嵌入python模块错误

发布于 2024-12-20 23:07:15 字数 833 浏览 1 评论 0原文

所以我有 ac 到 python 包装器，它接受输入字符串并将它们传递给 python 函数。我收到的错误是 python API 无法识别我的 python 文件...

PyObject *pName, *pModule, *pFunc;
QString pyFile="Test.py";
Py_Initialize();
pName = PyUnicode_FromString(pyFile.toAscii().data());
pModule = PyImport_Import(pName);

错误是“ImportError：没有名为 Test.py 的模块” 这是当我将 Test.py 放在与我的项目相同的目录中时

，当我将 Test.py 放在目录树中的上一层时，出现了另一个错误错误是“不支持按文件名导入”，

所以我猜绝对路径不起作用？但在我的示例中的第一种情况下，我明确地将 Test.py 放在与我的项目相同的目录中，为什么我会收到错误？蟒蛇代码是：

import sys
import os

def printFileClass(fileName, className):
    print ("The OMC CORBA File name is ", fileName,"\n")
    print ("The selected Modelica Class is ", className)
    return ("Done operations")


def main():
    print ("Hello! Here is testing script's main \n")



if __name__=='__main__':
main()

原文

So i have a c to python wrapper that takes input strings and pass them to a python function. the error im getting is that the python API is not recognizing my python file...

PyObject *pName, *pModule, *pFunc;
QString pyFile="Test.py";
Py_Initialize();
pName = PyUnicode_FromString(pyFile.toAscii().data());
pModule = PyImport_Import(pName);

error is "ImportError: No module named Test.py"
This is when i have my Test.py in the same directory as my project

when i placed my Test.py up one level in my directory tree, another error came up
error is "Import by filename is not supported"

so i guess absolute paths dont work? but in the first case in my example, i clearly placed my Test.py in the same directory as my project, why am i getting the error?
python code is:

import sys
import os

def printFileClass(fileName, className):
    print ("The OMC CORBA File name is ", fileName,"\n")
    print ("The selected Modelica Class is ", className)
    return ("Done operations")


def main():
    print ("Hello! Here is testing script's main \n")



if __name__=='__main__':
main()

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小矜持 2024-12-27 23:07:15

PYTHONPATH 环境变量可用于解决您的问题。

在代码中，您可以在 Py_Initialize() 之前的某个位置执行此操作：

setenv("PYTHONPATH", ".", 0); // #include <stdlib.h> to get the prototype

第三个参数 0 表示覆盖 - 它为零，因此您也可以从 shell 传递 PYTHONPATH。如果您想始终使用您编码的路径，可以将其设置为 1。

我不确定这不会给您带来其他问题，但对于简单的测试来说它是有效的。

另外，请勿在传递给 PyImport_Import 的模块名称中包含 .py 扩展名。

我在 Linux 系统上对此进行了测试。

The PYTHONPATH environment variable can be used to fix your problem.

In your code, you can do this somewhere before Py_Initialize():

setenv("PYTHONPATH", ".", 0); // #include <stdlib.h> to get the prototype

The third parameter, 0, means overwrite - it's zero so you can also pass PYTHONPATH from the shell. If you want to always use a path that you coded, you can set that to 1.

I'm not sure this doesn't expose you to other problems, but for a simple test it works.

Also, don't include the .py extension in the module name you pass to PyImport_Import.

I tested this on a Linux system.

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