给定字符串“ThisStringHasNoSpacesButItDoesHaveCapitals”,什么是在大写字母之前添加空格的最好方法。所以结尾字符串是"This string Has No space But It Does Have大写"
下面是我使用正则表达式的尝试
System.Text.RegularExpressions.Regex.Replace(value, "[A-Z]", " $0")
给定字符串“ThisStringHasNoSpacesButItDoesHaveCapitals”,什么是在大写字母之前添加空格的最好方法。所以结尾字符串是"This string Has No space But It Does Have大写"
下面是我使用正则表达式的尝试
System.Text.RegularExpressions.Regex.Replace(value, "[A-Z]", " $0")
当前回答
下面是在SQL中如何做到这一点
create FUNCTION dbo.PascalCaseWithSpace(@pInput AS VARCHAR(MAX)) RETURNS VARCHAR(MAX)
BEGIN
declare @output varchar(8000)
set @output = ''
Declare @vInputLength INT
Declare @vIndex INT
Declare @vCount INT
Declare @PrevLetter varchar(50)
SET @PrevLetter = ''
SET @vCount = 0
SET @vIndex = 1
SET @vInputLength = LEN(@pInput)
WHILE @vIndex <= @vInputLength
BEGIN
IF ASCII(SUBSTRING(@pInput, @vIndex, 1)) = ASCII(Upper(SUBSTRING(@pInput, @vIndex, 1)))
begin
if(@PrevLetter != '' and ASCII(@PrevLetter) = ASCII(Lower(@PrevLetter)))
SET @output = @output + ' ' + SUBSTRING(@pInput, @vIndex, 1)
else
SET @output = @output + SUBSTRING(@pInput, @vIndex, 1)
end
else
begin
SET @output = @output + SUBSTRING(@pInput, @vIndex, 1)
end
set @PrevLetter = SUBSTRING(@pInput, @vIndex, 1)
SET @vIndex = @vIndex + 1
END
return @output
END
其他回答
具有fold的实现,也称为Aggregate:
public static string SpaceCapitals(this string arg) =>
new string(arg.Aggregate(new List<Char>(),
(accum, x) =>
{
if (Char.IsUpper(x) &&
accum.Any() &&
// prevent double spacing
accum.Last() != ' ' &&
// prevent spacing acronyms (ASCII, SCSI)
!Char.IsUpper(accum.Last()))
{
accum.Add(' ');
}
accum.Add(x);
return accum;
}).ToArray());
除了请求之外,这个实现还正确地保存了开头、内部、结尾空格和首字母缩写,例如,
" SpacedWord " => " Spaced Word ",
"Inner Space" => "Inner Space",
"SomeACRONYM" => "Some ACRONYM".
仅由ASCII字符组成的输入字符串的c#解决方案。regex结合了反向回溯来忽略出现在字符串开头的大写字母。使用Regex.Replace()返回所需的字符串。
参见regex101.com演示。
using System;
using System.Text.RegularExpressions;
public class RegexExample
{
public static void Main()
{
var text = "ThisStringHasNoSpacesButItDoesHaveCapitals";
// Use negative lookbehind to match all capital letters
// that do not appear at the beginning of the string.
var pattern = "(?<!^)([A-Z])";
var rgx = new Regex(pattern);
var result = rgx.Replace(text, " $1");
Console.WriteLine("Input: [{0}]\nOutput: [{1}]", text, result);
}
}
预期的输出:
Input: [ThisStringHasNoSpacesButItDoesHaveCapitals]
Output: [This String Has No Spaces But It Does Have Capitals]
更新:这里有一个变种,也将处理首字母缩写(大写字母序列)。
参见regex101.com演示和ideone.com演示。
using System;
using System.Text.RegularExpressions;
public class RegexExample
{
public static void Main()
{
var text = "ThisStringHasNoSpacesASCIIButItDoesHaveCapitalsLINQ";
// Use positive lookbehind to locate all upper-case letters
// that are preceded by a lower-case letter.
var patternPart1 = "(?<=[a-z])([A-Z])";
// Used positive lookbehind and lookahead to locate all
// upper-case letters that are preceded by an upper-case
// letter and followed by a lower-case letter.
var patternPart2 = "(?<=[A-Z])([A-Z])(?=[a-z])";
var pattern = patternPart1 + "|" + patternPart2;
var rgx = new Regex(pattern);
var result = rgx.Replace(text, " $1$2");
Console.WriteLine("Input: [{0}]\nOutput: [{1}]", text, result);
}
}
预期的输出:
Input: [ThisStringHasNoSpacesASCIIButItDoesHaveCapitalsLINQ]
Output: [This String Has No Spaces ASCII But It Does Have Capitals LINQ]
这对聚合来说是个好机会。它被设计成可读的,但不一定特别快。
someString
.Aggregate(
new StringBuilder(),
(str, ch) => {
if (char.IsUpper(ch) && str.Length > 0)
str.Append(" ");
str.Append(ch);
return str;
}
).ToString();
下面是在SQL中如何做到这一点
create FUNCTION dbo.PascalCaseWithSpace(@pInput AS VARCHAR(MAX)) RETURNS VARCHAR(MAX)
BEGIN
declare @output varchar(8000)
set @output = ''
Declare @vInputLength INT
Declare @vIndex INT
Declare @vCount INT
Declare @PrevLetter varchar(50)
SET @PrevLetter = ''
SET @vCount = 0
SET @vIndex = 1
SET @vInputLength = LEN(@pInput)
WHILE @vIndex <= @vInputLength
BEGIN
IF ASCII(SUBSTRING(@pInput, @vIndex, 1)) = ASCII(Upper(SUBSTRING(@pInput, @vIndex, 1)))
begin
if(@PrevLetter != '' and ASCII(@PrevLetter) = ASCII(Lower(@PrevLetter)))
SET @output = @output + ' ' + SUBSTRING(@pInput, @vIndex, 1)
else
SET @output = @output + SUBSTRING(@pInput, @vIndex, 1)
end
else
begin
SET @output = @output + SUBSTRING(@pInput, @vIndex, 1)
end
set @PrevLetter = SUBSTRING(@pInput, @vIndex, 1)
SET @vIndex = @vIndex + 1
END
return @output
END
请确保您没有在字符串的开头放置空格,而是将它们放在连续的大写字母之间。这里的一些答案并没有解决其中的一个或两个问题。除了regex,还有其他方法,但如果你更喜欢使用它,试试这个:
Regex.Replace(value, @"\B[A-Z]", " $0")
\B是一个负的\B,所以它代表一个非单词边界。这意味着模式匹配XYzabc中的“Y”,但不匹配Yzabc或XYzabc。作为一个小奖励,你可以在一个有空格的字符串上使用它,它不会使它们加倍。