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获取%AppData%的路径返回错误的值(Getting path to %AppData% returns wrong value)

我有一个dotnet核心应用程序,我想获得本地%Appdata%目录的路径。 dotnet核心应用程序是从之前安装的Windows服务启动的。 我使用了这个问题所描述的方法,但它们都回到了错误的道路上。

代替

C:\ Users \用户MYUSER \应用程序数据\漫游\ MyApplication的\ file.txt的

我明白了

C:\ WINDOWS \ ServiceProfiles \本地服务\应用程序数据\漫游\ MyApplication的\ file.txt的

1. Environment.SpecialFolder.ApplicationData

Path.Combine(Environment.GetFolderPath(Environment.SpecialFolder.ApplicationData), "MyApplication", "file.txt")

2. Environment.GetEnvironmentVariable(“APPDATA”)

Path.Combine(Environment.GetEnvironmentVariable("APPDATA"), "MyApplication", "file.txt")

3. Environment.ExpandEnvironmentVariables(“%AppData%”)

Path.Combine(Environment.ExpandEnvironmentVariables("%AppData%"), "MyApplication", "file.txt")

所有方法都遗憾地回到了错误的道路上。 在使用Windows服务启动应用程序时,如何获得正确的%AppData%路径?


I have a dotnet core application and I want to get the path to the local %Appdata% directory. The dotnet core application is started from a Windows Service installed previously. I used the methods decribed under this question but they all return the wrong path.

Instead of

C:\Users\MyUser\AppData\Roaming\MyApplication\file.txt

I get

C:\WINDOWS\ServiceProfiles\LocalService\AppData\Roaming\MyApplication\file.txt

1. Environment.SpecialFolder.ApplicationData

Path.Combine(Environment.GetFolderPath(Environment.SpecialFolder.ApplicationData), "MyApplication", "file.txt")

2. Environment.GetEnvironmentVariable("APPDATA")

Path.Combine(Environment.GetEnvironmentVariable("APPDATA"), "MyApplication", "file.txt")

3. Environment.ExpandEnvironmentVariables("%AppData%")

Path.Combine(Environment.ExpandEnvironmentVariables("%AppData%"), "MyApplication", "file.txt")

All methods sadly returned the wrong path. How can I get the right %AppData% path, while the application is started with a Windows Service?


原文:https://stackoverflow.com/questions/50561685
更新时间:2024-01-24 17:01

最满意答案

可能有一种cumcount方法,但每组中的第一个元素是该组中的第0个元素,而cumcount为每个组中的元素编号。 所以:

In [19]: df.loc[df.groupby('series1').cumcount() == 0, 'series2'] = np.nan

In [20]: df
Out[20]: 
   series1  series2
0        A      NaN
1        A      1.0
2        A     10.0
3        A     99.0
4        B      NaN
5        B      9.0
6        B      0.0
7        C      NaN
8        C     20.0
9        C     10.0
10       C     10.0

There's probably a slicker way to do this, but the first element in each group is the 0th element in that group, and cumcount numbers the elements within each group. So:

In [19]: df.loc[df.groupby('series1').cumcount() == 0, 'series2'] = np.nan

In [20]: df
Out[20]: 
   series1  series2
0        A      NaN
1        A      1.0
2        A     10.0
3        A     99.0
4        B      NaN
5        B      9.0
6        B      0.0
7        C      NaN
8        C     20.0
9        C     10.0
10       C     10.0

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