个单词。 \1或$1取其支持你的编译器

演示在这里：http://regex101.com/r/jB5nN0

Answer 3

这可能会帮助您：

^[\\d]+\\sx\\s(.*?)\\s+.*?$ 

说明：

Assert position at the beginning of a line (at beginning of the string or after a line break character) «^» 
Match a single digit 0..9 «[\d]+» 
    Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+» 
Match a single character that is a “whitespace character” (spaces, tabs, and line breaks) «\s» 
Match the character “x” literally «x» 
Match a single character that is a “whitespace character” (spaces, tabs, and line breaks) «\s» 
Match the regular expression below and capture its match into backreference number 1 «(.*?)» 
    Match any single character that is not a line break character «.*?» 
     Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?» 
Match a single character that is a “whitespace character” (spaces, tabs, and line breaks) «\s+» 
    Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+» 
Match any single character that is not a line break character «.*?» 
    Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?» 
Assert position at the end of a line (at the end of the string or before a line break character) «$»