sample_string = "let's could've they'll you're won't"
sample_string.scan(/\w /)
以上給了我：
["let", "s", "could", "ve", "they", "ll", "you", "re", "won", "t"]
我想要的是：
["let", "could", "they", "you", "won"]
一直在https://rubular.com/ 上玩，并嘗試斷言，\w (?<=')但沒有運氣。
uj5u.com熱心網友回復：
您可以使用
sample_string.scan(/(?<![\w'])\w /)
sample_string.scan(/\b(?<!')\w /)
請參閱Rubular 演示。模式（它們是絕對同義詞）匹配
(?<![\w'])- 字串中沒有緊跟在單詞或'字符前面的位置\b(?<!')- 一個字邊界位置，前面沒有一個'字符\w- 一個或多個字字符。
請參閱Ruby 演示：
sample_string = "let's could've they'll you're won't"
p sample_string.scan(/(?<![\w'])\w /)
# => ["let", "could", "they", "you", "won"]
uj5u.com熱心網友回復：
鑒于：
> sample_string = "let's could've they'll you're won't"
您可以進行拆分和映射：
> sample_string.split.map{|w| w.split(/'/)[0]}
=> ["let", "could", "they", "you", "won"]
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