[name]: This field is missing for example recipe

Created on 9 May 2024, 2 months ago
Updated 24 May 2024, about 2 months ago


When running the Drupal recipe command php core/scripts/drupal recipe core/recipes/example -v, a RecipeFileException error is thrown indicating that the name field is missing in the recipe.yml file. This error prevents the recipe from being executed successfully.

Steps to reproduce

  1. Attempt to execute the example recipe with the command: php core/scripts/drupal recipe core/recipes/example -v.
  2. Observe that the execution fails with a validation error highlighting the absence of the name field in the recipe.yml file.

Proposed resolution

Add the name field to the example recipe.yml file to meet the required schema validation and allow successful execution of the recipe. The adjusted section should read:

type: 'Content type'
name: 'Example'

This ensures that the validation passes and the recipe can be processed as intended.

Remaining tasks

  • Review the updated recipe.yml file to ensure compliance with other schema requirements.
  • Test the execution of the recipe post-update to confirm no further issues.
  • Review documentation to ensure it clearly states all required fields in the recipe.yml schema.

User interface changes

No user interface changes are proposed with this update.

API changes

No API changes are proposed with this update.

Data model changes

No data model changes are proposed with this update.

Release notes snippet

Fixed an issue with the example Drupal recipe where the absence of the name field in the recipe.yml file prevented the recipe from executing. Users should now find that running the example recipe via the command line operates as expected.

πŸ› Bug report



10.3 ✨

ConfigurationΒ  β†’

Last updated about 6 hours ago

Created by

πŸ‡΅πŸ‡ͺPeru edutrul

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