My folder structure is:
\gitclone\SConstruct
\gitclone\level1\level2\SConscript.3dn
Now in SConscript.3dn I am creating empty folder: e.Execute(Mkdir('#/testDir'))
How to define the file path to create the folder in \gitclone\ (where root SConstruct file is)? I was reading manual, but somehow I can not manage it.
I have found solution here, so my code is:
e['SCONS_ROOT'] = Dir('#')
e.Execute(Mkdir('${SCONS_ROOT.abspath}/win_b64/code/bin/testDir'))
Related
I was using path.resolve but this command created a monster folder that can't be deleted called lib..
path.resolve(__dirname + "../assets/pic/" + `${fileName}.png`)
Question 1
What is the propper usage to create a folder 1 level up from the current path?
Question 2
How to remove the lib../assets/pic folder? Deleting the entire project or using git reset --hard, git stash doesn't work because Windows 10 says the folder doesn't exist.
Answer to Question 1:
tl;dr
const fs = require('fs')
const folderName1DirUp = '../SomeFolder'
try {
if (!fs.existsSync(folderName1DirUp)){
fs.mkdirSync(folderName1DirUp)
}
} catch (err) {
console.error(err)
}
The back story:
Two ways to reference the current folder in a Node.js script. You can use ./ or __dirname. Note in addition to ./ you can also use ../ which points to the parent folder.
The difference between the two.
__dirname in a Node script will return the path of the folder where the current JavaScript file resides.
./ will give you the current working directory (and ../ the parent). For ./ a call to process.cwd(). returns the same thing.
Initially the current working directory is the path of the folder where you ran the node command, but during the execution of your script it can change by calling process.chdir(...).
There is one place where ./ refers to the current file path which is in a require(...) call. There the ./, for convenience, refers to the JavaScript file path which lets you import other modules based on the folder structure.
Thus the call to fs.mkdirSync('../SomeFolder') with the given folder name makes a folder one level up from the current working directory.
Answer to Question 2:
From a PowerShell prompt Remove-Item './../Folder1UpToDelete' -Force The ./ is for the current folder. The ../ is for one folder up from the current. Finally the Folder1UpToDelete is the one folder up from the current that you want to delete. The -Force makes sure to delete all sub-folders/files under the folder you want deleted including hidden and/or read-only files.
To answer the first question, to create a path 1 level up, you can use path.join():
path.join(__dirname, "../assets/pics", `${fileName}.png`);
For the second question, if deleting it through the explorer doesn't work, you can try:
fs.rmdirSync("E:/path/to/broken/folder..");
Using Git Bash and running
cd /c/path/to/broken/
rmdir folder..
I'm trying to create a folder within the temp folder that doesn't have a random name.
Here is how I was trying to create the folder within the temp folder.
if not DirExists(ExpandConstant('{%tmp}\Utilities\SDK')) then
CreateDir(ExpandConstant('{%tmp}\Utilities\SDK'));
Log('Temp\Utilities\SDK Folder Has Been Created.');
I had a look at this thread, but even with the %, unfortunately, it still doesn't create the folder.The script compiles and runs as expected, however the folder doesn't create even though it says it has in the log file, I understand that the log file will say that because its told too, however, if the folder was unable to be created, wouldnt it crash? or return a false if an if statement was present?
With CreateDir() You must create dirs one after the other and not a dir structure at once.
if not DirExists(ExpandConstant('{tmp}\Utilities')) then
CreateDir(ExpandConstant('{tmp}\Utilities'));
if not DirExists(ExpandConstant('{tmp}\Utilities\SDK')) then
CreateDir(ExpandConstant('{tmp}\Utilities\SDK'));
if DirExists(ExpandConstant('{tmp}\Utilities\SDK')) then
Log('Temp\Utilities\SDK Folder Has Been Created.') else
Log('Temp\Utilities\SDK Folder ERROR : NOT Created.');
Inno Setup has a function to create a dir structure at once
function ForceDirectories(Dir: string): Boolean;
Example:
if not DirExists(ExpandConstant('{tmp}\Utilities\SDK')) then
ForceDirectories(ExpandConstant('{tmp}\Utilities\SDK'));
Also keep in mind :
{tmp} all is related to the Inno Setup Temp folder is-XXXXX.tmp
C:\Users\...\AppData\Local\Temp\is-XXXXX.tmp
{%temp} is the users Temp folder
C:\Users\...\AppData\Local\Temp
I think you want the Windows Temp and not the tmp from InnoSetup
{tmp}:
Temporary directory used by Setup or Uninstall. This is not the value of the user's TEMP environment variable. It is a subdirectory of the user's temporary directory which is created by Setup or Uninstall at startup (with a name like "C:\WINDOWS\TEMP\IS-xxxxx.tmp"). All files and subdirectories in this directory are deleted when Setup or Uninstall exits. During Setup, this is primarily useful for extracting files that are to be executed in the [Run] section but aren't needed after the installation.
So I think you want to do somethink like this:
if not DirExists(ExpandConstant('{%temp}\Utilities\SDK')) then
CreateDir(ExpandConstant('{%temp}\Utilities\SDK'));
Log('Temp\Utilities\SDK Folder Has Been Created.');
I have this Nodejs lambda function where some files are in a subfolder, like this:
- index.js
- connectors/
- affil.js
I have a Cannot find module error when trying to require the affil.js file. Trying to read it with fs.readFile returns an access denied error.
When I move the file to the root folder, it is accessible. Is there a requirement that Lambda functions files must all be at the root directory? How can I fix that?
Mostly it is because of the way zipping the files making the problem. Instead of zipping the root folder you have to select all files and zip it like below,
Please upload all files and subfolders like below. Please include node_modules folder as well in the zip.
As pointed by #Vijayanath Viswanathan, the issue is with how the zip file is created rather than Lambda.
I used to feed gulp-zip with this:
var src = gulp.src('src/**/*')
The correct way is to prevent folders from being included:
var src = gulp.src('src/**/*.js')
or (if you need to include file with other file extensions)
var src = gulp.src('src/**/*', {nodir: true})
I have the following project structure:
/prj
SConstruct
/src
/app
/lib1
/lib2
/...
'/prj/src/lib1' structure:
/lib1
/src
/test
SConscript
'lib1/SConscript':
SConscript('test/test1/SConscript',
exports = 'env',
variant_dir = '#build/release/lib1/test',
duplicate = 0)
and, finally, 'test' directory:
/test
/common
helpers.cpp
/test1
SConscript
main.cpp
In 'test/test1/SConscript' sources specified as:
Sources = ['../common/helpers.cpp', 'main.cpp']
Result:
scons: *** [build/release/lib1/common/helpers.o]
Source `build/release/lib1/common/helpers.cpp' not found,
needed by target `build/release/lib1/common/helpers.o'
As can be seen the problem is that scons tries to find out the source file 'helpers.cpp' in build directory, not in source one.
Some research shows that the problem raised when source file path begins with '../'. When all sources defined underneath 'SConscript' file all is Ok.
Scons v2.5.1 and v3.0.1 demonstrate the same behavior.
What I did wrong? I've found this answer where the author advised:
You could use ../test.cpp as filename
but I doing exactly the same. Is such scons behavior intended or this is a bug?
Your "code" in "lib1/SConscript":
SConscript('test/test1/SConscript',
exports = 'env',
variant_dir = '#build/release/lib1/test',
duplicate = 0)
uses the name of the given "test/test1/SConscript" and implicitly links the folder "test/test1" as "variant folder" (=variant_dir) to the target directory "#build/release/lib1/test". So if a required file can't be found in "#build/release/lib1/test", SCons tries to do an alternative "lookup" in "test/test1/". But this link is not automatically set for the "common" folder as well, and that's why the lookup of the "helpers.cpp" fails. This is the intended behaviour and correct in SCons.
A solution for your current problem would be to move the "test/test1/SConscript" one folder level higher, include the sources there as "common/helpers.cpp" and "test/main.cpp" and call this new SConscript from "lib1/SConscript" as:
SConscript('test/SConscript',
exports = 'env',
variant_dir = '#build/release/lib1/test',
duplicate = 0)
I have a file that is required in many other files, that are on different folders, inside the main directory.
Is there a way to just require the filename without having to write the relative path, or the absolute path? Like require('the_file'). And without having to go to npm and install it?
Create a folder inside your main directory , put the_file.js inside and set the NODE_PATH variable to this folder.
Example :
Let's say you create a ./libs folder within your main directory, you can just use :
export NODE_PATH = /.../main/lib
after that, you can require any module inside this directory using just :
var thefile = require('the_file')
To not have to do that every time, you'd have to add the variable to your .bashrc (assuming you're running a Unix system).
Or you can set a global variable inside your app.js file and store the path of your 'the_file' in it like so :
global.rootPath = __dirname;
Then you can require from any of your files using :
var thefile = require(rootPath+'/the_file')
These are the most convenient methods for me, short of creating a private npm, but there are a few other alternatives that I discovered when looking up an answer to your question, have a look here : https://gist.github.com/branneman/8048520