I am able to make an alias of an executable on the desktop using this code -How to change icon of alias created using applescript?
I want to copy/duplicate the alias to a folder inside the Applications directory ,
so I tried as follows
duplicate newAlias to folder "myFolder of folder Applications of startup disk" but it times out.
I'm a newbie to Applescript , so I'must be making some simple mistake.
What is the right way to do this ?
Note: I want to retain the same name of the alias at the copied destination
this worked for me on High Sierra and El Capitan as well
duplicate newAlias to folder "MyFolder" of folder "Applications" of startup disk
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The problem
In PhpStorm I have a style.css- and a app.js-file that I have to upload to a server over and over again. I'm trying to automate it.
They're compiled by Webpack, so they are generated/compiled. Which means that I can't simply use the 'Tools' >> 'Deployment' >> 'Upload to...' (since that file isn't and won't every be open).
What I currently do
At the moment, every time I want to see the changed I've done, then I do this (for each file):
Navigate to the files in the file-tree (using the mouse)
Select it
The I've set up a shortcut for Main menu >> Tools >> Deployment >> Upload to..., where-after I select the server I want to upload to.
I do this approximately 100+ times per day.
The ideal solution
The ideal solution would be, that if I pressed a shortcut like CMD + Option + Shift + G
That it then uploaded a selection of files (a scope?) to a predefined remote server.
Solution attempts
Open and upload.
Changing to those files (using CMD + p) and then uploading them (once they're open). But the files are generated, which means that it takes PhpStorm a couple of seconds to render the content (which is necessary before I can do anything with the file) - so that's not faster.
Macro.
Recording a macro, uploading the two files, looking like this:
If I go to the menu and trigger the Macro, then it works. So far so good.
But if I assign a shortcut key and trigger that shortcut while in a file, then it shows me this:
And if I press '1' (for it to upload to number 1 on the list), then it uploads the file that I'm currently in(!?), and not the two files from my macro.
I've tried several different shortcuts (to rule out some kind of keyboard-shortcut-clash):
CMD + Option + CTRL + 0
CMD + Shift 0
CMD + ;
... Same result.
And the PhpStorm Macro's doesn't seem to give me that many options anyways.
Keyboard Maestro.
I've tried doing it using Keyboard Maestro.
But I can't get it setup right. Because if it can't find the folders (if they're off-screen or if I'm in a different project and forgot to adjust they shortcuts), then it blasts through the rest of the recorded actions, resulting in chaos. Ideally it should stop, if it can't find the file on the screen.
Update1 - External program
Even if it's not possible to do in PhpStorm, - are there then another program that I could achieve this with?
Update2 - Automatic Deployment in PhpStorm
I've previously used this, - but I've had happen a few times that I started sync'ing waaaay to many files, overwriting critical core files. It seems smart, but can possibly tear down walls if I've forgotten to define an ignore properly.
I wish there was an 'Automatic Deployment for theses files'-function.
Update3 - File Watchers
I looked into file-watchers ( recommendation from #LazyOne ). Based on this forum thread, then file watchers cannot be used to upload files.
It is possible to accomplish it using external program scp (Secure Copy Protocol):
Steps:
1. Create a Scope (for compiled files app.js and style.css)
2. Create a Custom File Watcher with scp over that Scope
Start with Scope:
Create a Local Scope with name scp files for your compiled files directory (I will assume that your webpack compiles into dist directory):
Then, to add dist directory into Scope, select that folder and click on Include Recursively. Apply and Move to File Watchers
Create a custom template for File Watcher:
Choose a Name
Choose File type as Any
Choose Scope as scp files(created earlier)
Choose Program as scp
Choose Arguments as $FileName$ REMOTE_USER#REMOTE_HOST:/REMOTE_DIR_PATH/$FileName$
Choose Working directory as $FileDir$
That's it, basically what we have done is every time when a file in that scope changes, that file is copied with scp to the remote server to the corresponding path.
Voila. Apply Everything and recompile your project and you will see that everything is uploaded to the server.
(I assumed that you have already set up your ssh client; Generated public/private keys; Added a public key in your remote server; And, know ssh credentials to connect to your remote server)
I figured this out myself. I posted the answer here.
The two questions are kind of similar but not identical.
This way I found is also not the best, since it stores the server password in clean text. So I'll leave the question open, in case someone can come up with a better way to achieve this.
I am trying to insert an image into my MySQL server. I have done some research and it looks like the best way to do that is through LOAD_FILE(). However, LOAD_FILE() always returns null. I know there are 4 conditions for LOAD_FILE():
The file must be located on the server host
You must specify the full path name to the file, and you must have the FILE privilege.
The file must be readable by all and its size less than max_allowed_packet bytes.
If the secure_file_priv system variable is set to a nonempty directory name, the file to be loaded must be located in that directory.
I am currently using:
select LOAD_FILE('/Users/pricedb/Desktop/FolderName/imageName');
and it returns NULL
I have confirmed that all privileges are granted to the user. What does it mean that the file must be located on the server host? The server is being run off my local computer and and file is located there, so does that mean I am good on that?
Any advice would be greatly appreciated, I do not know why it's not returning a value.
I had the same issue.
Fond out that the file to be loaded, needs to be in the folder location where mysql/mariadb has privileges to read it.
It can be configured, BUT the data folder is already has access right.
In my case I copied my file to data folder: C:\Program Files\MariaDB 10.3\data
And than I just called it with full path:
select load_file('C:\\Program Files\\MariaDB 10.3\\data\\test.txt');
Here is my command on windows 7:
select load_file("C:/Program Files (x86)/MySQL/MySQL Workbench CE 6.0.7/images/hieuImages/a.jpg");
And it worked!
You maybe try to copy your images into that directory and select again in mysql.
"FILE" is an administrative privilege.
When you say that ALL privileges have been granted, does that include global administrative privileges too?
Here's the syntax http://dev.mysql.com/doc/refman/5.1/en/grant.html#grant-global-privileges, but I found "MySQL Workbench" GUI more helpful.
Check out OS file permissions, too.
Using MySQL Workbench 8.0:
In the Result Grid panel where you can see the rows and columns, right-click the cell that will hold the BLOB value (currently shown as NULL).
In the context menu that appears, select the first item: Open Value in Editor.
In the lower-left corner, there are two buttons. Click on: Load...
Browse to your image file.
The Binary tab is filled with the image's binary data.
Click: Apply. You'll return to the Result Grid panel again. Notice NULL has been replaced by BLOB.
Right-click BLOB and select Open Value in Editor once more.
You'll find a new tab at the top: Image. There you can preview the image you just added.
Greetings!
show variables like 'secure_file_priv';
Load file in this directory.
select load_file('directory_from_1');
I am using a cybersource web service (http://www.cybersource.com/developers/).I am using simple order api.I have installed that software and wse 3.0.But in their sample project, i am getting error like (The type or namespace name 'TransactionProcessorWse' could not be found)
Please help me, i have been trying to find a solution since yesterday.
You need to download the Clients Solution from here:
http://www.cybersource.com/developers/develop/integration_methods/simple_order_and_soap_toolkit_api/dot_net_2_0/
the zip file is called simapi-net-2.0-5.0.2.msi Once you download, unzip, and run the installer. The files should be located under Program Files (x68)/CyberSourceCorporation/simapi-net-2.0-5.0.2 folder.
Go then to the "lib" folder where you will find the CyberSource.Clients.dll, as well as all the other dll's needed, and import them into your solution.
CyberSource.Clients.dll has the TransactionProcessorWse client.
Good luck
I'm, porting a phone application written using MVVMCross to Windows RT. The application uses SQLite and has a database that is not empty at deploy so I want it to be packaged in the installation folder anc copied to LacalFolder when the application starts.
The database really belongs to the "Core" assembly shared between phone and WinRT but if I put the file in the Core project (in a custom directory called Data), define it as Content and set the Copy Always flag the file does not get copied to the application installation folder.
If I put the file in a directory under the UI project the file gets copied as expected. Any suggestion to avoid keeping two files for the same resource (and the troubles this will cause) ?
Thanks for help
Windows 8 WinRT projects by default use the Content type which does loose packing of files (they are not included in the assembly). If your Core assembly is part of your solution as a project - VS will properly package it copying all the content to the "Core" subfolder of the appx. If you simply have the dll file referenced in your solution - the resource files will not be packaged. The solution in that case is to do something to get these additional files to deploy with the dll. One option is to package the dll as a .vsix together with the assets as described in Tim Heuer's blog post. Another is to do what you did and put the file in the app project. You can add the file to the app project "As Link" to avoid having a copy of the file - it makes Visual Studio create a sort of logical/symbolic link to the file instead of creating a copy - simply right click a folder in the Solution Explorer/app project, select "Add/Existing Item", browse to the file and instead of hitting the "Add" button - use the dropdown triangle on the button to select "Add As Link".
I am writing a WIX installer and I have a following requirement:
During installation, I need to pass an absolute path to a file (lets call it A) included in my installer to a COM component, which already exists on the hard drive and is a part of another program. I have already written an appropriate Custom Action which expects a path to the file A. I don't want to include A as a file installed in the Program Files folder and removed during the uninstallation process. Instead, I would like to put A only temporary on the hard drive, call my Custom Action which will cause the COM component to use the content of A, and then remove A from disk. Is there an easy way to accomplish this goal?
I have tried to utilize the Binary Table and store A there, however I don't know how to reference A using absolute path. I know I could put A outside of MSI file but I would like to keep every file installer needs in a single MSI.
Any help would be appreciated.
Deleting a file that MSI installed means that MSI will consider it "broken" and try to auto-repair it if called on to do so. That happens automatically in several cases (e.g., advertised shortcuts and COM registration) so I'd recommend against it. Leave the file there instead -- it's done its job and there's no harm in leaving it there.
I would take this approach.
Install the file "A" into any directory. Run your custom action needed to update the COM component. Then run another custom action or modify the currently written one to remove the file after it is no longer in use. This would leave no trace of the file "A" and if you schedule the custom action to only run during the install you won't have to worry about it on uninstall.