![]() ![]() However, this breaks down in one situation: if you have two sets of children that you need to reorder, there's no way to put a key on each set without adding a wrapper element. Although it might still have some limitations it's still more than nothing. Keyed Fragments In most cases, you can use the key prop to specify keys on the elements you're returning from render. They have a jsx-key rule that might help you. However, this breaks down in one situation: if you have two sets of children that you need to reorder, thereâs no way to put a key on each. In most cases, you can use the key prop to specify keys on the elements youâre returning from render. What displays on my page is (So they are all unique, so WTF react) KEY: 2429 KEY: 2430 KEY: 3859 KEY: 2421 KEY: 2802 KEY: 2428. import createFragment from react-addons-create-fragment // ES6 var createFragment require. You could give eslint-plugin-react a try. Warning: Each child in a list should have a unique 'key' prop. Warn if an element that likely requires a key prop-namely, one present in an array literal or an arrow function expression. This might work but you are right: it's time consuming - then again, what isn't? :) Undoing changes chronologically until the error disappears. Having composed names for keys you can easily find the component that raised the warning - ex: find add some-element- in your code since the warning shows the name of the duplicate key. Using index as key is considered an anti-pattern but it can sure help you get passed this issue. How to add a new key value to react js state array How can I add a Placeholder in a React CKEDITOR React JSX: Unique Key Prop in Conditional Array React unique key prop Can we add events on react fragment How can I add computed state to graph objects in React Apollo Warning: Each child in a list should have a unique 'key' prop even tho it. What are effective strategies for finding the offensive tag? ![]() in react js Ask Question Asked 4 years, 5 months ago Modified 4 years, 5 months ago Viewed 5k times 5 I already provide different key.
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