在 VSCode 中调试 python 模块的问题 [英] Problem with debugging python modules in VSCode

问题描述

我的python练习项目的目录结构如下:

<预><代码>.├──数据├── ds-and-algo├── 练习|├── __init__.py│ ├── armstrong_number.py│ ├── extract_digits.py├── 输出

extract_digits.py 看起来像这样:

def extract_digits(n):经过

在 armstrong_number.py 中，我有以下内容:

from .extract_digits import extract_digits

如果我运行，则从项目根目录

python 练习/armstrong_number.py

我得到ModuleNotFoundError:没有名为练习的模块

使用 -m 标志运行以下命令可解决错误:

python -m Exercises.armstrong_number

但是使用 VSCode 来调试文件，我有以下调试配置 launch.json:

<代码>{版本":0.2.0"，配置":[{"name": "Python 模块",类型":蟒蛇"，请求":启动"，"program": "${file}","console": "integratedTerminal","pythonPath": "${config:python.pythonPath}","module": "exercises.${fileBasenameNoExtension}","cwd": "${workspaceRoot}","env": {"PYTHONPATH":"${workspaceRoot}"}}]}

但是这有几个问题:

1) 对于不同的文件夹，例如ds-and-algo，我要手动编辑launch.json文件中的模块入口调试配置为

"module" : "ds-and-algo.${fileBaseNameNoExtension}"

如果我有嵌套文件夹配置，例如:

练习├── 艰难||__init__.py|├──ex1.py|├──ex2.py├── 简单

我再次必须手动编辑 launch.json 文件中的调试配置:(考虑子文件夹 tough 的情况)

"module": "exercises.tough.${fileBaseNameNoExtension}"

我需要实现一个一般情况，根据被调试的文件，launch.json 文件中的 "module" 条目应该是:

"module": "folder1.folder2.folder3.....foldern.script"

就像 fileBaseNameNoExtension 一样，VSCode 有 一些其他预定义变量:

其中一个变量是relativeFile，它是当前打开的文件相对于workspaceFolder的路径因此对于文件 ex1.py，变量 relativeFile 将是 exercises/tough/ex1.py.

我需要操作这个字符串并将其转换为exercises.tough.ex1，如果我可以在"module"条目中编写和执行bash命令，这很简单在 launch.json 文件中.但我无法做到这一点.但是，链接VSCode 中的预定义变量 有一个关于命令的部分变量，其中说明:

如果上面的预定义变量不够用，您可以通过 ${command:commandID} 语法使用任何 VS Code 命令作为变量.

此链接包含许多可能有用的其他信息.我不是 python 专家，绝对不知道任何 javascript，如果这是解决这个问题所必需的.

解决方案

我无法重现您的错误:ModuleNotFoundError: no module named practice.

使用以下文件

练习/extract_digits.py

def extract_digits(n):返回 10

练习/armstrong_number.py

from extract_digits import extract_digitsdef armstrong_number(n):返回提取数字(n)如果 __name__ == "__main__":打印 armstrong_number(3)

如果您使用 Python3，请将打印语句更改为:print(armstrong_number(3)) 并更改使用的 Python 解释器.

如果你的当前目录是项目根目录并且你运行

python 练习/armstrong_number.py

您在控制台中得到数字 10

---

在 Visual Studio Code 中，您使用了错误的启动配置.

您只是在运行 Python 程序，因此您应该使用 Python: Current File 配置.launch.json

<代码>{版本":0.2.0"，配置":[{"name": "Python:当前文件",类型":蟒蛇"，请求":启动"，"program": "${file}","console": "integratedTerminal",}]}

1. 在任何目录中打开要运行/调试的示例 py 文件.
2. 或者将其设为当前文件
3. 选择调试侧栏
4. 选择Python: Current File配置
5. 点击绿色"三角形.

现在构建并运行一个复杂的命令.它会将 CWD 设置为当前文件的目录并在当前文件上启动调试器.

---

如果您确实需要启动模块，您可以在 VSC 中使用 Multi-Root Workspace 并为每个模块单独配置一个 launch.json根目录

---

如果您想使用 ${command:commandID} 语法，您可以构造一个使用 activeTextEditor 的简单扩展，并根据文件名构造一个字符串

---

编辑

另一种选择是使用多启动配置.

我删除了默认值有效或未使用的属性.

<代码>{版本":0.2.0"，配置":[{"name": "练习",类型":蟒蛇"，请求":启动"，"console": "integratedTerminal","module": "exercises.${fileBasenameNoExtension}",},{"name": "Exercise.Easy",类型":蟒蛇"，请求":启动"，"console": "integratedTerminal","module": "exercises.easy.${fileBasenameNoExtension}",},{"name": "DS",类型":蟒蛇"，请求":启动"，"console": "integratedTerminal","module": "ds.${fileBasenameNoExtension}",}]}

---

您可以使用扩展名 Command Variable 来获取带有点分隔符的相对目录.

<代码>{版本":0.2.0"，配置":[{"name": "Python: 模块 CmdVar",类型":蟒蛇"，请求":启动"，"console": "integratedTerminal","module": "${command:extension.commandvariable.file.relativeDirDots}.${fileBasenameNoExtension}",}]}

I have the following directory structure for my python practice project:

.
├── data
├── ds-and-algo
├── exercises
|   ├── __init__.py
│   ├── armstrong_number.py
│   ├── extract_digits.py
├── output

The extract_digits.py looks something like this:

def extract_digits(n):
    pass

In the armstrong_number.py I have the following:

from .extract_digits import extract_digits

From root project directory if I run

python exercises/armstrong_number.py

I get ModuleNotFoundError: no module named exercises

Running the following commad with -m flag resolves the error:

python -m exercises.armstrong_number

However using VSCode in order to debug the file, I have the following debug config launch.json:

{
"version": "0.2.0",
    "configurations": [
        {
            "name": "Python Module",
            "type": "python",
            "request": "launch",
            "program": "${file}",
            "console": "integratedTerminal",
            "pythonPath": "${config:python.pythonPath}",
            "module": "exercises.${fileBasenameNoExtension}",
            "cwd": "${workspaceRoot}",
            "env": {"PYTHONPATH":"${workspaceRoot}"}
        }
    ]
}

However this has a few problems:

1) For a different folder, for e.g. ds-and-algo, I have to manually edit the module entry in the launch.json file debug configuration to

"module" : "ds-and-algo.${fileBaseNameNoExtension}"

In case I have nested folder configuration like:

exercises
├── tough
|   | __init__.py
|   ├──ex1.py
|   ├──ex2.py
├── easy

I again have to manually edit the debug config in launch.json file to: (considering the case for the sub-folder tough)

"module": "exercises.tough.${fileBaseNameNoExtension}"

I need to implement a general case where depending on the file being debugged, the "module" entry in the launch.json file should be:

"module": "folder1.folder2.folder3.....foldern.script"

Just like fileBaseNameNoExtension, VSCode has some other predefined variables:

One of the variables is relativeFile, which is the path of the current opened file relative to workspaceFolder So for the file ex1.py, the variable relativeFile will be exercises/tough/ex1.py.

I need to manipulate this string and convert this to exercises.tough.ex1, which is trivial if I can write and execute bash command inside the "module" entry in launch.json file. But I am unable to do that. However, the link predefined variables in VSCode has a section on Command variables, which states:

if the predefined variables from above are not sufficient, you can use any VS Code command as a variable through the ${command:commandID} syntax.

This link has a bunch of other information that may be helpful. I am no expert in python, and definitely don't know any javascript, if that is what is required to solve this problem.

解决方案

I can't reproduce your error: ModuleNotFoundError: no module named exercises.

Using the following files

exercises/extract_digits.py

def extract_digits(n):
  return 10

exercises/armstrong_number.py

from extract_digits import extract_digits

def armstrong_number(n):
  return extract_digits(n)

if __name__ == "__main__": 
  print armstrong_number(3)

If you use Python3 change the print statement to: print(armstrong_number(3)) and change the used Python interpreter.

If your current directory is the project root directory and you run

python exercises/armstrong_number.py

You get the number 10 in the console

---

In Visual Studio Code you use the wrong Launch configuration.

You are just running python programs so you should use the Python: Current File configuration. launch.json

{
  "version": "0.2.0",
  "configurations": [
    {
      "name": "Python: Current File",
      "type": "python",
      "request": "launch",
      "program": "${file}",
      "console": "integratedTerminal",
    }
  ]
}

1. Open the example py file you want to run/debug, in whatever directory it is.
2. Or make it the current file
3. Select the Debug Side Bar
4. Select the Python: Current File configuration
5. Click the "green" triangle.

Now a complicated command is constructed and run. It will set the CWD to the directory of the current file and start the debugger on the current file.

---

If you really need the module launch you can use a Multi-Root Workspace in VSC and have a separate configured launch.json for each of the root directories

---

If you want to use the ${command:commandID} syntax you can construct a simple extension that uses the activeTextEditor and construct a string based on the file name

---

Edit

Another option is to use Multiple Launch configurations.

I have removed the properties where the default value works or that are not used.

{
  "version": "0.2.0",
  "configurations": [
    {
      "name": "Exercise",
      "type": "python",
      "request": "launch",
      "console": "integratedTerminal",
      "module": "exercises.${fileBasenameNoExtension}",
    },
    {
      "name": "Exercise.Easy",
      "type": "python",
      "request": "launch",
      "console": "integratedTerminal",
      "module": "exercises.easy.${fileBasenameNoExtension}",
    },
    {
      "name": "DS",
      "type": "python",
      "request": "launch",
      "console": "integratedTerminal",
      "module": "ds.${fileBasenameNoExtension}",
    }
  ]
}

---

You can use the extension Command Variable to get this relative directory with dot separator.

{
  "version": "0.2.0",
  "configurations": [
    {
      "name": "Python: Module CmdVar",
      "type": "python",
      "request": "launch",
      "console": "integratedTerminal",
      "module": "${command:extension.commandvariable.file.relativeDirDots}.${fileBasenameNoExtension}",
    }
  ]
}

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