[英]How to split string in a variable without loosing spaces?
how can i split the string 我怎么能分裂字符串
" This is a text with spaces "
that is in the variable "string" into text parts without loosing the spaces ? 那是在变量“string”中的文本部分而没有丢失空格?
set string="# This is a text with spaces #"
should be split into 应分成
"# This"
" is"
" a"
" text"
" with"
" spaces #"
Using For /F "delims= " ...
doesn't work because it eliminates all spaces. 使用For /F "delims= " ...
不起作用,因为它消除了所有空格。
Is there a 'simple' solution or can anyone explain how to parse the string character by character so i can count up spaces to first character, then read all characters until next space and write the counted spaces and the read characters together to a new/temp variable ?? 有一个'简单'的解决方案或任何人都可以解释如何逐字符解析字符串,这样我可以计算第一个字符的空格,然后读取所有字符直到下一个空格,并将计数的空格和读取的字符一起写入一个新的/临时变量??
thanks 谢谢
Yeah, I don't really understand the #
either. 是的,我不太了解#
也。 What is it about " spaces #" that makes it hold onto the trailing spaces, while all other elements keep the preceding but not the proceeding spaces? 什么是“空间#”使它保持在尾随空格,而所有其他元素保持前面但不是前进的空格?
Oh, well, effort spent in asking = effort spent in answering. 哦,好吧,努力花在询问=花在回答上的努力。 Do with this what you will. 做到这一点,你会做什么。
@if (@a==@b) @end /*
:: batch portion
@echo off
setlocal
call :split "# This is a text with spaces #"
exit /b
:split <string>
cscript /nologo /e:jscript "%~f0" "%~1"
goto :EOF
:: JScript portion */
WSH.Echo(WSH.Arguments(0).match(/\s*\S+/g).join('\n'));
Output: 输出:
#
This
is
a
text
with
spaces
#
If you want the first + second, and the penultimate + ultimate elements joined, modify the JScript portion of the above script as follows: 如果您想要第一个+第二个,并且倒数第二个+终极元素已连接,请修改上述脚本的JScript部分,如下所示:
:: JScript portion */
var m = WSH.Arguments(0).match(/\s*\S+/g);
m[0] = m.shift() + m[0];
m[m.length - 2] += m.pop();
WSH.Echo(m.join('\n'));
Output: 输出:
# This
is
a
text
with
spaces #
And if you want each element enclosed in quotation marks, change the last line as follows: 如果您希望每个元素都用引号括起来,请更改最后一行,如下所示:
WSH.Echo('"' + m.join('"\n"') + '"');
Output: 输出:
"# This"
" is"
" a"
" text"
" with"
" spaces #"
I don't see a simple solution in batch, though of course if you can consider powershell or javascript you'll be working with a more appropriate toolset for string manipulation. 我没有批量看到一个简单的解决方案,当然如果你可以考虑使用powershell或javascript,你将使用更合适的字符串操作工具集。
Sticking with the batch requirement, you can loop through character by character and "collect" your words with something like this: 坚持批量要求,你可以逐个字符地循环,并用这样的东西“收集”你的单词:
@echo off
setlocal enabledelayedexpansion
set "string= This is a text with spaces "
set idx=0
set "word="
set "char="
set "lastchar= "
:loop
if "!string:~%idx%!" equ "" goto :eof
set char=!string:~%idx%,1!
if "%char%" equ " " (
if "%lastchar%" neq " " (
echo [%word%]
set word=%char%
) else (
set word=%word%%char%
)
) else (
set word=%word%%char%
)
set lastchar=%char%
set /a idx=%idx%+1
goto loop
This script uses batch's substring feature !string:~%idx%,1
to grab a single character from the string, incrementing idx
with each loop. 此脚本使用批处理的子字符串功能!string:~%idx%,1
从字符串中获取单个字符,每个循环递增idx
。 Then it's just a matter of processing the word ( echo
in this example) when the previous character was not a space and the current one is. 然后,当前一个字符不是空格而当前字符不是空格时,只需要处理单词(本例中为echo
)。
This writes out: 这写道:
[ This]
[ is]
[ a]
[ text]
[ with]
[ spaces]
Note that I'm ignoring the #
you had in your example because I don't understand where they fit in. 请注意,我忽略了您在示例中的#
,因为我不明白它们适合的位置。
the trick is substituting the contiguous spaces by just one space and the rest by some arbitrary character. 诀窍是用一个空格代替连续的空格,其余用一些任意的字符代替。 Assuming your string does not contain #
s and that there are no more than 9 contiguous spaces, you can try this 假设你的字符串不包含#
s以及有不超过9个连续的空间,你可以试试这个
set st=%st: = ########%
set st=%st: = #######%
set st=%st: = ######%
set st=%st: = #####%
set st=%st: = ####%
set st=%st: = ###%
set st=%st: = ##%
set st=%st: = #%
then you may parse with for /f
and substitute back your #
s by spaces 然后你可以用解析for /f
和替换回你#
通过空间S.
setlocal enabledelayedexpansion
for /f %%a in ("%st%") do (
set ss= %%a
echo !ss:#= !
)
note that set
inside the parentheses block requires you to enable delayed expansion and to use the !
请注意,在括号内set
要求您启用延迟扩展并使用!
syntax (see HELP SET
) 语法(参见HELP SET
)
But this technique will only extract the first substring. 但是这种技术只会提取第一个子串。 To generalize, you need another trick, that is substituting the spaces into newlines so that the for /f
will loop kinda line by line 为了概括,你需要另一个技巧,即将空格替换为换行符,以便for /f
将逐行循环
note that in order to obtain a newline char you need to preserve the two blank lines after the set
command 请注意,为了获取换行符,您需要在set
命令后保留两个空行
set nl=^
rem continue two lines down....
for /f %%a in ("%st: =!nl!%") do (
set ss= %%a
set ss=!ss:#= !
echo [!ss!]
)
Try this: 尝试这个:
@echo off &setlocal enabledelayedexpansion
set "string=# This is a text with spaces #"
set string1=%string%
for %%i in (%string%) do (
set string1=!string1: %%i = "%%i" !
set /a strings+=1
)
set string1=#"%string1:~1,-1%"#
set string1=%string1:"= "%
for %%i in (%string1%) do (
set /a count+=1
set string2=%%i
set string2=!string2: "=!
set string2=!string2:"=!
if !count! equ 2 (
set $s1=!$s1!!string2!
)else if !count! equ %strings% (
set /a count-=1
call set $s!count!=%%$s!count!%%!string2!
) else set $s!count!=!string2!
)
for /f "tokens=1*delims==" %%i in ('set "$s"') do echo "%%j"
Output: 输出:
"# This"
" is"
" a"
" text"
" with"
" spaces #"
If I had to accomplish this obscure task, I would use a hybrid JScript/batch technique like in rojo's answer . 如果我必须完成这个不起眼的任务,我会使用像rojo的答案那样的混合JScript /批处理技术。 However, I would use a REPL.BAT utility that I have already written . 但是,我会使用我已编写的REPL.BAT实用程序 。 Assuming my REPL.BAT is in either the current folder, or else somewhere in the PATH, then the following will work: 假设我的REPL.BAT位于当前文件夹中,或者位于PATH中的某个位置,则以下内容将起作用:
@echo off
setlocal enableDelayedExpansion
set "string=# This is a text with spaces #"
:: Build an "array" of text parts
set cnt=0
for /f delims^=^ eol^= %%A in ('repl "([^ ])(?= )" "$1\n" xs string') do (
set /a cnt+=1
set "string!cnt!=%%A"
)
:: Print the array values
for /l %%N in (1 1 %cnt%) do echo string%%N=[!string%%N!]
But if I wanted a pure batch solution, I would use the fairly efficient method below: 但如果我想要一个纯批处理解决方案,我会使用下面相当有效的方法:
@echo off
setlocal enableDelayedExpansion
set "string=# This is a text with spaces #"
:: Define LF to contain a single line feed character (0x0A)
set LF=^
:: Above 2 blank lines are critical - DO NOT REMOVE
:: Insert a line feed before every space
for %%n in ("!LF!") do set "string=!string: =%%~n !"
:loop Remove line feeds sandwiched by spaces
for %%n in ("!LF!") do set "string2=!string: %%~n = !"
if "!string2!" neq "!string!" (
set "string=!string2!"
goto :loop
)
:: Build an "array" of text parts: FOR /F splits the string at line feeds
set /a cnt=0
for /f delims^=^ eol^= %%A in ("!string!") do (
set /a cnt+=1
set "string!cnt!=%%A"
)
:: Print out the array values
for /l %%N in (1 1 %cnt%) do echo string%%N=[!string%%N!]
Both solutions above give the following output: 以上两种解决方案都提供以下输出:
string1=[#]
string2=[ This]
string3=[ is]
string4=[ a]
string5=[ text]
string6=[ with]
string7=[ spaces]
string8=[ #]
Note that the FOR loop %%A
expansion will corrupt the results if the string contains !
请注意,如果字符串包含FOR循环%%A
扩展将破坏结果!
due to delayed expansion. 由于推迟扩张。 This limitation can be eliminated with additional coding. 通过额外的编码可以消除这种限制。 All the other posted solutions that use a FOR loop suffer from this same limitation. 所有其他使用FOR循环的已发布解决方案都受到同样的限制。 (at least they did when I wrote this) (至少我写这篇文章的时候他们这么做了)
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