Python SUBPROCESS.RUN说文件不在适当的位置

Question

因此，当我尝试打开.bat文件时：

Traceback (most recent call last):
  File "accorddownload.py", line 31, in <module>
    subprocess.run(["Tools\aAIvwb.bat"], creationflags=0x08000000)
  File "C:\Program Files\Python37\lib\subprocess.py", line 472, in run
    with Popen(*popenargs, **kwargs) as process:
  File "C:\Program Files\Python37\lib\subprocess.py", line 775, in __init__
    restore_signals, start_new_session)
  File "C:\Program Files\Python37\lib\subprocess.py", line 1178, in _execute_child
    startupinfo)
FileNotFoundError: [WinError 2] The system cannot find the file specified

我正在使用它运行

subprocess.run(["tools\aAIvwb.bat"], creationflags=0x08000000)

但是问题是我检查了该位置20次并继续尝试修复，但是同一件事正在发生。这是文件结构

AB
|
|--Tools--|
|         |--aaitp.bat
|         |--aAIvwb.bat
|         |--mdfa.bat
|
|--basefile.py

编辑1：当前工作目录从不更改在整个文件中更改

Answer 1

我怀疑\转义。您是否考虑过像这样打电话：

subprocess.run([r"tools\aAIvwb.bat"], creationflags=0x08000000)

请注意，我在双引号前加上了'r'。