从变量中未知数量的定界符中提取某些标记的批处理文件
Batch file extracting certain tokens from unknown number of delimiters in variable
第三，希望是对问题的最后修改...
批处理文件用 for 循环逐行读取文本文件到变量中。所述文本文件的每一行的格式都可以与下一行完全不同。唯一常见的分隔符是每行某处的四位数（年份）。目标是 return 通过 echo 将任何文本跟在上述每行的四位数字后面。
文本文件示例：
Monday, January 1, 1900 there was an event-e6718
On this day in 1904 nothing occurred
Wednesday, March 3, 1908 an error occurred when attempting to access the log
Thursday, , 1911 - access denied
Friday, in whatever month, on whatever day, in 1938, nothing happened
因此，根据上面的文本文件示例，return 就像...
there was an event-e6718
nothing occurred
an error occurred when attempting to access the log
- access denied
nothing happened
截至 1318 PST，我已经尝试了以下评论中的每个代码片段，但其中 none 能够 return 我需要 return 的数据。
但是，这些评论与我最初的问题相关，该问题已得到显着改进。
我什至试过正则表达式“^[1-9][0-9][0-9][0-9]$”，然而我是正则表达式的新手，所以我确定我错了。
这可能吗？
提前致谢。
如果字符串中确实没有共同点或令牌计数不一致，请调整以下 for 循环中的迭代次数以匹配最大可能的令牌。
@Echo off & Setlocal EnableDelayedexpansion
Set "event=Monday, January 1, 1900 there was an event-e6718"
For /L %%i in (1 1 10) Do (
Set "event=!event:*, =!"
)& rem // arbitrary number of iterations that should be adjusted to match the maximum expected tokens
Set "event=%event:~5,100%"& rem // remove the year[space] from the string - final string maximum length is also arbitrary and may need adjusting.
Echo/%event%
** 更新**
使用上述方法的宏版本获取 for 循环中最后一个标记的宏示例：
注意：您需要调整输入文件的文件路径。
@Echo off
(Set \n=^^^
%=Newline Var=%
)
Set Gettoken=For %%n in (1 2) Do if %%n==2 (%\n%
For /F "Delims=" %%G in ("!string!") Do (%\n%
Set "output=%%G"%\n%
For %%D in ("!Delim!") Do For /L %%i in (1 1 10) Do Set "output=!output:*%%~D=!"%\n%
Set "output=!output:~5,100!"%\n%
)%\n%
Set output%\n%
) Else Set string=
Setlocal EnableDelayedExpansion
Set "Delim=, "&& For /F "Delims=" %%I in (inputfile.txt) Do %GetToken%%%I
编辑：
批改题实际要求的解决方案。
@Echo off & CD "%~dp0"
Setlocal Enabledelayedexpansion
rem // replace inputfile.txt with your filepath
For /F "Delims=" %%L in (inputfile.txt) Do (
Call :sub "%%L"
rem // the below for loop will remove everything up to including the first year from the string
rem // as well as traling coma[space] / [space]
For %%E in (!Errorlevel!) Do (
If Not "%%E"=="0" (
Set "String=!String:*%%E=####!"
Set "String=!String:####, =!"
Set "String=!String:#### =!"
Set "String=!String:####=!"
)
)
rem // output only if a year "delimiter" was encountered
If not "%%~L"=="!String!" Echo/!String!
)
Exit /b
:sub
Set "String=%~1"
rem // adjust for loop %%I for valid year range and %%# for maximum expected string length
For /L %%I in (1899 1 2050) Do (For /L %%# in (0 1 100) Do (If "!String:~%%#,4!"=="%%I" (Exit /B %%I)))
Exit /B 0
试一试：
@echo off & setlocal enabledelayedexpansion
for /f "delims=" %%i in ('type "C:\textfile.txt" ^| findstr /IRC:"there was an event"') do (
set "event=%%i"
echo "!event:*there was an event=there was an event!"
)
textfile.txt
Monday, January 1, 1900 there was an event-e6718
On this day in 1904 nothing occurred
Wednesday, March 3, 1908 an error occurred when attempting to access the log
Thursday, , 1911 - access denied
Monday, January 1, 1910 there was an event-dsfd318
Friday, in whatever month, on whatever day, in 1938, nothing happened
结果：
批处理是一项可怕的任务。 REGEX 是一个很好的工具，但 cmd 不支持它（findstr 的一个非常残缺的子集除外）。如果你愿意使用 external tool，这很容易：
<old.txt call jrepl ".*(\d{4})\D\ *(.*$)" "" >new.txt
搜索四位数字 \d{4}，然后是 non-digit \D 和零个或多个空格，直到“EndOfLine” .*$。 （括号）标记匹配项，由 $x 引用。您想要的字符串在 </code>.</p>
<p>使用您的示例文件输出：</p>
<pre><code>there was an event-e6718
nothing occurred
an error occurred when attempting to access the log
- access denied
there was an event-dsfd318
nothing happened
如果您决定包括年份，您可以在 </code>:</p> 中找到它
<pre><code><old.txt call jrepl ".*(\d{4})\D\ *(.*$)" ": " >new.txt
给出：
1900: there was an event-e6718
1904: nothing occurred
1908: an error occurred when attempting to access the log
1911: - access denied
1910: there was an event-dsfd318
1938: nothing happened
call 是批处理文件所必需的，因为 jrepl 是一个批处理文件，因此没有 call 就不会 return。
（REGEX 模式可能需要改进；我还没有太多经验。）
jrepl.bat 由 dbenham 编程。
第三，希望是对问题的最后修改...
批处理文件用 for 循环逐行读取文本文件到变量中。所述文本文件的每一行的格式都可以与下一行完全不同。唯一常见的分隔符是每行某处的四位数（年份）。目标是 return 通过 echo 将任何文本跟在上述每行的四位数字后面。
文本文件示例：
Monday, January 1, 1900 there was an event-e6718
On this day in 1904 nothing occurred
Wednesday, March 3, 1908 an error occurred when attempting to access the log
Thursday, , 1911 - access denied
Friday, in whatever month, on whatever day, in 1938, nothing happened
因此，根据上面的文本文件示例，return 就像...
there was an event-e6718
nothing occurred
an error occurred when attempting to access the log
- access denied
nothing happened
截至 1318 PST，我已经尝试了以下评论中的每个代码片段，但其中 none 能够 return 我需要 return 的数据。
但是，这些评论与我最初的问题相关，该问题已得到显着改进。
我什至试过正则表达式“^[1-9][0-9][0-9][0-9]$”，然而我是正则表达式的新手，所以我确定我错了。
这可能吗？
提前致谢。
如果字符串中确实没有共同点或令牌计数不一致，请调整以下 for 循环中的迭代次数以匹配最大可能的令牌。
@Echo off & Setlocal EnableDelayedexpansion
Set "event=Monday, January 1, 1900 there was an event-e6718"
For /L %%i in (1 1 10) Do (
Set "event=!event:*, =!"
)& rem // arbitrary number of iterations that should be adjusted to match the maximum expected tokens
Set "event=%event:~5,100%"& rem // remove the year[space] from the string - final string maximum length is also arbitrary and may need adjusting.
Echo/%event%
** 更新** 使用上述方法的宏版本获取 for 循环中最后一个标记的宏示例：
注意：您需要调整输入文件的文件路径。
@Echo off
(Set \n=^^^
%=Newline Var=%
)
Set Gettoken=For %%n in (1 2) Do if %%n==2 (%\n%
For /F "Delims=" %%G in ("!string!") Do (%\n%
Set "output=%%G"%\n%
For %%D in ("!Delim!") Do For /L %%i in (1 1 10) Do Set "output=!output:*%%~D=!"%\n%
Set "output=!output:~5,100!"%\n%
)%\n%
Set output%\n%
) Else Set string=
Setlocal EnableDelayedExpansion
Set "Delim=, "&& For /F "Delims=" %%I in (inputfile.txt) Do %GetToken%%%I
编辑：
批改题实际要求的解决方案。
@Echo off & CD "%~dp0"
Setlocal Enabledelayedexpansion
rem // replace inputfile.txt with your filepath
For /F "Delims=" %%L in (inputfile.txt) Do (
Call :sub "%%L"
rem // the below for loop will remove everything up to including the first year from the string
rem // as well as traling coma[space] / [space]
For %%E in (!Errorlevel!) Do (
If Not "%%E"=="0" (
Set "String=!String:*%%E=####!"
Set "String=!String:####, =!"
Set "String=!String:#### =!"
Set "String=!String:####=!"
)
)
rem // output only if a year "delimiter" was encountered
If not "%%~L"=="!String!" Echo/!String!
)
Exit /b
:sub
Set "String=%~1"
rem // adjust for loop %%I for valid year range and %%# for maximum expected string length
For /L %%I in (1899 1 2050) Do (For /L %%# in (0 1 100) Do (If "!String:~%%#,4!"=="%%I" (Exit /B %%I)))
Exit /B 0
试一试：
@echo off & setlocal enabledelayedexpansion
for /f "delims=" %%i in ('type "C:\textfile.txt" ^| findstr /IRC:"there was an event"') do (
set "event=%%i"
echo "!event:*there was an event=there was an event!"
)
textfile.txt
Monday, January 1, 1900 there was an event-e6718
On this day in 1904 nothing occurred
Wednesday, March 3, 1908 an error occurred when attempting to access the log
Thursday, , 1911 - access denied
Monday, January 1, 1910 there was an event-dsfd318
Friday, in whatever month, on whatever day, in 1938, nothing happened
结果：
批处理是一项可怕的任务。 REGEX 是一个很好的工具，但 cmd 不支持它（findstr 的一个非常残缺的子集除外）。如果你愿意使用 external tool，这很容易：
<old.txt call jrepl ".*(\d{4})\D\ *(.*$)" "" >new.txt
搜索四位数字 \d{4}，然后是 non-digit \D 和零个或多个空格，直到“EndOfLine” .*$。 （括号）标记匹配项，由 $x 引用。您想要的字符串在 </code>.</p>
<p>使用您的示例文件输出：</p>
<pre><code>there was an event-e6718
nothing occurred
an error occurred when attempting to access the log
- access denied
there was an event-dsfd318
nothing happened
如果您决定包括年份，您可以在 </code>:</p> 中找到它
<pre><code><old.txt call jrepl ".*(\d{4})\D\ *(.*$)" ": " >new.txt
给出：
1900: there was an event-e6718
1904: nothing occurred
1908: an error occurred when attempting to access the log
1911: - access denied
1910: there was an event-dsfd318
1938: nothing happened
call 是批处理文件所必需的，因为 jrepl 是一个批处理文件，因此没有 call 就不会 return。
（REGEX 模式可能需要改进；我还没有太多经验。）
jrepl.bat 由 dbenham 编程。