I have a robot to find a file of the given name at a particular location in a system but now I want to find all the text files at that particular location. I have tried to use "*.txt", but it didn't worked out. Is there a way to do that?
file.exists ♥environment⟦USERPROFILE⟧\Documents\t.txt errormessage ‴Sorry, I could not find a file‴
dialog ‴File exists‴
You can use the directory command. The pattern arguments allows you to filter out files of a particular extension.
directory path ♥environment⟦USERPROFILE⟧\Desktop pattern *.txt result ♥files
dialog ♥files⟦count⟧
The above code should let you know how many files of the given extension exist in the given directory.
You could take values from the returned list and use it with file.exists command.
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 am executing a Pig script, which reads files from a directory, performs some operation and stores to some output directory. In output directory I'm getting one or more "part" files, one _SUCCESS file and one _logs directory. My questions are:
Is there any way to control the name of files generated (upon execution of STORE command) in output directory. To be specific, I don't want the names to be "part-.......". I want Pig to generate files according to the file name pattern I specify.
Is there any way to suppress the _SUCCESS file and the _log directory? Basically I don't want the _SUCCESS and _logs to be generated in the output directory.
Regards
Biswajit
See this post.
To remove _SUCCESS, use SET mapreduce.fileoutputcommitter.marksuccessfuljobs false;. I'm not 100% sure how to remove _logs but you could try SET pig.streaming.log.persist false;.
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.
I would like to parse the file location information in an M3U playlist into fully qualified paths. The possible formats in M3U files seem to be:
c:\mydir\songs\tune.mp3
\songs\tune.mp3
..\songs\tune.mp3
For the first example, just leave it alone. For the second add the directory that the playlist resides in so it would become c:\playlists\songs\tune.mp3 and the same for the third case so it would also become: c:\playlists\songs\tune.mp3.
I'm using vb under VS2008 and I can't find a way to recognise each of the potential location formats in the M3U file. System.IO.Path offers no solution that I can find. I've searched extensively for terms like "convert relative path to absolute" but no luck.
Any advice appreciated.
Thanks.
Write a batch script that just reads the m3u file line by line, and then just parse each line looking for ":" , and for "..", and edit the string as needed. You can then just write the "converted" strings to another file...
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