I am using plugins in cypress and referred to https://docs.cypress.io/api/plugins/configuration-api.html#Usage. When we deploy them into jenkins, I am getting
`pluginsFile` is set to `/e2e/cypress/plugins/index.js`, but either the file is missing, it contains a syntax error, or threw an error when required. The `pluginsFile` must be a `.js` or `.coffee` file.
Please fix this, or set `pluginsFile` to `false` if a plugins file is not necessary for your project.[39m
Error: Cannot find module 'fs-extra'
I did go through a few threads which manually asks you to download the fs-extra in node_module. I did that and the dependency has been automatically added into the package.json file. However, the build fails. The code runs perfectly when you run locally and all the test passes. However, this fails when the deployed into jenkins.
// promisified fs module
const fs = require('fs-extra')
const path = require('path')
function getConfigurationByFile (file) {
const pathToConfigFile = path.resolve('cypress', 'config', `${file}.json`)
return fs.readJson(pathToConfigFile)
}
module.exports = (on, config) => {
// `on` is used to hook into various events Cypress emits
// `config` is the resolved Cypress config
// accept a configFile value or use development by default
const file = config.env.configFile || 'environment-2'
return getConfigurationByFile(file)
}
The code should deploy successfully on Jenkins, however, evening installing it locally on node_module doesn't work. Can somebody help in figuring out what I am missing, please?
This issue has been resolved. Thanks to #Mr.J. This had nothing to do with fs-extra. The entry point in docker file was not correct and I had to modify that. After modifying that, it worked fine.
If at all you have this issue,
1. Try installing fs-extra inside node_module.
2. please check the path in the cypress config file.
3. Check the path in docker file.
I have an installation file created by electron-builder and faced issue that can't find a way how to run powershell script after installation.
The idea is to make some changes in windows registry and set permission for application folder.
As far as I understand it should be configured in build section in package.json. In api I found that it is exist afterPack method, but I can't figure it out how to execute powershell file through it.
Thank you.
First, create an afterPack.js file, containing this code:
exports.default = async function () {
const { exec } = require('child_process');
const bat = exec('powershell "& ""path/to/powershell/file.ps1"""');
}
Edit the path to the .ps1 file accordingly. You may need to add a '.\' to the start of the path. The console readout will tell you if you do.
Next, edit your package.json file to include an afterPack parameter in your build settings, like so:
Now when you run your build script, it will run your .ps1 file.
The file structure looks like this:
= (main-folder)
- package.json
- ...
= server
- index.js
- ...
= deploy-abc // new server
= src
- index.js
= src
- index.js
- ...
I am trying to install 'deploy-abc' as a module within the main-folder app. So, I ran : yarn add "/var/www/main-folder/deploy-abc". It installed correctly & I can see the 'deploy-abc' dependency listed in package.json.
However, when I try to access the deploy-abc's exported object, I get node error Error: Cannot find module 'deploy-abc'. E.g: In some file in the main folder:
const deployAbc = require("deploy-abc");
console.log(deployAbc); // error
Where am I going wrong?
As per the node docs: https://nodejs.org/api/modules.html#modules_loading_from_node_modules_folders
If the module identifier passed to require() is not a core module, and does not begin with '/', '../', or './', then Node.js starts at the parent directory of the current module, and adds /node_modules, and attempts to load the module from that location.
So, seeing as your module is now in the main folder, how you require it will depend on the relative location. If you are requiring it from say /src/index.js, then you will need:
const deployAbc = require('../deploy-abc')
You also don't need to specify the actual file in this case, as it defaults to index.js. If however the entry file was say, dabc.js or something else, you would need to also specify that in the location.
You might have to use the exact relative path. For example, const deployAbc = require("../deploy-abc")
I keep getting this error in the aws-lambda console when uploading code from a zip file. I have tried uploading other zip files and they work correctly. The .js file is named "CreateThumbnail.js" in the zip file. I believe the handler is also named properly "CreateThumbnail.handler". the node_modules subdirectory is also setup. Anyone have any idea?
{
"errorMessage": "Cannot find module 'CreateThumbnail'",
"errorType": "Error",
"stackTrace": [
"Function.Module._resolveFilename (module.js:338:15)",
"Function.Module._load (module.js:280:25)",
"Module.require (module.js:364:17)",
"require (module.js:380:17)"
]
}
The way I was able to get this to work was:
Name the file exports.js
Name the handler, within the file, exports.handler
Set the handler in the lambda config to exports.handler
Zip up only the contents of the folder, not the folder itself (as mentioned above) and rename the zip file exports.zip
Ok, I did this myself, just make sure that you make the zip such that the .js file doesn't end up inside a folder, because AWS would unzip the file you upload and tries to find a .js file by the name of handler you gave, and if its inside a folder it won't help you.
One possible problem is if you upload the lambda as a zip file created via PowerShell Compress-Archive. Compress-Archive has a bug which causes AWS to extract the files into a flat tree (no subdirectories), with backslashes in filenames:
This exact error can show up if your zipped file(s) do not have world-wide read permission. (chmod -R ugo+r).
Check the file permissions before they are zipped. This is not emphasized enough unfortunately by AWS and it caused a lot of headaches for many.
If you are using AWS Lambda Layers you need to validate if your directory structure is on the needed structure for a layer:
For example for the moment.js node.js module you need the following structure:
aws-lambda-layer.zip
│ nodejs
│ nodejs/node_modules
└ nodejs/node_modules/moment
So to create a layer zip file with the correct structure we can use the following command on the root of our project:
mkdir -p nodejs && cp -r node_modules nodejs/ && zip -r aws-lambda-layer.zip nodejs
Some library files might not have global Read so lambda will not be able to read to content and build the content.
Make sure all files in node_modules are readable before packaging:
chmod -R +r node_modules
Then zip and upload.
This is the instruction from https://docs.aws.amazon.com/lambda/latest/dg/nodejs-package.html that I have followed and it works.
To update a Node.js function with dependencies
Open a command line terminal or shell. Ensure that the Node.js version in your local environment matches the Node.js version of your function.
Create a folder for the deployment package. The following steps assume that the folder is named my-function.
Install libraries in the node_modules directory using the npm install command.
npm install the_package_that_is_missing
Create a .zip file that contains the contents of your project folder. Use the r (recursive) option to ensure that zip compresses the subfolders.
zip -r function.zip .
Upload the package using the update-function-code command.
aws lambda update-function-code --function-name my-function --zip-file fileb://function.zip
Now your function is ready to run!
I had this problem on a custom module I had built that was in the node_modules dir. Everything ran fine in testing on my Win10 machine, but when uploaded I kept getting that same "Cannot find module 'modulename'" error.
It turns out that I had a mismatch; here's the package.json line from the module that couldn't be found:
"main": "./build/modulename.js",
and here's the actual filename:
Modulename.js
Case-sensitive; Windows isn't, linux (and thus AWS) is.
This is unrelated but google brought me here, so:
AWS will give you an error:
Unable to import module '<myfile>': Error
What was really happening for me, was that was requiring an unexisting JS file. The error is a bit misleading.
I ran into this same scenario, solved it by using these specific steps to create a Layer, then hook that up to the Lambda function.
make a new empty directory:
mkdir newdir && cd newdir
install whatever npm things:
npm install --save xyz
make a directory skeleton that matches the expected Lambda structure for Node14 (there's a different structure for Node12, or various other languages; see https://docs.aws.amazon.com/lambda/latest/dg/configuration-layers.html?icmpid=docs_lambda_help):
mkdir -p nodejs/node14
copy the "node_modules" directory into that newly made directory skeleton:
cp -R node_modules nodejs/node14
zip the whole thing up (name it whatever you want):
zip -r custom-drivers-node14.zip nodejs
from there, go to AWS console, Lambda, then "Layers" and create a new layer. In the dialog, upload your .zip file ("custom-drivers-node14.zip").
finally, edit your Lambda function in AWS console, and add a new Layer – the interface might change, but as of now, this is under the main screen for a single function, then scroll way down to the bottom. Follow the "Add a layer" flow, choose the Layer you made, and then try your code.
One final note, this code structure worked:
const xyz = require('xyz');
exports.handler = async (event) => {
xyz.doSomething();
}
AWS Lambda uses the name of the file and the name of the handler function, so if you defined your handler like this: exports.myHandler = function(event, context) in a file named index.js, your handler is index.myHandler.
This turned out to be a simple one for me.
I was getting, cannot create index. in my case, my main lambda file with the exports.handler in had to be called index.js
Try calling your main file CreateThumbnail.js
The tutorial tells you to include the following items in your zip file:
CreateThumbnail.js
/node_modules/gm
/node_modules/async
What it fails to consider is that there are dependencies of the two packages (gm, async) that also need to be part of the package.
So here's what you have to do:
Change directory to node_modules folder in your project folder, and run the command 'npm install gm async'. This will install gm, async and all their dependencies in this folder.
Now package the 'CreateThumbnail.js' file and the complete 'node_modules' folder into a zip file and upload it. It should work now.
So your complete package should look something like this:
CreateThumbnail.js
/node_modules/.bin
/node_modules/array-parallel
/node_modules/array-series
/node_modules/async
/node_modules/cross-spawn
/node_modules/debug
/node_modules/gm
/node_modules/isexe
/node_modules/lodash
/node_modules/lru-cache
/node_modules/ms
/node_modules/pseudomap
/node_modules/which
/node_modules/yallist
File Name:
app.js
Lambda Function in "app.js":
exports.handler = function(event, context)...
Lambda Handler on Amazon Console:
app.handler ({app}.js + exports.{handler} = app.handler)
When you unzip the folder, you should see:
app.js
node_modules
If I have Grunt installed in some folder /foo, but my current folder is /foo/bar/baz, and I run "grunt sometask" from within my current folder, how can I get Grunt (or NodeJS for that matter) to determine my current path? That is to say, how can I programmatically GET the folder I was in when I called grunt?
When I use process.cwd(), I get the path of the gruntfile, ie, "foo", which is not what I want.
I don't have to do this in Grunt specifically, any nodejs-based solution would work.
According to the source code:
By default, all file paths are relative to the Gruntfile
And, voilá, this line of code shows how grunt actually changes the current directory to the path of the Gruntfile:
process.chdir(grunt.option('base') || path.dirname(gruntfile));
However, option --base is there for just that. See docs: http://gruntjs.com/api/grunt.file
If you don't need to do it from inside the Gruntfile, simply run a script that captures the process.cwd() and then execs grunt.
See: https://www.npmjs.com/package/exec
var exec = require('exec');
process.cwd(); // Will have your current path
exec(['grunt', 'mytask'], function(err, out, code) {
if (err instanceof Error)
throw err;
process.stderr.write(err);
process.stdout.write(out);
process.exit(code);
});
in the Mac or Linex, you can get this by
process.env.PWD
in the windows, unknown
You can edit the grunt-cli to get finish this.
grunt-cli/bin/grunt
require(gruntpath).cli({_originDir:basedir});
then in the gruntfile.js, you can follows:
grunt.option('_originDir')