2

I would like to add a unique key to the returned elements this function provides.

function RowList() {
  const rows = [<Row0 />, <Row1 />, <Row2 />, <Row3 />];
  return (
    <>
      {rows.map(row => (
        <tr key={?}>{row}</tr>
      ))}
    </>
  );
}

I have tried:

function Rows() {
  const rows = [<Row0 />, <Row1 />, <Row2 />, <Row3 />];
  return (
    <>
      {rows.map(row => (
        <tr key={row}>{row}</tr>
      ))}
    </>
  );
}

But [object Object] is returned as the key.

I also will not be able to do something like

let x = 0
function Rows() {
  const rows = [<Row0 />, <Row1 />, <Row2 />, <Row3 />];
  return (
    <>
      {rows.map(row => (
        <tr key={x = x + 1}>{row}</tr>
      ))}
    </>
  );
}

As I will need to be able to remove and add back to the array later.

1 Answer 1

1

As you know, you can't just do this:

// DON'T DO THIS
{rows.map((row, index) => (
  <tr key={index}>{row}</tr>
))}

As the documentation says, that's a "last resort" and really only useful for static lists. You've said your list won't be static.

It's fairly unusual to have an array of already-created elements like that rather than an array of the data for the elements. If you can avoid it, I would, and give the data entries enduring ID values that you can use as keys, e.g. (name is obviously a stand-in for actual data):

class RowInfo {
  static id = 0;
  constructor(name) {
    this.name = name;
    this.id = RowInfo.id++;
  }
}

function RowList() {
  const rows = [new RowInfo("one"), new RowInfo("two"), new RowInfo("three"), new RowInfo("four")];
  return (
    <>
      {rows.map(({id, name}) => (
        <tr key={id}><Row name={name}/></tr>
      ))}
    </>
  );
}

That assumes that they should all be the same type of component, of course, which may not be true.

If you can't do that and must pre-create the actual elements, I'd probably create wrapper objects:

class RowInfo {
   static id = 0;
   constructor(element) {
     this.element = element;
     this.id = RowInfo.id++;
   }
}
function RowList() {
  const rows = [new RowInfo(<Row0 />), new RowInfo(<Row1 />), new RowInfo(<Row2 />), new RowInfo(<Row3 />)];
  return (
    <>
      {rows.map(({id, element}) => (
        <tr key={id}>{element}</tr>
      ))}
    </>
  );
}

Or if they don't have any props you need to specify, you can let React keep track of them, as that's part of its job:

class RowInfo {
   static id = 0;
   constructor(Comp) {
     this.Comp = Comp;
     this.id = RowInfo.id++;
   }
}
function RowList() {
  const rows = [new RowInfo(Row0), new RowInfo(Row1), new RowInfo(Row2), new RowInfo(Row3)];
  return (
    <>
      {rows.map(({id, Comp}) => (
        <tr key={id}><Comp/></tr>
      ))}
    </>
  );
}

Here's a live example of that one:

const Row0 = () => <div>Row 0</div>;
const Row1 = () => <div>Row 1</div>;
const Row2 = () => <div>Row 2</div>;
const Row3 = () => <div>Row 3</div>;

const {Fragment} = React;

class RowInfo {
   static id = 0;
   constructor(Comp) {
     this.Comp = Comp;
     this.id = RowInfo.id++;
   }
}
// Have to use <Fragment></Fragment> in the below instead of <></> because
// Stack Snippet's version of Babel is out of date and
// doesn't understand <></>.
function RowList() {
  const rows = [new RowInfo(Row0), new RowInfo(Row1), new RowInfo(Row2), new RowInfo(Row3)];
  return (
    <Fragment>
      {rows.map(({id, Comp}) => (
        <tr key={id}><Comp/></tr>
      ))}
    </Fragment>
  );
}

ReactDOM.render(<RowList/>, document.getElementById("root"));
<div id="root"></div>

<script src="https://cdnjs.cloudflare.com/ajax/libs/react/16.8.6/umd/react.production.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/react-dom/16.8.6/umd/react-dom.production.min.js"></script>

Sign up to request clarification or add additional context in comments.

2 Comments

Absolutely incredible answers. Thank you.
@Neil - No worries! Glad that helped!

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.