C#端点以JSON形式返回目录和文件的嵌套列表 [英] C# endpoint giving back a nested list of directories and files as JSON
问题描述
我的目标是实现一个端点,该端点从某个根目录(例如,根目录)开始返回所有文件和目录的嵌套列表. C:\Temp
.我编写了以下代码:
My goal is to implement an endpoint which gives back a nested list of all files and directories starting from a certain root directory, e.g. C:\Temp
. I have written the following code:
namespace API.Controllers
{
public class UploadController : BaseController
{
[Route("api/Uploaded", Order = -1)]
[ResponseType(typeof(IEnumerable<string[]>))] // <- this has to be adjusted, I guess.
public IHttpActionResult AutoUpload()
{
string[] entries = Directory.GetFileSystemEntries("C:\Temp", "*", SearchOption.AllDirectories);
// <-- Here should come some conversion to a nested JSON.
return Ok(entries);
}
}
}
当我查询端点时,例如使用curl -X POST --header 'Accept: application/json' 'http://localhost:63291/api/Uploaded'
时,响应是某种情况
When I am querying the endpoint, e.g. with curl -X POST --header 'Accept: application/json' 'http://localhost:63291/api/Uploaded'
the response is something
[
"C:\\\\Temp\\file1",
"C:\\\\Temp\\dir1\\file2",
"C:\\\\Temp\\dir1\\file3"
]
我想拥有的东西类似于以下内容
What I would rather like to have is something like the following
[
{ "~":
[ "file1" ]
},
{
"~/dir1":
[
"file2" ,
"file3"
]
}
]
我确信将列表转换为嵌套JSON不会那么复杂,但是我不知怎么做.我担心,我缺少适当的搜索词.请帮忙!
I am sure it cannot be so complicated to convert my list into a nested JSON, but I somehow do not manage to do it. I fear, I am missing the appropriate search terms. Please help!
推荐答案
首先属性 [ResponseType]
不会影响您的响应,仅仅是元数据,例如大张旗鼓地用来记录您的端点.
First of all Attribute [ResponseType]
does not affect your response, it's just metadata that is used by for example swagger to document your endpoints.
Directory.GetFileSystemEntries
如Greg所说,返回表示路径的集合(数组)od字符串.您需要将结果解析为所需的格式,以便让我看到类似于File
类和包含File
的Directory
之类的东西,但更好的方法是使用诸如FileInfo
的内建类型.
Directory.GetFileSystemEntries
as Greg said returns collection (array) od strings representing paths. You need to parse your results to format you want i see that is something like File
class and Directory
containing File
but better approach it would be to use build in types like FileInfo
.
DirectoryInfo
类提供了一些可以使用的有用方法:
DirectoryInfo
class provides some useful methods that you could use:
Get all files in directory: https://docs.microsoft.com/en-us/dotnet/api/system.io.directoryinfo.getfiles?view=netframework-4.8
获取目录中的所有目录: https://docs. microsoft.com/en-us/dotnet/api/system.io.directoryinfo.getdirectories?view=netframework-4.8
Get all directories in directory: https://docs.microsoft.com/en-us/dotnet/api/system.io.directoryinfo.getdirectories?view=netframework-4.8
Or also get both: https://docs.microsoft.com/en-us/dotnet/api/system.io.directoryinfo.getfilesysteminfos?view=netframework-4.8
了解该方法返回的数据类型并充分利用它;)
Read about data types returned by that methods and use the best of it ;)
我认为您需要进行一些递归才能遍历所有子目录.
I think you will need to do some recursion to go through all subdirectories.
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