如何批量移动句子中的空格

Question

这就是我的问题。我做了一个批次，想要从句子中移动单词之间的所有空格。我用一种方法做到了这一点。但我想将space设置为变量，并在set命令中使用该变量。在检查了变量的定义之后，即使这个空格变量得到了很好的响应，我也不能在最后一句话中移动它。谁能告诉我为什么？

@echo off

::why this doesnt work???


(set /p word2=i will make )<nul>textfile.txt

<nul (set/p word3=all strings be )>>textfile.txt

<nul (set/p word4=one longstring)>>textfile.txt

echo(

type textfile.txt

echo(

echo(

::seting space

set sp=a b

set sa=%sp:a=%

set space=%sa:b=%

if defined space (echo space is defined) else (echo space is not defined)

echo(

echo lets see:%space%%space%%space%something%space%something%space%%space%, ...works

echo(

set /p inputalltext=<textfile.txt

echo %inputalltext%...input from textfile

set noplaceforspace=%inputalltext: =% ...as you see as normal,no space

echo %noplaceforspace%

::so, problem starts here. why are here  places if space is defined and good echoing

::of space variable??

setlocal enabledelayedexpansion

set nogoodresult=%inputalltext:!space!=%

echo %nogoodresult%%space%%space% here i dont want space between words

echo(

pause

Answer 1

请看我的例子：

@echo off&setlocal
set "space= "
set "sentence=this is a long sentence with many words "

setlocal enabledelayedexpansion
set "sentence=!sentence:%space%=!"
echo "%sentence%"

Answer 2

如果你正在运行一个常规的批处理，你应该能够做“句子”，只要你在它周围有“”。系统应输入一个空格