Question

我有以下字符串

This is some testing dbo.GetPersonData(this_.PersonId,'Date contract recieved from the client') cs.dbo.Person 'this is test'

我想要它通用并获得以下结果

This
is
some
testing
dbo.GetPersonData(This_.PersonId,
'DateContract received from the client'
)
cs.dbo.person
'this is test'

我在这里想要实现的是用空格分隔单引号之间的任何内容

我试过使用

'(.*?)'|\S+

然而，它忽略了：<'Date contract recieved from the client'>

Result

Answer 1

<块引用>

我试过使用 '(.*?)'|\S+

你的尝试还不错，我们只需要从第二个选项中排除引用：

'.*?'|[^\s']+ regex101

Answer 2

\S 匹配任何非空白字符，' 也匹配。

使用

'([^']*)'|[^\s']+

或者，如果您有 PCRE：

(?|'([^']*)'|([^\s']+))

见proof。

说明

--------------------------------------------------------------------------------
  '                        '\''
--------------------------------------------------------------------------------
  (                        group and capture to \1:
--------------------------------------------------------------------------------
    [^']*                    any character except: ''' (0 or more
                             times (matching the most amount
                             possible))
--------------------------------------------------------------------------------
  )                        end of \1
--------------------------------------------------------------------------------
  '                        '\''
--------------------------------------------------------------------------------
 |                        OR
--------------------------------------------------------------------------------
  [^\s']+                  any character except: whitespace (\n, \r,
                           \t, \f, and " "), ''' (1 or more times
                           (matching the most amount possible))

Answer 3

您应该使用以下模板：

(\n.*?)*)\n\s*\n

除单引号内的字符串外，正则表达式按空格分割

3 个答案: