Re: [racket-users] Working out which directory the current code was loaded from?

[Peter W A Wood]

Many thanks Philip

The resulting code is much neater and, to me, more readable:

(define-runtime-path foo.txt "foo.txt")
(define-runtime-path bar.txt "bar.txt")
(define-runtime-path outfile.txt "outfile.txt")
(define-runtime-path Data/portfolio.csv "../../Data/portfolio.csv")
(define-runtime-path Data/portfolio.csv.gz "../../Data/portfolio.csv.gz”)

Peter

> On 28 Jul 2020, at 09:40, Philip McGrath  wrote:
> 
> For this particular purpose, you want `define-runtime-path`: 
> https://docs.racket-lang.org/reference/Filesystem.html#%28part._runtime-path%29
>  
> 
> 
> -Philip
> 
> 
> On Mon, Jul 27, 2020 at 9:38 PM Peter W A Wood  > wrote:
> I have a short racket program/script that reads a file from the directory in 
> which it is stored. The directory structure is something like this:
> 
> a/
> b/
> c/
> my-racket.rkt 
> my-data-file.txt
> 
> I want to be able to run the program from the command line no matter what is 
> the current working directory. E.G.:
> 
> a> racket b/c/my-racket.rkt
> a/b> racket c/my-racket.rkt
> a/b/c> racket my-racket.rkt
> 
> In order to do so, I need to provide the correct path to my-data-file.txt 
> depending on from where the script was launched. I haven’t learnt about 
> Racket modules yet so I resorted to searching Stack Overflow. I found a code 
> snippet that I used which worked:
> 
> (define script-dir (path-only (resolved-module-path-name
>   (variable-reference->resolved-module-path
>(#%variable-reference)
> 
> Is this the best way to ascertain the directory of the “current module”?
> 
> Thanks in advance
> 
> Peter 
> 
> 
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>  
> .

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Re: [racket-users] Working out which directory the current code was loaded from?

[Philip McGrath]

For this particular purpose, you want `define-runtime-path`:
https://docs.racket-lang.org/reference/Filesystem.html#%28part._runtime-path%29

-Philip


On Mon, Jul 27, 2020 at 9:38 PM Peter W A Wood 
wrote:

> I have a short racket program/script that reads a file from the directory
> in which it is stored. The directory structure is something like this:
>
> a/
> b/
> c/
> my-racket.rkt
> my-data-file.txt
>
> I want to be able to run the program from the command line no matter what
> is the current working directory. E.G.:
>
> a> racket b/c/my-racket.rkt
> a/b> racket c/my-racket.rkt
> a/b/c> racket my-racket.rkt
>
> In order to do so, I need to provide the correct path to my-data-file.txt
> depending on from where the script was launched. I haven’t learnt about
> Racket modules yet so I resorted to searching Stack Overflow. I found a
> code snippet that I used which worked:
>
> (define script-dir (path-only (resolved-module-path-name
>   (variable-reference->resolved-module-path
>(#%variable-reference)
>
> Is this the best way to ascertain the directory of the “current module”?
>
> Thanks in advance
>
> Peter
>
>
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> "Racket Users" group.
> To unsubscribe from this group and stop receiving emails from it, send an
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> To view this discussion on the web visit
> https://groups.google.com/d/msgid/racket-users/92DACE01-60C8-445A-A07E-A4E6A6F5F684%40gmail.com
> .
>

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[racket-users] Working out which directory the current code was loaded from?

[Peter W A Wood]

I have a short racket program/script that reads a file from the directory in 
which it is stored. The directory structure is something like this:

a/
b/
c/
my-racket.rkt 
my-data-file.txt

I want to be able to run the program from the command line no matter what is 
the current working directory. E.G.:

a> racket b/c/my-racket.rkt
a/b> racket c/my-racket.rkt
a/b/c> racket my-racket.rkt

In order to do so, I need to provide the correct path to my-data-file.txt 
depending on from where the script was launched. I haven’t learnt about Racket 
modules yet so I resorted to searching Stack Overflow. I found a code snippet 
that I used which worked:

(define script-dir (path-only (resolved-module-path-name
  (variable-reference->resolved-module-path
   (#%variable-reference)

Is this the best way to ascertain the directory of the “current module”?

Thanks in advance

Peter 


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