求助：编写Batch脚本生成符合规则的6位随机数CSV文件

需求概述 #

• 生成包含300组数据的CSV文件
• 每组为1-39范围内的6个不重复随机数
• 任意两行的重复数字不得超过2个

当前脚本问题 #

• 输出行内容持续变长，无法固定为6个数字
• 未实现行内去重及跨行重复数字校验功能
• 临时文件逻辑混乱，导致数据累积

当前脚本代码 #

@echo off
setlocal enabledelayedexpansion 

for /L %%j in (1 1 10) do (
    call:get_rand
    echo %%j  ;!NUM[1]!;!NUM[2]!;!NUM[3]!;!NUM[4]!;!NUM[5]!;!NUM[6]! >> test.txt

    sort "get_six.txt" >> "get_sort.txt"
    for /f %%b in (get_sort.txt) do findstr "%%~b" "get_tmp.txt" >nul 2>&1 || echo %%b>>"get_tmp.txt"
    call:get_horiz
rem call:del_get_files
)
goto:EOF


rem delete extra files:
rem :del_get_files
rem for %%z in (get_six.txt,get_sort.txt,get_tmp.txt) do (
rem     if exist "%%z" del /q /f "%%z"
rem )
rem goto:EOF


rem get_tmp.txt vertical list to csv horizontal list.txt
:get_horiz
set var=
for /f "tokens=*" %%c in (get_tmp.txt) do (
    call set var=%%var%%,%%c
)
SET var=%var:~1%
echo !var! >> list.txt
goto:EOF


rem get first set of six random numbers
:get_rand
for /L %%i in (1 1 6) do (
    call:get_n %%i
    rem get random number, 1 is the min, 39 is the max
    echo !num[%%i]! >> get_six.txt
)
goto:EOF


rem get random num[%%i] between 1 and 39
:get_n
set /a num[%1] = %RANDOM% * 39 / 32768 + 1
goto:EOF

期望输出示例（list.txt） #

03,11,17,24,28,37
11,13,25,26,36,39
02,09,15,23,28,37
03,13,14,26,28,32
08,11,21,22,34,36
02,13,16,26,27,31
01,06,14,20,27,31
07,09,20,23,33,34
...

当前实际输出问题示例 #

10,28,39 
10,28,39,14,20,22,36 
10,28,39,14,20,22,36,25,34,35 
10,28,39,14,20,22,36,25,34,35,18,31,32 
10,28,39,14,20,22,36,25,34,35,18,31,32,19,29,7 
10,28,39,14,20,22,36,25,34,35,18,31,32,19,29,7,11,17,21,23,33 
10,28,39,14,20,22,36,25,34,35,18,31,32,19,29,7,11,17,21,23,33,12,37 
10,28,39,14,20,22,36,25,34,35,18,31,32,19,29,7,11,17,21,23,33,12,37,24 
10,28,39,14,20,22,36,25,34,35,18,31,32,19,29,7,11,17,21,23,33,12,37,24,15 
10,28,39,14,20,22,36,25,34,35,18,31,32,19,29,7,11,17,21,23,33,12,37,24,15,27

修复后的完整脚本 #

@echo off
setlocal enabledelayedexpansion

:: 初始化输出文件
if exist "list.txt" del "list.txt"
set "total_lines=300"

:: 生成第一组符合要求的6个不重复数字
call :generate_valid_line
echo !current_line! >> list.txt
set "line_count=1"

:: 生成剩余299组
:generate_next_line
if !line_count! geq !total_lines! goto :cleanup

call :generate_valid_line
call :check_against_all_lines
if !is_valid! equ 1 (
    echo !current_line! >> list.txt
    set /a line_count+=1
)
goto :generate_next_line

:: 生成一行6个1-39的不重复数字，格式为两位补零的CSV
:generate_valid_line
set "current_line="
set "used_nums="
set "count=0"

:pick_num
if !count! equ 6 goto :format_line
set /a rand_num=%RANDOM% * 39 / 32768 + 1
:: 补零为两位
if !rand_num! lss 10 set "rand_num=0!rand_num!"
:: 检查是否已使用
echo !used_nums! | findstr /c:"!rand_num!," >nul
if not errorlevel 1 goto :pick_num
set "used_nums=!used_nums!!rand_num!,"
set /a count+=1
goto :pick_num

:format_line
:: 移除末尾的逗号
set "current_line=!used_nums:~0,-1!"
goto :eof

:: 检查当前行与已生成所有行的重复数字是否不超过2个
:check_against_all_lines
set "is_valid=1"
for /f "tokens=*" %%l in (list.txt) do (
    set "match_count=0"
    :: 将当前行拆分为单个数字
    for %%n in (!current_line:,= !) do (
        echo %%l | findstr /c:"%%n" >nul
        if not errorlevel 1 set /a match_count+=1
        if !match_count! gtr 2 goto :invalid_line
    )
)
goto :eof

:invalid_line
set "is_valid=0"
goto :eof

:: 清理临时变量并退出
:cleanup
endlocal
echo 生成完成，共!total_lines!组数据。
goto :eof

关键逻辑说明 #

• 行内去重：生成随机数时通过findstr检查是否已加入当前行，确保6个数字无重复
• 跨行重复校验：新生成的行与已存在的每一行逐一比对重复数字数量，超过2个则重新生成
• 格式规范：所有数字补零为两位，输出为标准CSV格式
• 循环控制：直到生成满300组符合要求的数据为止

内容的提问来源于stack exchange，提问作者Rene