将es5重构为es6以避免声明临时变量

Question

I have a codebase where tempVar are everywhere, just like below example

const tempQuestions = this.state.questions
    const result = tempQuestions.filter(function(value, index1) {
      return index !== index1
    })
    this.setState({
      questions: result
    })

How to avoid such case?

Answer 1

Edit:

Since you know which index value you want to remove. As @undefined mentioned you can use Array.prototype.splice() method to remove the value rather than iterating the entire array

Check below solution which will remove one index value

const { questions } = this.state;
const result = questions.slice();
result.splice(index, 1);
this.setState({
  questions: result
})

Answer 2

This looks pretty straightforward:

this.setState({
  questions: this.state.questions.filter((q, i) => index !== i)
})

Using splice will require cloning the array first, otherwise setState won't notify a state change. So an iteration is required either way, for that filter makes more sense than splice .