搜索一個字符串並返回特定文本android

Question

嗨如何從下面的字符串中返回Grapes，我想搜索一個字符串並在四個字符後返回字符串中間的文本，並丟棄文本的其餘部分。

String grapes = "2 x Grapes @Walmart"; 

Answer 1

感謝您幫助我的傢伙下面的代碼工作

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

Answer 2

我的建議是不會爲此使用正則表達式。但是爲了以防萬一，你發現沒有倒過來，用這個：

(\w+\s){3} 

，你會在第一時間拿到反向引用的第三個單詞。 \1或$1取其支持你的編譯器

演示在這裏：http://regex101.com/r/jB5nN0

Answer 3

這可能會幫助您：

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

說明：

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) «$»