I want to make make cab file and the cab file should contain more then one files. (eg:- test.dll,ATL80.dll,Microsoft.VC80.ATL.manifest etc).
I tried below command
makecab Test.dll Test.cab
it works fine and adds only test.dll in the cab file.
How can we add more then one file in the cab file.
Is there any specific command or process to add more then one file in the cab.
Please note that i do not want to use cabarc.exe to do this.
Thank you.
It is possible using makecab and a .ddf file
1. Create a ddf file with your files in it (make.ddf):
;*** Sample Source Code MakeCAB Directive file example
;
.OPTION EXPLICIT ; Generate errors
.Set CabinetNameTemplate=ThyName.cab
.Set Cabinet=on
.Set Compress=on
"test.dll"
"ATL80.dll"
"Microsoft.VC80.ATL.manifest"
;*** <the end>
run the makecab:
makecab.exe /F make.ddf
Related
I am new to programming and I am running into a issue. I am calling a table and need to put my results into a csv file in a certain path.
This is what I am doing and the error I get.
dbuser#cbos1:/var/lib/dbspace/bosarc/testing/Abe_Lincoln> cd dbaccess labor32<<?
> UNLOAD TO '/var/lib/dbspace/bosarc/Active_Sites/Cronos_test/Position7'
> select * from informix.position;
> ?
-bash: cd: dbaccess: No such file or directory
dbuser#cbos1:/var/lib/dbspace/bosarc/testing/Abe_Lincoln>
the file path exist but keeps getting message.
Using just $ as the command line prompt, you should be using just:
$ dbaccess labor32 <<?
> UNLOAD TO '/var/lib/dbspace/bosarc/Active_Sites/Cronos_test/Position7'
> select * from informix.position;
> ?
…message(s) from dbaccess
$
This will run the dbaccess program (usually from $INFORMIXDIR/bin) against the database labor32, and generate an UNLOAD format file in the given file name.
The cd command is for changing directory; you don't have a directory called dbaccess (and probably shouldn't), and even if you did have such a directory, you shouldn't provide more options to the cd command, or a here document as standard input — it will ignore them.
Note that the file generated (Position7 will be the base name of the file) will be in Informix's UNLOAD format (pipe delimited fields by default), not CSV. It's certainly possible to convert between the two; I have Perl scripts that can do the conversions — last modified about a decade ago, but not much has changed in the interim. You could also consider using SQLCMD (available as open source from the IIUG Software Repository) which does have support for CSV load and unload formats. (This is the original SQLCMD — or at least an original SQLCMD — and is not Microsoft's Johnny-come-lately program of the same name.)
Create a file unload-table.sh containing:
#!/bin/sh
dbaccess labor32 <<EOF
UNLOAD TO '/var/lib/dbspace/bosarc/Active_Sites/Cronos_test/Position7'
SELECT * FROM informix.position;
EOF
You can then run this as bash unload-table.sh, or make it executable and install it in your $HOME/bin directory (which is on your PATH, isn't it?) so that you can simply run unload-table.sh. Or you can arrange to 'compile' (copy) the file to unload-table (no .sh suffix) so you don't have to type it to execute it: unload-table. You can enhance the script to allow the program (dbacess), database (labor32), table (informix.position) and file (/var/lib/dbspace/bosarc/Active_sites/Cronos_test/Position7) to be set as command line arguments or via environment variables. That requires a bit of fiddling in the script, but nothing outrageous. I'd probably allow the file name to be specified separately from the directory where the file is to be stored so that it is easier to configure on the command line.
Is it possible to rename a file after building it in Sublime Text 3? By default, the output is the same as the input; so filename.scss is built to filename.css. But what if I want filename.scss.css by default (to indicate that this file is based off of a scss file). Is this possible?
Yes. Read through the Build Systems Reference for details. First, there are several variables you can use. $file is the reference to the full path of the current file, say /home/foo/test.php. There is also $filepath (/home/foo), $file_name (test.php), $file_extension (php), $file_base_name (test), and some others. You can also use regexes just about anywhere inside curly braces ($file is the same as ${file}). For example, ${file/\.php/\.txt/} will rename its suffix from .php to .txt (/home/foo/test.txt). ${filepath/testing/production} changes the directory. Here are several combined in a contrived example:
"cmd": ["myprocessor", "--infile", "$file", "--outfile", "/mnt/${project_name}/var/www/assets/${file_base_name/test/final}.css"],
...
For your particular case, this should work:
"cmd": ["myprocessor", "--infile", "$file", "--outfile", "$filepath/$file_name.css"],
should take /path/to/yourfile.scss and spit out the processed /path/to/yourfile.scss.css if that's what you want.
1) I have a folder called CCBuilds containing couple of files in this path: E:\Testing\Builds\CCBuilds.
2) I have written C# code (Process.Start) to Rar this folder and save it in E:\Testing\Builds\CCBuilds.rar using the following command
"C:\program files\winrar\rar.exe a E:\Testing\Builds\CCBuilds.rar E:\Testing\Builds\CCBuilds"
3) The problem is that, though the rar file gets created properly, when I unrar the file to CCBuilds2 folder (both through code using rar.exe x command or using Extract in context menu), the unrared folder contains the full path, ie. extracting E:\Testing\Builds\CCBuilds.rar ->
E:\Testing\Builds\CCBuilds2\Testing\Builds\CCBuilds\<<my files>>
Whereas I want it to be something like this: E:\Testing\Builds\CCBuilds2\CCBuilds\<<my files>>
How can I avoid this full path persistence while adding to rar / extracting back from it. Any help is appreciated.
Use the -ep1 switch.
More info:
-ep = Files are added to an archive without including the path information. Could result in multiple files existing in the archive
with same name.
-ep1 = Do not store the path entered at the command line in archive. Exclude base folder from names.
-ep2 = Expand paths to full. Store full file paths (except drive letter and leading backslash) when archiving.
(source: http://www.qa.downappz.com/questions/winrar-command-line-to-add-files-with-relative-path-only.html)
Just in case this helps: I am currently working on an MS Access Database project (customer relations management for a small company), and one of the tasks there is to zip docx-files to be sent to customers, with a certain password encryption used.
In the VBA procedure that triggers the zip-packaging of the docx-files, I call WinRAR as follows:
c:\Programme\WinRAR\winrar.exe a -afzip -ep -pThisIsThePassword "OutputFullName" "InputFullName"
-afzip says: "Create a zip file (as opposed to a rar file)
-ep says: Do not include the paths of the source file, i.e. put the file directly into the zip folder
A full list of such switches is available in the WinRAR Help, section "Command line".
x extracts it as E:\Testing\Builds\CCBuilds2\Testing\Builds\CCBuilds\, because you're using full path when declaring the source. Either use -ep1 or set the default working dir to E:\Testing\Builds.
Use of -ep1 is needed but it's a bit tricky.
If you use:
Winrar.exe a output.rar inputpath
Winrar.exe a E:\Testing\Builds\CCBuilds.rar E:\Testing\Builds\CCBuilds
it will include the input path declared:
E:\Testing\Builds\CCBuilds -> E:\Testing\Builds\CCBuilds.rar:
Testing\Builds\CCBuilds\file1
Testing\Builds\CCBuilds\file2
Testing\Builds\CCBuilds\folder1\file3
...
which will end up unpacked as you've mentioned:
E:\Testing\Builds\CCBuilds2\Testing\Builds\CCBuilds\
There are two ways of using -ep1.
If you want the simple path:
E:\Testing\Builds\CCBuilds\
to be extracted as:
E:\Testing\Builds\CCBuilds2\CCBuilds\file1
E:\Testing\Builds\CCBuilds2\CCBuilds\file2
E:\Testing\Builds\CCBuilds2\CCBuilds\path1\file3
...
use
Winrar.exe a -ep1 E:\Testing\Builds\CCBuilds.rar E:\Testing\Builds\CCBuilds
the files inside the archive will look like:
CCBuilds\file1
CCBuilds\file2
CCBuilds\folder1\file3
...
or you could use ep1 to just add the files and folder structure sans the base folder with the help of recursion and defining the base path as the inner path of the structure:
Winrar.exe a -ep1 -r E:\Testing\Builds\CCBuilds.rar E:\Testing\Builds\CCBuilds\*
The files:
E:\Testing\Builds\CCBuilds\file1
E:\Testing\Builds\CCBuilds\file2
E:\Testing\Builds\CCBuilds\folder1\file3
...
inside the archive will look like:
file1
file2
folder1\file3
...
when extracted will look like:
E:\Testing\Builds\CCBuilds2\file1
E:\Testing\Builds\CCBuilds2\file2
E:\Testing\Builds\CCBuilds2\folder1\file3
...
Anyway, these are two ways -ep1 can be used to exclude base path with or without the folder containing the files (the base folder / or base path).
I am creating a batch file to consolidate some hard coded text with a few of other existing text files.
for this I am using the below.
set "txtFile=.\text.txt"
call:Append "C:\test 123\test.txt" %textFile%
over here, when I execute it, it thros an error as it is not able to proceed with the path as it has spaces.
how should this be addressed.
I have no idea what your append batch file is doing, but you can simply use copy to concatenate two files.
It's not clear to me what the needs to be appended to what, but the following will append the contents of text.txt to C:\test 123\test.txt by writing everything to C:\test 123\test.txt.
set txtFile=.\text.txt
copy "C:\test 123\test.txt" /a + %txtFile% /a "C:\test 123\test.txt"
If you want a different output file, just change the last parameter.
Btw: it's better to not rely on a specific working directory
The following:
set txtFile=%~dp0text.txt
makes sure that the text.txt is used that is in the same directory as your batch file.
Is it possible to use xcopy to copy files from several directories into one directory using only one xcopy command?
Assuming that I have the directory tree
root\Source\Sub1\Sub2
I want to copy all .xml files from the directory root\Source including sub folder to root\Destination. I don't want to copy the folder structure, just the files.
As DandDI said, you don't need xcopy. for statement helps much. However, you don't need to state process outcome of dir command as well, this command helps better
for /R c:\source %f in (*.xml) do copy "%f" x:\destination\
By the way, when you use it from a batch file, you need to add spare % in front of variable %f hence your command line should be;
for /R c:\source %%f in (*.xml) do copy %%f x:\destination\
when you use it within a batch
Should surround %f with double quotes otherwise it will fail copying file names with spaces
You don't need xcopy for that.
You can get a listing of all the files you want and perform the copy that way.
For example in windows xp command prompt:
for /f "delims==" %k in ('dir c:\source\*.xml /s /b') do copy "%k" x:\destination\
The /s goes into all subdirectories and the /b lists only the files name and path. Each file inturn is assigned to the %k variable, then the copy command copies the file to the destination. The only trick is making sure the destination is not part of the source.
The Answer to this problem which I think is "How to gather all your files out of all the little subdirectories into one single directory" is to download a piece of software called XXCOPY. This is freely available via XXCOPY.COM and there's a free non-commercial version fortunately. One of the Frequently Asked Questions on the help facility on XXCOPY.COM is effectively "How do I gather all my files into one directory" and it tells you which switch to use. XXCOPY is though a surefire way of doing this and it comes in a .zip archive so unzipping it can be not that straightforward but it's not particularly difficult either. There is an unzipping program called ZipGenius available through the ZipGenius.it website so maybe before you download XXCOPY then download ZipGenius then it's a smallpart smalltime double wammy(!)
Might not be the exact answer but if anyone would like to do this without coding.
You can search the name of the item inside a specific folder, and then you can copy the results and later paste it into your desired folder. It will rename the same file to be the folder I believe as the prefix and then the repeated name.