I'm relatively new to Netlogo and already struggling ;)
I have the following problem: I want my program to open a folder, check a file in that folder and afterwards remove that file from that folder. I figured the best way to do this is via a while loop, but I'm struggling to find the right syntax. Hope you all can help!
The command 'file-open' will open a file using the path provided (the string after file-open: e.g. file-open "C:\Documents\model-out.txt" will open a file titled model-out.txt in the Documents folder on the C drive.)
You can then use 'file-read' or 'file-write' to read or write to the file respectively.
The command 'file-close' will close the file, which then can be deleted with 'file-delete'.
You can also check if a file exists in a folder using the command if file-exists? "C:\Documents\model-out.txt", and if true, the file can be deleted using file-delete.
Also check the command 'set-current-directory'.
Best,
Related
For example, I want to open a PDF file in the browser from the command line (just because it's much faster and I need to open many files at once) and when I use the command start [file name] from its directory it try to open it as a executable, so I need to open the browser and type the full path of the file as an attribute, is there a way to call the full path without typing it?
what I exactly need is I need the full path of a file to convert it to string (for example in the browser)
Using tab completions may help. For example, if your target file is named thisPDFisTotallyBananas.pdf and you have another file in the same folder named thisOtherPDFisNot.pdf, you could type thisP then TAB to complete the file name in the command prompt without needing to type the whole filename.
I have received a file from a customer. The file is said to be
SQL code (application/sql)
However, this has turned out to be wrong: nothing could open it. It turns out it was secretely a .zip file. By renaming it to '.zip' and manually extracting it I was able to get the files contained in it. I would like to do a similar process in python.
So far I've renamed the file:
file_name_zip = file_name.replace('.sql', '.zip')
os.rename(file_name, file_name_zip)
And I've tried extracting it:
zip_ref = zipfile.ZipFile(file_name_zip, 'r')
zip_ref.extractall(extracted_file)
However, this failed because
zipfile.BadZipFile: File is not a zip file
I've googled, and apparently this can sometimes be fixed using:
zip_file_name_2 = zip_file_name.replace('.zip', '2.zip')
os.system(f'zip -FF {zip_file_name} --out {zip_file_name_2}')
This required me to put in a bunch of settings, which I wasn't able to figure out. There must be a better way to go about this.
Does anybody know how to parse such an .sql file?
From the screenshot below: I did a command-C copy of a scala source file - seen in the screenshot as BpmSpecs . Then went to the containing folder eventapi , right clicked and selected paste.
Well .. Intellij did a rather boneheaded thing here: it created a new directory called .. wait for it .. LocalEngineCycles.scala .. and proceeded to copy the original file into that directory under the same name as the original file.
So .. how to copy a scala source file into the same directory - and having intelllij actually help us by allowing us to supply the new filename ?
Re: possible duplicate: it is not. This problem relates to a File. I clone classes successfully all the time: IJ asks me to change the ClassName and I do it. For the file it gets confused and instead creates a new directory and places a copy of the original file inside this new directory.
is it possible to load module from file with extension other than .lua?
require("grid.txt") results in:
module 'grid.txt' not found:
no field package.preload['grid.txt']
no file './grid/txt.lua'
no file '/usr/local/share/lua/5.1/grid/txt.lua'
no file '/usr/local/share/lua/5.1/grid/txt/init.lua'
no file '/usr/local/lib/lua/5.1/grid/txt.lua'
no file '/usr/local/lib/lua/5.1/grid/txt/init.lua'
no file './grid/txt.so'
no file '/usr/local/lib/lua/5.1/grid/txt.so'
no file '/usr/local/lib/lua/5.1/loadall.so'
no file './grid.so'
no file '/usr/local/lib/lua/5.1/grid.so'
no file '/usr/local/lib/lua/5.1/loadall.so'
I suspect that it's somehow possible to load the script into package.preaload['grid.txt'] (whatever that is) before calling require?
It depends on what you mean by load.
If you want to execute the code in a file named grid.txt in the current directory, then just do dofile"grid.txt". If grid.txt is in a different directory, give a path to it.
If you want to use the path search that require performs, then add a template for .txt in package.path, with the correct path and then do require"grid". Note the absence of suffix: require loads modules identified by names, not by paths.
If you want require("grid.txt") to work should someone try that then yes, you'll need to manually loadfile and run the script and put whatever it returns (or whatever require is documented to return when the module doesn't return anything) into package.loaded["grid.txt"].
Alternatively, you could write your own loader just for entries like this which you set into package.preload["grid.txt"] which finds and loads/runs the file or, more generically, you could write yourself a loader function, insert it into package.loaders, and then let it do its job whenever it sees a "*.txt" module come its way.
By default AutoCAD installs a text based file called acad2010.lsp at the set location below
Dim FILE_NAME As String = "C:\Program Files\AutoCAD 2010\Support\acad2010.lsp"
However it my be that the user/ administrator/ or third party has changed the location of this file. Is it possible to then locate it using the following
Dim FILE_NAME As String = "C:\*\acad2010.lsp"
In other words search the entire c:\ drive for file acad2010.lsp?
If this doesn't work can you please let me know what would?
You could search for it with an FSO. It's not going to be fast however you do it but this is the fastest way I can think of.
http://www.microbion.co.uk/developers/fso.htm should give you a rough idea of how it's done.
Your solution will not work. Is not possible to locate it using *. (BTW is possible in ms-builds scripts). The only way of doing it is:
1- Create a FindFile function (check for example
http://xlvba.3.forumer.com/index.php?showtopic=125)
2- Use it to locate the exact path of the file. (It could be really time
consuming)
3- From this point your code is the same...
Unfortunately, you can't use wildcards in a filepath. You have two options:
Prompt the user for the file location using the "Open File" dialog. The code to do this varies based on which Office product you are using. In Excel, you would use the Application.FindFile method (more info here).
Write your own function to search the filesystem for the file. Microsoft provides an example here.
If that file is used by internal functions of the application, the installer will have recorded a registry key for the file's location.
Open regedit.exe and search for the file name and path.
You can read a registry entry using this VBA one-liner:
CreateObject("WScript.Shell").RegRead(strRegPath)
You may need a terminating backslash on the key address, but that's a safe and simple registry access method. More details on the MSDN site:
https://msdn.microsoft.com/en-us/library/x05fawxd%28v=vs.84%29.aspx