Question

我试图跳过字符串中的第一个分号，并使用正则表达式对分号的其余部分进行拆分：

lines = </li> <li> Urinary tract infection </li> <li> Respiratory infection </li> <li> Sinus problems; and </li> <li> Ear infections; <li> Some more info </li>

我使用此代码在除第一个分号之外的每个分号处拆分：

lines.split(/(?<!\\\\);/)

我的预期输出是：

["</li> <li> Urinary tract infection </li> <li> Respiratory infection </li> <li> Sinus problems; and </li> <li> Ear infections","<li> Some more info </li>" ]

请注意，字符串可以是任意数量的分号，但我想防止分裂只发生在第一个分号上。

Answer 1

我必须使用正则表达式吗？

str = "Now is; the time; for all good; people; to party" 

f, sc, l = str.partition(';')
af, *a = str[str.index(';')+1..-1].split(';')
[f+sc+af, *a]
  #=> ["Now is; the time", " for all good", " people", " to party"]

步骤：

f, sc, l = str.partition(';')
   #=> ["Now is", ";", " the time; for all good; people; to party"] 
f  #=> "Now is" 
sc #=> ";" 
l  #=> " the time; for all good; people; to party" 
idx = str.index(';')+1
   #=> 7 
b = str[idx..-1]
   #=> " the time; for all good; people; to party" 
af, *a = b.split(';')
af #=> " the time" 
a  #=> [" for all good", " people", " to party"] 
[f+sc+af, *a]
   #=> ["Now is; the time", " for all good", " people", " to party"]

对于OP的例子：

f, sc, l = lines.partition(';')
af, *a = lines[lines.index(';')+1..-1].split(';')
[f+sc+af, *a]
  #=> ["</li> <li> Urinary tract infection </li> <li> Respiratory infection\    
         </li> <li> Sinus problems; and </li> <li> Ear infections",
  #    " <li> Some more info </li>"]

另一种方式：

b = str.sub(';', 0.chr).split(';')
  #=> ["Now is\u0000 the time", " for all good", " people", " to party"] 
a[0].sub!(0.chr, ';')
  #=> "Now is; the time" 
a #=> ["Now is; the time", " for all good", " people", " to party"]

使用正则表达式跳过第一次出现的分号

1 个答案: