I run my .fsx file like
>fsi A.fsx
In this file I read csv with CsvProvider that has to have path to csv data.
type Data = CsvProvider<"my_data.txt", ";", Schema
I need to pass file name as command line argument and it is possible
>fsi A.fsx my_data.txt
I can read it like
let originalPath = fsi.CommandLineArgs.ElementAt(1)
Problem is, that file name used in CsvProvider constructor needs to be constant and command line argument is not. How I can initialize CsvProvider from command line argument?
The value inside the angle brackes <"my_data.txt"...> specifies an example format file and is checked at compile time, hence the need for it to be a constant string. Assuming your .fsx script merely wants to load a different CSV file of the same general format, you would use
let contents = Data.Load(originalPath)
Related
I'm using Lua in Scite on Windows, but hopefully this is a general Lua question.
Let's say I want to write a temporary string content to a temporary file in Lua - which I want to be eventually read by another program, - and I tried using io.tmpfile():
mytmpfile = assert( io.tmpfile() )
mytmpfile:write( MYTMPTEXT )
mytmpfile:seek("set", 0) -- back to start
print("mytmpfile" .. mytmpfile .. "<<<")
mytmpfile:close()
I like io.tmpfile() because it is noted in https://www.lua.org/pil/21.3.html :
The tmpfile function returns a handle for a temporary file, open in read/write mode. That file is automatically removed (deleted) when your program ends.
However, when I try to print mytmpfile, I get:
C:\Users\ME/sciteLuaFunctions.lua:956: attempt to concatenate a FILE* value (global 'mytmpfile')
>Lua: error occurred while processing command
I got the explanation for that here Re: path for io.tmpfile() ?:
how do I get the path used to generate the temp file created by io.tmpfile()
You can't. The whole point of tmpfile is to give you a file handle without
giving you the file name to avoid race conditions.
And indeed, on some OSes, the file has no name.
So, it will not be possible for me to use the filename of the tmpfile in a command line that should be ran by the OS, as in:
f = io.popen("python myprog.py " .. mytmpfile)
So my questions are:
Would it be somehow possible to specify this tmpfile file handle as the input argument for the externally ran program/script, say in io.popen - instead of using the (non-existing) tmpfile filename?
If above is not possible, what is the next best option (in terms of not having to maintain it, i.e. not having to remember to delete the file) for opening a temporary file in Lua?
You can get a temp filename with os.tmpname.
local n = os.tmpname()
local f = io.open(n, 'w+b')
f:write(....)
f:close()
os.remove(n)
If your purpose is sending some data to a python script, you can also use 'w' mode in popen.
--lua
local f = io.popen(prog, 'w')
f:write(....)
#python
import sys
data = sys.stdin.readline()
How do I replace the new File method with a secure one? Is it possible to create a python script and connect it?
Part of the code where I have a problem:
def template Name = new File(file: "${template}").normalize.name.replace(".html", "").replace(".yaml", "")
But when I run my pipeline, I get the error
java.lang.SecurityException: Unable to find constructor: new java.io .File java.util.LinkedHashMap
This method is prohibited and is blacklisted. How do I replace it and with what?
If you're reading the contents of the file, you can replace that "new File" with "readFile".
See https://www.jenkins.io/doc/pipeline/steps/workflow-basic-steps/#readfile-read-file-from-workspace
readFile: Read file from workspace
Reads a file from a relative path (with root in current directory, usually > workspace) and returns its content as a plain string.
file : String
Relative (/-separated) path to file within a workspace to read.
encoding : String (optional)
The encoding to use when reading the file. If left blank, the platform default encoding will be used. Binary files can be read into a Base64-encoded string by specifying "Base64" as the encoding.
trying to deploy resource group but getting below error
Error: Unsupported argument
on auto.tfvars.tf line 1:
1: resource_group = "india1111"
An argument named "resource_group" is not expected here.
code
code attached
You could specfiy the variable values with file .auto.tfvars instead of auto.tfvars.tf.
For example, the content of the variables.tf file.
variable "resource_group" {
type = string
}
The content of the .auto.tfvars file.
resource_group = "trafff"
To specify variable values in a variable definitions file (with a filename ending in either .tfvars or .tfvars.json) and then specify that file on the command line with -var-file:
terraform apply -var-file="testing.tfvars"
Terraform also automatically loads a number of variable definitions files if they are present:
Files named exactly terraform.tfvars or terraform.tfvars.json.
Any files with names ending in .auto.tfvars or .auto.tfvars.json.
Ref: https://www.terraform.io/docs/language/values/variables.html#variable-definitions-tfvars-files
I am doing my project on incremental deep drawing using ABAQUS.
I am trying to import a text file of loop program into abaqus script so that there is no need of entering amplitude values manually.
But I am getting an error when trying to import the data using the following code
f = open('data_x', 'r')
values=f.read()
values=f.readline()
Error:
data_x is not defined
Error NameError: name 'data_x' is not defined points that you are using data_x as a name in your code, not as a string (with quotes).
This means that in your code, you probably have something like
f = open(data_x)
Python is trying to figure out which value is associated with data_x, which is a Python name, not a string. Since it's not defined before getting to that line, you are getting an error.
If you want to store the name of a file and then open a file, write
data_x = 'data_x.txt'
f = open(data_x)
You could also directly write
f = open('data_x.txt')
Whichever solution you adopt, make sure that a correct path to the file is passed to the function open, so that it could find the file.
I am having some trouble writing a file to a specific path taking the file name from excel. Here is the code which I am using
out_file = File.new (#temp_path/ "#{obj_info[3].to_s}","w")
"#{obj_info[3].to_s}" = sample.txt
The value sample.txt comes from Excel during run time
#temp_path = "C:/Users/Somefolder/"
The error displayed is:
NoMethodError: undefined method `[]' for nil:NilClass
However, if the code is:
out_file = File.new ("#{obj_info[3].to_s}","w")
it successfully creates a file called sample.txt in the default directory. However, I want it to be stored in a specific directory and the file name needs to be passed from Excel.
Any help is much appreciated.
I believe your problem is because there a space between / and "
#temp_path/ "#{obj_info[3].to_s}
and I guess you want to build a path.
My advice is that you use File.join
f_path = File.join(#temp_path,obj_info[3].to_s)
out_file = File.new (f_path,"w")
Let me know if that solved the problem
You have 2 problems:
obj_info is nil so you make an error reading the value from excel, the error you get indicates it is on an array, in the code you published the only thing that's an array is what you read from excel.
Print the contents with p obj_info right before your code to check.
#temp_path and {obj_info[3].to_s} need to be concatenated to make a path.
You can do that with File.join like Mauricio suggests or like this
out_file = File.new ("#{#temp_path}/#{obj_info[3]}","w")
You can drop the to_s in this case.
It would be better if you publish the whole of your script that is meaningful.