add task solution #3778
add task solution #3778
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Thank you for your hard work on this task! 😊 While your implementation is mostly on the right track, there are a few critical issues that need to be addressed before we can approve the pull request:
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Iteration Methods: The task requires the use of iteration methods like
maporfilterinstead of loops. Currently, aforloop is used, which does not meet the task requirements. Please refactor this part to use the appropriate iteration methods. -
Handling Missing Style Values: The code assumes that every style has a value, which can lead to errors if a style is missing a value (e.g., 'color:'). It's important to add a check to handle such cases gracefully.
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Handling Multiple Colons: Although not explicitly required, the current implementation does not account for multiple colons within a style string's property values. This could lead to incorrect parsing. Adding a check to ensure the split results in exactly two parts would make the code more robust.
Once these issues are addressed, your code will be in great shape! Keep up the good work, and don't hesitate to reach out if you have any questions or need further guidance. 🌟
✨ Please rate this review in the Platform's notification center. I will be grateful for your feedback! ✨
🔄 Fix the issues, commit the changes, and then re-request my review using the Re-request review button in the Reviewers tab.
src/convertToObject.js
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| const filtredStylesList = stylesList.filter((val) => val.trim().length > 0); | ||
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| for (let i = 0; i < filtredStylesList.length; i++) { | ||
| filtredStylesList[i] = filtredStylesList[i].split(':'); |
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The code assumes that every style in filtredStylesList contains a colon and splits it into exactly two parts. If a style does not contain a colon or contains multiple colons, this could lead to unexpected behavior. Consider adding a check to ensure that the split results in exactly two parts before proceeding.
src/convertToObject.js
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| for (let i = 0; i < filtredStylesList.length; i++) { | ||
| filtredStylesList[i] = filtredStylesList[i].split(':'); | ||
| filtredStylesList[i][0] = filtredStylesList[i][0].trim(); | ||
| filtredStylesList[i][1] = filtredStylesList[i][1].trim(); |
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This line assumes that filtredStylesList[i][1] exists. If a style does not have a value (e.g., 'color:'), this will result in an error. Consider checking if filtredStylesList[i][1] is undefined before attempting to trim it.
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Thank you for your submission! 🎉 Your implementation of the convertToObject function is mostly correct, but there are a few critical issues that need to be addressed to enhance the robustness and correctness of your code:
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Iteration Methods: The task requires using iteration methods like
maporfilterinstead of loops. Currently,forloops are used in thesegregateandmakeSegregatedObjectfunctions. Please refactor these parts to use the appropriate iteration methods. -
Handling Missing Style Values: The code assumes that every style has a value, which can lead to errors if a style is missing a value (e.g., 'color:'). It's important to add a check to handle such cases gracefully.
-
Handling Multiple Colons: Although not explicitly required, the current implementation does not account for multiple colons within a style string's property values. Adding a check to ensure the split results in exactly two parts would make the code more robust.
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Variable Naming: While the variable names
segregatedListandstylesObjectare acceptable, consider improving them for clarity. Avoid generic names likearray,object,arr,obj,result.
These issues should be addressed to ensure the code meets the task requirements. However, the decision to approve your pull request is made to help you move forward with your learning. Please review the comments and make the necessary improvements to your code. Keep up the good work and continue learning from your experiences! 🚀
✨ Please rate this review in the Platform's notification center. I will be grateful for your feedback! ✨
| for (let i = 0; i < splited.length; i++) { | ||
| if (splited[i].length > 0) { | ||
| segregatedList.push(splited[i].trim()); | ||
| } | ||
| } |
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The use of a for loop here does not meet the task requirement of using iteration methods like map or filter. Consider refactoring this logic to use map or filter instead.
| for (let i = 0; i < arr.length; i++) { | ||
| if (i % 2 === 0) { | ||
| stylesObject[arr[i]] = arr[i + 1].replaceAll(', ', ',\n'); | ||
| } | ||
| } |
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The use of a for loop here does not meet the task requirement of using iteration methods like map or filter. Consider refactoring this logic to use map or filter instead.
| function makeSegregatedObject(el, index, arr) { | ||
| for (let i = 0; i < arr.length; i++) { | ||
| if (i % 2 === 0) { | ||
| stylesObject[arr[i]] = arr[i + 1].replaceAll(', ', ',\n'); |
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The code assumes that every style has a value. To handle cases where a style might not have a value (e.g., 'color:'), add a check to ensure that arr[i + 1] is not undefined before assigning it to stylesObject[arr[i]].
| const stylesObject = {}; | ||
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| function segregate(el) { | ||
| const splited = el.split(':'); |
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The current implementation does not handle multiple colons within a style string's property values. Consider adding a check to ensure the split results in exactly two parts to make the code more robust.
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