我想我会通过将其分成两部分来解决这个问题。首先，我们需要一种方法来解析目录/文件路径数组并将其放入分层结构中。其次，我们需要采用该结构并将其转换为 JSON。 （从你的问题中我不能完全确定你想使用哪个序列化器，所以对于这个答案我会假设Json.Net http://james.newtonking.com/json is OK.)
对于第一部分，我将创建一个Dir类，它有一个名称、一个子目录字典（以便于查找）和一组文件。我们可以在此类中创建一个方法，它将解析路径并查找或添加适当的子对象。
class Dir
{
public string Name { get; set; }
public Dictionary<string, Dir> Dirs { get; set; }
public HashSet<string> Files { get; set; }
public Dir(string name)
{
Name = name;
Dirs = new Dictionary<string, Dir>();
Files = new HashSet<string>();
}
public Dir FindOrCreate(string path, bool mightBeFile = true)
{
int i = path.IndexOf('/');
if (i > -1)
{
Dir dir = FindOrCreate(path.Substring(0, i), false);
return dir.FindOrCreate(path.Substring(i + 1), true);
}
if (path == "") return this;
// if the name is at the end of a path and contains a "."
// we assume it is a file (unless it is "." by itself)
if (mightBeFile && path != "." && path.Contains("."))
{
Files.Add(path);
return this;
}
Dir child;
if (Dirs.ContainsKey(path))
{
child = Dirs[path];
}
else
{
child = new Dir(path);
Dirs.Add(path, child);
}
return child;
}
}
使用这个类，我们可以轻松地循环dirArray在你的问题中给出并建立目录层次结构：
Dir root = new Dir("");
foreach (string dir in dirArray)
{
root.FindOrCreate(dir);
}
所以此时，root现在拥有整个目录层次结构。如果您愿意，可以直接使用 Json.Net 序列化该对象，以获得合理的 JSON 结构。但是，它会比您在问题中描述的更详细一些。以下是将生成的 JSON：
{
"Name": "",
"Dirs": {
".": {
"Name": ".",
"Dirs": {
"proc": {
"Name": "proc",
"Dirs": {
"15": {
"Name": "15",
"Dirs": {
"task": {
"Name": "task",
"Dirs": {
"15": {
"Name": "15",
"Dirs": {
"exe": {
"Name": "exe",
"Dirs": {},
"Files": []
},
"mounts": {
"Name": "mounts",
"Dirs": {},
"Files": [
"mounts.xml"
]
},
"mountinfo": {
"Name": "mountinfo",
"Dirs": {},
"Files": [
"mountinfo.xml",
"moremountinfo.xml"
]
},
"clear_refs": {
"Name": "clear_refs",
"Dirs": {},
"Files": [
"clear_ref.xml"
]
}
},
"Files": []
}
},
"Files": []
}
},
"Files": []
},
"14": {
"Name": "14",
"Dirs": {
"loginuid": {
"Name": "loginuid",
"Dirs": {},
"Files": [
"loginuid.xml"
]
},
"sessionid": {
"Name": "sessionid",
"Dirs": {},
"Files": [
"sessionid.xml"
]
},
"coredump_filter": {
"Name": "coredump_filter",
"Dirs": {},
"Files": [
"coredump_filter.xml"
]
},
"io": {
"Name": "io",
"Dirs": {},
"Files": [
"io.xml"
]
}
},
"Files": []
}
},
"Files": []
}
},
"Files": []
}
},
"Files": []
}
为了获得您想要的 JSON，我们需要一个JsonConverter class:
class DirConverter : JsonConverter
{
public override bool CanConvert(Type objectType)
{
return (objectType == typeof(Dir));
}
public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
{
Dir dir = (Dir)value;
JObject obj = new JObject();
if (dir.Files.Count > 0)
{
JArray files = new JArray();
foreach (string name in dir.Files)
{
files.Add(new JValue(name));
}
obj.Add("list_of_files", files);
}
foreach (var kvp in dir.Dirs)
{
obj.Add(kvp.Key, JToken.FromObject(kvp.Value, serializer));
}
obj.WriteTo(writer);
}
public override bool CanRead
{
get { return false; }
}
public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
{
throw new NotImplementedException();
}
}
我们可以使用转换器序列化目录层次结构，如下所示：
JsonSerializerSettings settings = new JsonSerializerSettings();
settings.Converters.Add(new DirConverter());
settings.Formatting = Formatting.Indented;
string json = JsonConvert.SerializeObject(root, settings);
这是输出。请注意，我将原始 JSON 中的“文件”属性更改为数组，并将其重命名为“list_of_files”，以根据您的评论容纳每个目录多个文件的可能性。我还假设永远不会有一个名为“list_of_files”的实际目录。如果可能的话，您需要将文件数组的名称更改为不会与任何目录名称冲突的其他名称。 （如果您遇到错误“Can not add property list_of_files to Newtonsoft.Json.Linq.JObject. Property with the same name already exists on object“这意味着您的数据中有一个名为“list_of_files”的目录。）
{
".": {
"proc": {
"15": {
"task": {
"15": {
"exe": {},
"mounts": {
"list_of_files": [
"mounts.xml"
]
},
"mountinfo": {
"list_of_files": [
"mountinfo.xml"
]
},
"clear_refs": {
"list_of_files": [
"clear_ref.xml"
]
}
}
}
},
"14": {
"loginuid": {
"list_of_files": [
"loginuid.xml"
]
},
"sessionid": {
"list_of_files": [
"sessionid.xml"
]
},
"coredump_filter": {
"list_of_files": [
"coredump_filter.xml"
]
},
"io": {
"list_of_files": [
"io.xml"
]
}
}
}
}
}
Fiddle: https://dotnetfiddle.net/ConJiu https://dotnetfiddle.net/ConJiu
