command line - create shortcut to a exe file on the desktop - command-line

i need to create shortcuts to many exe files on my desktop.
i used this command line: mklink /d Name c:\folder\subfolder\action.exe
it created a folder instead of an exe shortcut.
Plse advise how to do this. thks

C:\Windows\System32>mklink Creates a symbolic link.
MKLINK [[/D] | [/H] | [/J]] Link Target
/D Creates a directory symbolic link. Default is a file
symbolic link.
/H Creates a hard link instead of a symbolic link.
/J Creates a Directory Junction.
Link specifies the new symbolic link name.
Target specifies the path (relative or absolute) that the new link
refers to.

Related

Powershell vs GUI shortcuts

I have a file named new.txt. Using GUI I can create a shortcut, for example, "new.lnk". When I click on "new.lnk" file, Notepad is opened with the contents of "new.txt" file. When I create the shortcut using PowerShell
NEW-ITEM -TYPE SYMBOLICLINK -TARGET "NEW.TXT" "NEW.LNK"
I can see the contents of the file using
CAT "NEW.LNK"
but the shortcut file is not working in the GUI: it does nothing.
I expect to see the contents in Notepad editor. The properties of the file created using GUI and PowerShell are the same, except for "Start in" information: blank when the short cut is created using PowerShell and with the path file directory when using GUI.
Symbolic link (symlink) is not the same as Windows shortcut. A symbolic link is created on file system level - it says "here's a file with such filename, but the content is actually in this other file". It's size is 0 Bytes, as it just points to other file.
It would be more proper to name the file "new-linked.txt" instead of "new.lnk".
A shortcut ".lnk" is a separate file that is interpreted by Windows shell. It contains a path to the target file (among other additional properties). If you create shortcut from UI and then try cat my.lnk, you'll see the content of the shortcut file itself, not the target file.
For creating a Windows shortcut from Powershell, see How to create a shortcut using PowerShell.

ssh-keygen: The System Cannot Find the Path Specified

ssh-agent, ssh-add all works on Cmder running PowerShell/Posh-Git
I have added C:\Program Files\Git\usr\bin Environment Variables
ssh-keyen works fine on Git Bash,
how can I make ssh-keygen to also work with Powershell/Posh-Git ?
EDIT:
It turns out that the error is due to the fact that ssh-keygen in PowerShell is running the ssh-keygen.bat file instead of ssh-keygen.exe
So setting an alias as mention below is the way to go.
I use Set-Alias instead of New-Alias because New-Alias requires me to reset my $profile every now and then, which is weird.
To reset, type . $profile
One way to do this is to add it to your path.
Unfortunately, there appears to be no way to add only the executable. In other words, you will be required to add the whole bin folder (namely C:\Program Files\Git\usr\bin) to your path.
I can show you how to add the whole directory to the path, but I don't think this is what you want. Instead, below is an alternative method that only adds the ssh-keygen.exe executable.
Navigate to $env:homepath\Documents\WindowsPowerShell (create it if you don't have it).
Create a file called profile.ps1.
Add to the file the following line of code.
New-Alias Ssh-Keygen "C:\Program Files\Git\usr\bin\ssh-keygen.exe"
Now, each time you launch Powershell, Ssh-Keygen will be available. It even works with tab completion (e.g. type ssh- and press tab, then it automatically becomes Ssh-Keygen).
Actually, you can add single executables from the git bin folder to your path by symlinking them into a folder that is contained in your path variable.
I also wanted to use some of the git tools inside powershell but I didn't want to load the git-bin-folder into path every time I wanted to use one of the tools and I also didn't want to overwrite tools like cp, find, ls and so on...
Create a folder that you can add to path. I created mine in C:\dev\bin and added it to the path variable.
Then you need to symlink all the dll-files from the git bin folder to your new bin folder because the tools you'll link need them in the same directory (the runpath won't be the git-bin-dir but the new bin-dir). I'll add a Script to do that below.
And finally you just need to create symlinks for all tools you'd like to use in powershell as well (hint: ssh.exe is really nice to use in powershell ;))
Here are my powershell functions that'll help you set up your new bin-folder with all the nice tools from git:
# generic symlink function
function Create-Symlink {
param(
[string]$link,
[string]$target
)
& cmd.exe /c mklink "$link" "$target"
}
# symlink all teh git dlls
function Create-GitDllSymlinks {
param(
[string]$newBinDir='C:\dev\bin',
[string]$gitBinDir='C:\Program Files (x86)\Git\bin'
)
$dlls = gci $gitBinDir -Filter *.dll
$dlls | foreach {
Create-Symlink -link (join-path $newBinDir $_.Name) -target $_.FullName
}
}
# to easily link your git tools
function Create-GitSymlink {
param(
[string]$executable,
[string]$newBinDir='C:\dev\bin',
[string]$gitBinDir='C:\Program Files (x86)\Git\bin'
)
if (-not $executable.EndsWith('.exe')) {
$executable = ($executable + '.exe')
}
Create-Symlink -link (join-path $newBinDir $executable) -target (join-path $gitBinDir $executable)
}
# create all dll symlinks needed
Create-GitDllSymlinks
# link ssh and ssh-keygen to use in powershell
Create-GitSymlink -executable ssh-keygen
Create-GitSymlink -executable ssh
Another nice thing to do is this:
create a bash.bat file in your new bin-folder
write this into that .bat file:
"C:\Program Files (x86)\Git\bin\sh.exe" --login -i
quickly switch between powershell and bash:
when in powershell, type bash and git-bash will start inside the same window with access to all git tools in the git-bin-folder.
want to go back to powershell? just type exit!
Edit:
I reinstalled my computer a few days ago (with Win 10) and ran into some issues with my solution above. Apparently, git and/or the cygwin environment in git has been updated and the cygwin applications now search for their unix-environment-paths differently.
For example, ssh always said that it couldn't find my home path:
Could not create directory '/home/myuser/.ssh'.
The authenticity of host 'hostx (IP)' can't be established.
ECDSA key fingerprint is SHA256:xxxxxx.
Are you sure you want to continue connecting (yes/no)?
Using procmon I found out, that it was looking for files in c:\etc and c:\home\myuser. Setting environment variables like HOME or USERPROFILE didn't work, so I just made two additional symlinks to satisfy my cygwin tools:
PS bin # cmd /c mklink /D C:\etc "C:\Program Files\Git\etc"
symbolic link created for C:\etc <<===>> C:\Program Files\Git\etc
PS bin # cmd /c mklink /D C:\home "C:\Users"
symbolic link created for C:\home <<===>> C:\Users
I solved the problem by generating my own ssh key pair using command
$ ssh-keygen -t rsa -C "your_email#mail.com"
in powershell. If you already have a SSH key, then don't a generate new key, as they will be overwritten. You can use ssh-keygen command, only if you have installed Git with Git Bash.
When you run the above command, it will create 2 files in the ~/.ssh directory.
~/.ssh/id_rsa − It is private key or identification key.
~/.ssh/id_rsa.pub − public key
See link generate ssh key

Why is the "dir" command not showing me certain folders?

I used the dir command in the Windows command prompt to display the list of files/folders in a directory. I noticed that it did not display a folder named tmp. However, I tried running dir in Powershell, and it did display the tmp folder in the output. Why did the Windows command prompt hide this folder from me?
You need to add the "show hidden" option to dir:
dir /a
this should do the trick.
source
Try attrib /?. EG attrib c:\folder\tmp /d.
A leading dot means nothing in Windows.

Compressing only files using 7z without preserving the path

I am using 7z command line executable to zip files, but I see that while adding to an archive the path of the files is preserved in the archive.
So if I do
7z a -tzip myzip.zip dir1\dir2\*
the archive myzip.zip will contain the path dir1\dir2. I do not want this, rather I want only the files to be added to the zip file without the paths being preserved.
I searched quite a bit but do not seem to find any way of doing this, maybe I am missing something obvious?
Thanks
Just add a dot before the path, i.e.
7z a -tzip -r myzip.zip .\Relative\Dir\*
Give the full path. That should work. Not the relative path from the current location.
For example, I give the below, where I want the files in the man5 folder to be archived.
$ 7z a -tzip myzip.zip /home/pradeeban/Desktop/man4/man5/*
The zip contained only the files, without the directories.
Then I gave only the relative path. It had the directories, inside the zip.
$ 7z a -tzip myzip.zip Desktop/man4/man5/*
Tried with Linux (Ubuntu 12.04). Not sure whether that differs from Windows.
I discovered a way to do this by using a relative path:
7z a -tzip myzip.zip %CD%\dir1\dir2\*
%CD% is how you get the current path in a Windows batch file, but it also works from the command line. More info about Capturing the current directory from a batch file.
As explained in related question in 7-zip user FAQ, 7z stores paths relative to working directory, so you will need to first cd to desired top-level directory for archive and run 7-zip from here.
cd dir1\dir2\
7z a -tzip myzip.zip *
If you run it from script and don't want to affect it with changed directory, use directory push/pop facilities available in your shell of choice or run cd+7-zip in spawned process to avoid affecting your entire script with changed directory. For example, using Windows' start that would be:
start /D dir1\dir2\ /wait 7z a -tzip myzip.zip *
This worked for me
Consider folder structure like C:\Parent\SubFolders..... And you want to create parent.zip which will contain all files and folders C:\Parent without parent folder [i.e it will start from SubFolders.....]
cd /D "C:\Parent"
"7z.exe" a Parent.zip "*.*" -r
This will create Parent.zip in C:\Parent

xcopy file, rename, suppress "Does xxx specify a file name..." message

This seems pretty simple and maybe I'm just overlooking the proper flag, but how would I, in one command, copy a file from one directory to another and rename it in the destination directory? Here's my command:
if exist "bin\development\whee.config.example"
if not exist "TestConnectionExternal\bin\Debug\whee.config"
xcopy "bin\development\whee.config.example"
"TestConnectionExternal\bin\Debug\whee.config"
It prompts me with the following every time:
Does TestConnectionExternal\bin\Debug\whee.config specify a file name
or directory name on the target (F = file, D = directory)?
I want to suppress this prompt; the answer is always F.
I use
echo f | xcopy /f /y srcfile destfile
to get around it.
Don't use the xcopy, use copy instead, it doesn't have this issue.
xcopy is generally used when performing recursive copies of multiple files/folders, or when you need the verification/prompting features it offers. For single file copies, the copy command works just fine.
Another option is to use a destination wildcard. Note that this only works if the source and destination filenames will be the same, so while this doesn't solve the OP's specific example, I thought it was worth sharing.
For example:
xcopy /y "bin\development\whee.config.example" "TestConnectionExternal\bin\Debug\*"
will create a copy of the file "whee.config.example" in the destination directory without prompting for file or directory.
Update: As mentioned by #chapluck:
You can change "* " to "[newFileName].*". It persists file extension but allows to rename. Or more hacky: "[newFileName].[newExt]*" to change extension
There is some sort of undocumented feature in XCOPY. you can use:
xcopy "bin\development\whee.config.example" "c:\mybackup\TestConnectionExternal\bin\Debug\whee.config*"
i tested it just today. :-)
Just go to http://technet.microsoft.com/en-us/library/bb491035.aspx
Here's what the MAIN ISSUE is "... If Destination does not contain an existing directory and does not end with a backslash (), the following message appears: ...
Does destination specify a file name
or directory name on the target
(F = file, D = directory)?
You can suppress this message by using the /i command-line option, which causes xcopy to assume that the destination is a directory if the source is more than one file or a directory.
Took me a while, but all it takes is RTFM.
So, there is a simple fix for this. It is admittedly awkward, but it works.
xcopy will not prompt to find out if the destination is a directory or file IF the new file(filename) already exists. If you precede your xcopy command with a simple echo to the new filename, it will overwrite the empty file. Example
echo.>newfile.txt
xcopy oldfile.txt newfile.txt /Y
I met same issue when try to copy file with new name only if file does not exist in destination or exist (with new name), but is older. The solution is to add * char at end of destination file name. Example:
xcopy "C:\src\whee.config.txt" "C:\dest\bee.config.txt*" /D /Y
This is from Bills answer.
Just to be really clear for others.
If you are copying ONE file from one place to another AND you want the full directory structure to be created, use the following command:
xcopy "C:\Data\Images\2013\08\12\85e4a707-2672-481b-92fb-67ecff20c96b.jpg" "C:\Target Data\\Images\2013\08\12\85e4a707-2672-481b-92fb-67ecff20c96b.jpg\"
Yes, put a backslash at the end of the file name and it will NOT ask you if it's a file or directory. Because there is only ONE file in the source, it will assume it's a file.
xcopy src dest /I
REM This assumes dest is a folder and will create it, if it doesnt exists
XCOPY with * at the end of the target to copy files whether they exist or not in destination
XCOPY with \ at the end of the target to copy folders and contents whether exist or not in destination
Alternatively
RoboForm SOURCE DEST FILE for files
RoboForm SOURCE DEST for folders
I had a similar issue and both robocopy and xcopy did not help, as I wanted to suppress the comments and use a different destination filename. I found
type filename.txt > destfolder\destfilename.txt
working as per my requirements.
Back to the original question:
xcopy "bin\development\whee.config.example" "TestConnectionExternal\bin\Debug\whee.config"
could be done with two commands eg:
mkdir "c:\mybackup\TestConnectionExternal\bin\Debug\whee.config\.."
xcopy "bin\development\whee.config.example" "c:\mybackup\TestConnectionExternal\bin\Debug\whee.config\"
By simply appending "\.." to the path of the destination file the destination directory is created if it not already exists. In this case
"c:\mybackup\TestConnectionExternal\bin\Debug\"
which is the parent directory of
the non-existing directory
"c:\mybackup\TestConnectionExternal\bin\Debug\whee.config\.."
At least for WIN7 mkdir does not care if the directory
"c:\mybackup\TestConnectionExternal\bin\Debug\whee.config\"
really exists.
The right thing to do if you wanna copy just file and change it's name at destination is :
xcopy /f /y "bin\development\example.exe"
"TestConnectionExternal\bin\Debug\NewName.exe*"
And it's Gonna work fine
I suggest robocopy instead of copy or xcopy. Used as command or in GUI on clients or servers. Tolerant of network pauses and you can choose to ignore file attributes when copying of copy by file attributes. Oh, and it supports multi-core machines so files are copied much faster in "parallel" with each other instead of sequentially. robocopy can be found on MS TechNet.
For duplicating large files, xopy with /J switch is a good choice. In this case, simply pipe an F for file or a D for directory. Also, you can save jobs in an array for future references. For example:
$MyScriptBlock = {
Param ($SOURCE, $DESTINATION)
'F' | XCOPY $SOURCE $DESTINATION /J/Y
#DESTINATION IS FILE, COPY WITHOUT PROMPT IN DIRECT BUFFER MODE
}
JOBS +=START-JOB -SCRIPTBLOCK $MyScriptBlock -ARGUMENTLIST $SOURCE,$DESTIBNATION
$JOBS | WAIT-JOB | REMOVE-JOB
Thanks to Chand with a bit modifications:
https://stackoverflow.com/users/3705330/chand
Place an asterisk(*) at the end of the destination path to skip the dispute of D and F.
Example:
xcopy "compressedOutput.xml" "../../Execute
Scripts/APIAutomation/Libraries/rerunlastfailedbuild.xml*"
Use copy instead of xcopy when copying files.
e.g.
copy "bin\development\whee.config.example"
"TestConnectionExternal\bin\Debug\whee.config"
Work Around, use ReName... and Name it some Cryptic Name, then ReName it to its Proper Name
C:
CD "C:\Users\Public\Documents\My Web Sites\AngelFire~Zoe\"
XCopy /D /I /V /Y "C:\Users\Public\Documents\My Web Sites\HostGator ~ ZoeBeans\cop.htm"
Ren "cop.htm" "christ-our-passover.htm"
xcopy will allow you to copy a single file into a specifed folder it just wont allow you to define a destination name. If you require the destination name just rename it before you copy it.
ren "bin\development\whee.config.example" whee.config
xcopy /R/Y "bin\development\whee.config"
"TestConnectionExternal\bin\Debug\"
When working with single files , I use both commands.
To copy a file to another existing directory, use copy
copy srcPath\srcFile existingDir\newFile
To copy an existing file to and create new directories, use xcopy
xcopy srcPath\srcFile newDirectoryPath\newFile
To suppress the xcopy 'file or directory' prompt, echo in the response. So for a file copy echo in f.
echo f | xcopy srcPath\srcFile newDirectoryPath\newFile
Note flag /y works in both commands to suppress the confirmation to overwrite the existing destination file.
MS Docs: copy, xcopy
Since you're not actually changing the filename, you can take out the filename from the destination and there will be no questions.
xcopy bin\development\whee.config.example TestConnectionExternal\bin\Debug\ /Y
This approach works well when the destination directory is guaranteed to exist, and when the source may equally be a file or directory.
You cannot specify that it's always a file. If you don't need xcopy's other features, why not just use regular copy?
Does xxxxxxxxxxxx specify a file name
or directory name on the target
(F = file, D = directory)? D
if a File : (echo F)
if a Directory (echo D)