Query mysql database before webpack - mysql

I have info in mysql. In simple terms I want to get that info, create a bundled file.js using webpack and finally run that script in frontend browser.
Currently I get that info, create a new temp.js file with that info and then run webpack getting the info from that file, but I dont like the idea of creating a new temporal file when running webpack.

Related

preserving a folder from being removed at build by github-action

adonis js cleans build folder every time you trigger a new build as a result public upload folder will be removed this created a lot of issue for me and Im trying diffrent methods to solve this.
Im not currently using github actions to build my project and I was wondering if it can help me on this matter by this order or somthing like this on every commit:
copy build/tmp folder
build project by running yarn build command
past the copied folder from step 1 back to build/tmp
for those who came here like me, I found the configuration. You can change local disk configuration.
/config/drive.ts
{ local: { root: Env.get('STORAGE_ROOT'), } }
In this case I created a variable in the .env with the absolute path

Cypress BDD - Unable to populate log.json file & messages.ndjson using the latest boiler plate code

I'm using the new boiler plate code present here - https://github.com/JoanEsquivel/cypress-cucumber-boilerplate on a Windows machine to generate a log.json file, which in turn makes use of the "cucumber-json-formatter.exe" to format the json file and generate a cucumber-html report. Seem to have followed all the steps correctly, but the log.json file is not getting populated with any data and in turn no cucumber-html report.
Steps followed:
Cloned the project
Performed npm commands to install all latest packages (not required but as a double-check)
Downloaded cucumber-json-formatter-windows-386 from https://github.com/cucumber/json-formatter/releases/tag/v19.0.0 , renamed to cucumber-json-formatter.exe and included in the project folder
Performed "npm run cypress:execution" command - This comes from the script in package.json file. Able to see the feature files getting executed in the terminal. This creates the json logs folder with the 2 json files (log.json, messages.ndjson)
Performed "node .\cucumber-html-report.js" command. This generates the cucumber-html report which is empty, because it should be the formatted version of the log.json file. The formatting is done by the cucumber-json-formatter.exe.
Reaching out, if anyone else also came across the same issue. If yes, require some guidance here please.

##[error]Error: NO JSON file matched with specific pattern: **/appsettings.json

I'm trying to deploy an Azure App Service using Azure Devops.
I'm using the Task Azure App Service deploy version 4.*
I started noticing the following error in the log recently with the deployment failing (saw it first on 24th September)
Applying JSON variable substitution for **/appsettings.json
##[error]Error: NO JSON file matched with specific pattern: **/appsettings.json.
In the pipeline I use the task Extract files to extract *.zip, then use the result to search for **/appsettings.json.
The same task was running fine till a few days ago.
I tried redeploying an old release which was successful earlier, but it failed now with the same above error.
I double checked, there was no changes done in the pipeline recently for this to break.
How can I fix this.
Turns out my issue was not with the task Azure App Service deploy, but with the task Extract Files.
A rough look on my pipeline is as below:
Before the fix
Extract files
Deploy Azure App Service
The JSON variable substitution failed because the Extract files task was not able to find *.zip files in the root folder and hence extracted nothing. So, there was no appsettings.json file in the folder structure at all.
The Fix
Update the Extract files task search pattern as **/*.zip
Now my pipeline looks like below.
Extract files
Deploy Azure App Service
It now works fine for me.

How can jenkins find junit.xml in a zip file?

I am running Jenkins on a script, that generates a junit.xml report file and other files. However, all those files are zipped by the script, hence Jenkins cannot find it.
Is there a way to make Jenkins unzip the .zip file , find my particular junit file and generate the run results ?
All this is in Linux.
Thanks
Jenkins has the ability to execute arbitrary shell commands as a build step, just add the 'Execute Shell' step to your build and put in the commands you want (presumably 'unzip' would be among them).
Once you've extracted the xml, provided your internal tool generates it using this schema the JUnit plugin just needs the path you extracted to and it will show the results on the build page.
If you have the option, I would really suggest executing your tests via gradle or maven, as outputs from those tasks will produce a report that Jenkins (and other tools) can already read, and can streamline the job setup process for your users. But, if you can't get away with that, the above should work for you.

How do I use html5bolierplate build script?

Okay, so.
I've installed Ant.
Downloaded the build script. Extracted to Ant installation location.
When I run the 'runbuildscript' a cmd interface opens for around a second, and stops.
If I navigate to the location of files through CMD, then do ant minify it doesn't work.
I then thought to place and extract the build script in the location of the files.
Upon doing so, an error came -
BUILD FAILED
E:\NamanyayG\wamp\www\namanyayg\build.xml:150: The following error occurred whil
e executing this line:
E:\NamanyayG\wamp\www\namanyayg\build.xml:416: E:\NamanyayG\wamp\www\${dir.source} does not exist.
Please help me out, I'm thoroughly confused.
Build script expects some external process (in this case you) to set dir.source property.
The most sure way to set it is to define it on ant command line, e.g.
ant -Ddir.source=/path/to/dir.source