I have a simple .csv file.
Is it possible to convert it to .xls using the command line tool ssconvert?
I would also need to specify the name of the sheet.
ln -s input.csv MySheetName
ssconvert MySheetName output.xls
The OP asked how to convert csv to xls while controlling the sheet name in the output.
The generated .xls file will use the name of the input CSV file as the sheet name, so you can symlink the .csv to anything you want (or rename the input file) to produce the desired result.
The previous answer implies that --list-exporters leads to a solution, but it merely lists exporter names with no information about their options, and no options are documented in the man page for xls-exporters. Experimentally, none of the exporters which can create .xls accept options (they fail with "The file saver does not take options" if you use -O).
Yes, it is possible.
You must specify names with extensions as input and output files.
For example:
ssconvert in.csv out.xls
Using --list-importers and --list-exporters options can take a look to available formats.
Related
i have a problem, i used "everything" to extract every txt file from a specific directory so that i can merge them. But on emeditor i don't find a way to merge file from a list of localisation.
Here what the everything file look like:
E:\Main directory\subdirectory 1\file.txt
E:\Main directory\subdirectory 2\file.txt
E:\Main directory\subdirectory 3\file.txt
E:\Main directory\subdirectory 4\file.txt
The list goes over 40k location. is there a way to use a program to read all the location in the text file and combine them ?
Also, the subdirectory has other txt file that i don't want to so i can't just merge all txt file from the main. Another thing is that there are variation of the "file.txt" like "Files.txt" for example.
Oracle's sqlldr defaults to a .dat extension. That I want to override. I don't like to rename the file. When googled get to know few answers to use . like data='fileName.' which is not working. Share your ideas, please.
Error message is fileName.dat is not found.
Sqlloder has default extension for all input files data,log,control...
data= .dat
log= .log
control = .ctl
bad =.bad
PARFILE = .par
But you have to pass filename without apostrophe and dot
sqlloder pass/user#db control=control data=data
sqloader will add extension. control.ctl data.dat
Nevertheless i do not understand why you do not want to specify extension?
You can't, at least in Unix/Linux environments. In Windows you can use the trailing period trick, specifying either INFILE 'filename.' in the control file or DATA=filename. on the command line. WIndows file name handling allows that; you can for instance do DIR filename. at a command prompt and it will list the file with no extension (as will DIR filename). But you can't do that with *nix, from a shell prompt or anywhere else.
You said you don't want to copy or rename the file. Temporarily renaming it might be the simplest solution, but as you may have a reason not to do that even briefly you could instead create a hard or soft link to the file which does have an extension, and use that link as the target instead. You could wrap that in a shell script that takes the file name argument:
# set variable from correct positional parameter; if you pass in the control
# file name or other options, this might not be $1 so adjust as needed
# if the tmeproary file won't be int he same directory, need to be full path
filename=$1
# optionally check file exists, is readable, etc. but overkill for demo
# can also check temporary file does not already exist - stop or remove
# create soft link somewhere it won't impact any other processes
ln -s ${filename} /tmp/${filename##*/}.dat
# run SQL*Loader with soft link as target
sqlldr user/password#db control=file.ctl data=/tmp/${filename##*/}.dat
# clean up
rm -f /tmp/${filename##*/}.dat
You can then call that as:
./scriptfile.sh /path/to/filename
If you can create the link in the same directory then you only need to pass the file, but if it's somewhere else - which may be necessary depending on why renaming isn't an option, and desirable either way - then you need to pass the full path of the data file so the link works. (If the temporary file will be int he same filesystem you could use a hard link, and you wouldn't have to pass the full path then either, but it's still cleaner to do so).
As you haven't shown your current command line options you may have to adjust that to take into account anything else you currently specify there rather than in the control file, particularly which positional argument is actually the data file path.
I have the same issue. I get a monthly download of reference data used in medical application and the 485 downloaded files don't have file extensions (#2gb). Unless I can load without file extensions I have to copy the files with .dat and load from there.
Suppose i have a folder structure which has a path like this..
C:\Fold1\Fold2\Fold3\Sample.xls
I want to display the File name i.e Sample.xls in a separate text file along with the Last folder name. That is Fold3
Output in Text file should be like this..
Fold 3
Sample.xls
Please need the code ASAP...
I would do a pure Dir /B in a temp file and then use some awk or grep solution to extract the part I need with regular expressions.
There are many free tools to allow this (GNU for Windows).
I want to write an asciidoc document and convert it into a pdf document. However, I want to use a format style different than the default ones. To do so I convert the txt file to docbook using asciidoc and then try to convert the resulting docbook xml to a pdf file using dblatex.
The idea is to set a particular tex style for dblatex to obtain the desired pdf result. I've copied the existing docbook.sty style as it is recommended here to do a small style modification. The only change done in the ./docbook file is \setlength{\textwidth}{18cm} to \setlength{\textwidth}{12cm}. However, when I run the command
dblatex --texstyle=./docbook.sty test.txt
Or the command
dblatex -s ./docbook.sty test.txt
Both produce the same result in the style change: none. I mean, no matter which modification I do to ./docbook.sty file, these modifications are not applied to the output. I obtain always the same result, a pdf with the default formatting. Do you guys have any idea where is the problem?
Thanks in advance.
I would recommend:
Copy the Dblatex docbook.sty to a new filename in your working directory which is "obviously yours" (e.g., mydbstyle.sty).
Continue to supply a full or relative path argument to the --texstyle option (e.g., /path/to/mydbstyle.sty or ./mydbstyle.sty). Failing to do so requires that mydbstyle.sty be in a directory enumerated by the TEXINPUTS environment variable (which you likely have not explicitly set).
Within mydbstyle.sty, use the following directives to initialize your style:
\NeedsTeXFormat{LaTeX2e}
\ProvidesPackage{mydbstyle}[2013/02/15 DocBook Style]
\RequirePackageWithOptions{docbook}
% ...
% your LaTeX commands here
Pass a DocBook 4.5 XML file as an argument to Dblatex (in your example you are passing test.txt which makes me uncertain whether you're passing an AsciiDoc source file).
dblatex --texstyle=./mydbstyle.sty mybook.xml
I would like to scan text of textfiles in Matlab with the textscan function. Before I can open the textfile with fid = fopen('C:\path'), I need to unzip the files first. The files have the extension: *.gz
There are thousands of files which I need to analyze and high performance is important.
I have two ideas:
(1) Use an external program an call it from the command line in Matlab
(2) Use a Matlab 'zip'toolbox. I have heard of gunzip, but don't know about its performance.
Does anyone knows a way to unzip these files as quick as possible from within Matlab?
Thanks!
You could always try the Matlab unzip() function:
unzip
Extract contents of zip file
Syntax
unzip(zipfilename)
unzip(zipfilename, outputdir)
unzip(url, ...)
filenames = unzip(...)
Description
unzip(zipfilename) extracts the archived contents of zipfilename into the current folder and sets the files' attributes, preserving the timestamps. It overwrites any existing files with the same names as those in the archive if the existing files' attributes and ownerships permit it. For example, files from rerunning unzip on the same zip filename do not overwrite any of those files that have a read-only attribute; instead, unzip issues a warning for such files.
Internally, this uses Java's zip library org.apache.tools.zip. If your zip archives each contain many text files it might be faster to drop down into Java and extract them entry by entry, without explicitly unzipped files. look at the source of unzip.m to get some ideas, and also the Java documentation.
I've found 7zip-commandline(Windows) / p7zip(Unix) to be somewhat speedier for this.
[edit]From some quick testing, it seems making a system call to gunzip is faster than using MATLAB's native gunzip. You could give that a try as well.
Just write a new function that imitates basic MATLAB gunzip functionality:
function [] = sunzip(fullfilename,output_dir)
if ~exist('output_dir','var'), output_dir = fileparts(fullfilename); end
app_path = '/usr/bin/7za';
switches = ' e'; %extract files ignoring directory structure
options = [' -o' output_dir];
system([app_path switches options '_' fullfilename]);
Then use it as you would use gunzip:
sunzip('/data/time_1000.out.gz',tmp_dir);
With MATLAB's toc timer, I get the following extraction times with 6 uncompressed 114MB ASCII files:
gunzip: 10.15s
sunzip: 7.84s
worked well, just needed a minor change to Max's syntax calling the executable.
system([app_path switches ' ' fullfilename options ]);