Why Does .then() Chained to a Promise Return Undefined?-JS Tutorial-php.cn

Why Does .then() Chained to a Promise Return Undefined?

Susan Sarandon

Release： 2024-10-19 22:17:29

Original

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Why Does .then() Chained to a Promise Return Undefined?

Why .then() Chained to a Promise Returns Undefined

Consider the code snippet below:

<code class="js">function doStuff(n) {
  return new Promise((resolve, reject) => {
    setTimeout(() => {
      resolve(n * 10);
    }, Math.floor(Math.random() * 1000));
  })
  .then((result) => {
    if (result > 100) {
      console.log(`${result} is greater than 100`);
    } else {
      console.log(`${result} is not greater than 100`);
    }
  });
}

doStuff(9)
.then((data) => {
  console.log(data); // undefined, why?
});</code>

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Why is the value of data undefined in the second .then() callback?

Answer:

When you chain a .then() callback to a Promise, it returns a new Promise that resolves to the return value of the callback. However, in the provided code, no value or Promise is returned from the first .then().

Solution:

To fix the issue, you need to return a value or invoke another function that returns a value or a Promise from the first .then(). Here's an updated version of the code:

<code class="js">function doStuff(n) {
  return new Promise((resolve, reject) => {
    setTimeout(() => {
      resolve(n * 10);
    }, Math.floor(Math.random() * 1000));
  })
  .then((result) => {
    if (result > 100) {
      console.log(`${result} is greater than 100`);
    } else {
      console.log(`${result} is not greater than 100`);
    }

    // Return `result` here to avoid undefined at chained `.then()`.
    return result;
  });
}

doStuff(9)
.then((data) => {
  console.log(`data is: ${data}`); // data is not undefined
});</code>

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In this updated code, the first .then() returns the value of result (n multiplied by 10), which ensures that the second .then() receives a defined value as its argument.

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