```JSON
{
  "model_name": "alexnet",
  "message": "Deep Learning FTW!"
}
```

## Parameter path berisi path

Misalkan terdapat *operasi path* dengan path `/files/{file_path}`.

Tetapi anda memerlukan `file_path` itu berisi *path*, seperti like `home/johndoe/myfile.txt`.

Sehingga URL untuk file tersebut akan seperti: `/files/home/johndoe/myfile.txt`.

### Dukungan OpenAPI

```JSON
{
  "model_name": "alexnet",
  "message": "Deep Learning FTW!"
}
```

## Pfad Parameter die Pfade enthalten

Angenommen, Sie haben eine *Pfadoperation* mit einem Pfad `/files/{file_path}`.

Aber `file_path` soll selbst einen *Pfad* enthalten, etwa `home/johndoe/myfile.txt`.

Sprich, die URL für diese Datei wäre etwas wie: `/files/home/johndoe/myfile.txt`.

### OpenAPI Unterstützung

  "message": "Deep Learning FTW!"
}
```

## Path parameters containing paths { #path-parameters-containing-paths }

Let's say you have a *path operation* with a path `/files/{file_path}`.

But you need `file_path` itself to contain a *path*, like `home/johndoe/myfile.txt`.

So, the URL for that file would be something like: `/files/home/johndoe/myfile.txt`.

### OpenAPI support { #openapi-support }

```JSON
{
  "model_name": "alexnet",
  "message": "Deep Learning FTW!"
}
```

## Parámetros de path conteniendo paths

Imaginemos que tienes una *path operation* con un path `/files/{file_path}`.

Pero necesitas que `file_path` en sí mismo contenga un *path*, como `home/johndoe/myfile.txt`.

Entonces, la URL para ese archivo sería algo como: `/files/home/johndoe/myfile.txt`.

### Soporte de OpenAPI

```JSON
{
  "model_name": "alexnet",
  "message": "Deep Learning FTW!"
}
```

## Yol İçeren Yol Parametreleri

Farz edelim ki elinizde `/files/{file_path}` isminde bir *yol operasyonu* var.

Fakat `file_path` değerinin `home/johndoe/myfile.txt` gibi bir *yol* barındırmasını istiyorsunuz.

Sonuç olarak, oluşturmak istediğin URL `/files/home/johndoe/myfile.txt` gibi bir şey olacaktır.

```JSON
{
  "model_name": "alexnet",
  "message": "Deep Learning FTW!"
}
```

## Path-параметри, що містять шляхи

Припустимо, у вас є *операція шляху* з маршрутом `/files/{file_path}`.

Але вам потрібно, щоб `file_path` містив *шлях*, наприклад `home/johndoe/myfile.txt`.

Отже, URL для цього файлу виглядатиме так: `/files/home/johndoe/myfile.txt`.



### Підтримка OpenAPI

						if name == "null" {
							name = v0
						}

						f := filemap[file][name]
						if f != "" {
							fn = f + ".data"
							err = os.MkdirAll(filepath.Dir(fn), os.ModePerm)
							if err != nil {
								fmt.Println("MkdirAll:", filepath.Dir(fn), err)
							}
							err = os.WriteFile(fn+".json", []byte(filemap[file][name+".json"]), os.ModePerm)
							combineFiles[name] = append(combineFiles[name], fn)

        responseData.setMethod(Constants.GET_METHOD);
        final String filePath = preprocessUri(uri);
        responseData.setUrl(filePath);

        SmbFile file = null;
        final SmbAuthentication smbAuthentication = smbAuthenticationHolder.get(filePath);
        if (logger.isDebugEnabled()) {
            logger.debug("Creating SmbFile: {} - {}", filePath, smbAuthentication);
        }
        try {

            throw new CrawlerSystemException("The uri is empty.");
        }

        String filePath = uri;
        if (!filePath.startsWith("file:")) {
            filePath = "file://" + filePath;
        }

        final StringBuilder buf = new StringBuilder(filePath.length() + 100);
        try {
            for (final char c : filePath.toCharArray()) {
                if (c == ' ') {
                    buf.append("%20");

        String filePath = directoryPath;
        if (!filePath.endsWith("/")) {
            filePath += "/";
        }
        filePath += fileInfo.getName();

        // Ensure proper SMB URL format
        if (!filePath.startsWith("smb://")) {
            if (filePath.startsWith("/")) {
                filePath = "smb://localhost" + filePath;
            } else {