CODEWITHSUNDEEP
javascriptreactjs
stu
stu

Reputation: 49

How to count multiple objects inside nested array sequentially in react js

I have an array structure like this :

const array = [
    [
        {
            "item": { "name": "item1" },
            "property": { "name": "property1" }
        },
        {
            "item": { "name": "item2" },
            "property": { "name": "property2" }
        }
    ],
    [
        {
            "item": { "name": "item3" },
            "property": { "name": "property3" }
        }
    ]
]

If I run this command in pure Javascript:

let count = 0
for(let i = 0; i < array.length; i++) {
    for(let j = 0; j < array[i].length; j++) {
        count ++
        console.log(count)
    }
}

Then the output will be like this :

1,2,3

I want to be able to generate the output like above using array.map() in ReactJS and keep those sections. This is the code I'm currently using:

return (
    <div className="container">
    {array.map((nestedArray, arrayIndex) => (
        <div key={arrayIndex} className="array-list">
            <h1>Group: {arrayIndex+1}</h1>
            {nestedArray.map((items, nestedArrayIndex) => (
                <p>Item: {nestedArrayIndex+1}</p>
            ))}
        </div>
    ))}
    </div>
)

And this is the render output (using my css styles):

What is an effective and short way to do it?

I hope I have explained everything clearly

Upvotes: 3

Views: 1656

Answers (3)

Andy
Andy

Reputation: 63524

If you want to keep those sections I think the easiest thing would be to use a variable to keep track of the count. You can then use map to iterate over array and its arrays, updating count as each new array is encountered.

To do that you should separate out the code into a function that you call from your component's return.

const array=[[{item:{name:"item1"},property:{name:"property1"}},{item:{name:"item2"},property:{name:"property2"}}],[{item:{name:"item3"},property:{name:"property3"}}]];

function Example({ data }) {

  function createJSX(data) {

    let count = 0;

    return array.map((arr, i) => {
      return (
        <div>
          <h3>Group: {i + 1}</h3>
          {arr.map(obj => {
            ++count;
            return <p>Item: {count}</p>;
          })}
        </div>
      )
    });

  }

  return (
    <div>
      {createJSX()}
    </div>
  );

};

// Render it
ReactDOM.render(
  <Example data={array} />,
  document.getElementById("react")
);
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/17.0.1/umd/react.production.min.js"></script>
    <script src="https://cdnjs.cloudflare.com/ajax/libs/react-dom/17.0.1/umd/react-dom.production.min.js"></script>
    <div id="react"></div>

And then call the function from your render:

return (
  <div className="container">
    {createJSX()}
  </div>
);

Upvotes: 5

Ryan Le
Ryan Le

Reputation: 8412

Another option if you want to keep track of those sections:

render() {
  let indexCount = 0;
  return (
    <div className="container">
      {array.map((nestedArray, arrayIndex) => (
        <div key={arrayIndex} className="array-list">
          {nestedArray.map((items, nestedArrayIndex) => (
            <p key={(indexCount += 1)}>Item: {indexCount}</p>
          ))}
      </div>
    ))}
  </div>
);

}

You will need a variable to save the current indexCount then just simply increase it each time you render the final index. Also, I added a missing key to your child nestedArray render.

Just a side note is that it's not recommended to use array indexes as render list keys.

Upvotes: 1

Ryan Le
Ryan Le

Reputation: 8412

You could do something like this.

const array = [
  [
    {
      item: { name: "item1" },
      property: { name: "property1" }
    },
    {
      item: { name: "item2" },
      property: { name: "property2" }
    }
  ],
  [
    {
      item: { name: "item3" },
      property: { name: "property3" }
    }
  ]
];

 <div className="container">
    {array.flat().map((item, arrayIndex) => (
       <p>Item: {arrayIndex + 1}</p>
    ))}
 </div>

You will need to flatten your array first, then all children will be in one array.

Upvotes: 3

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