Array.splice() 结果值 'undefined'

Question

目前，我正在研究应该从列表中删除用户选择的选项的代码。当我使用 .splice() 方法时，它会产生值 undefined。

一开始以为是async的问题，所以加了async的代码。

app.intent("Optionpicker", (conv, input) => {
  conv.data.list = ['optionA', 'optionB', 'optionC', 'optionD']

  let choice = conv.parameters.choice

  async function listremover() {
    let index = conv.data.list.indexOf(choice);
    if (index > -1) {
      conv.data.list = conv.data.list.splice(index, 1);

      return conv.data.list
    }
  }
  listremover().then()

  conv.add(`Ok, your choice is ${choice}. The remaining options are: ${conv.data.list[0]}, ${conv.data.list[1]} and ${conv.data.list[2]}. Which one is next?`)
})

我的预期结果是函数 returns 一个没有用户选择选项的数组。此时值为'undefined'，crshes代码

Answer 1

我不得不对一些值进行硬编码并删除你的 app.intent("Optionpicker", (conv, input) => { 因为应用程序、转换和输入在此示例中未定义。

let conv = {};
//let choice = 2; // conv.parameters.choice;
// had to hard code values b/c app and conv are undefined in this example
conv.data = {};
conv.data.list = ['optionA', 'optionB', 'optionC', 'optionD'];
let selected = {};

async function listremover() {
  let index = conv.data.list.indexOf('optionB');
  //if (index > -1) {
  //conv.data.list = conv.data.list.splice(index, 1);
  //return conv.data.list;
  //}

  if (index > -1) {
    selected = conv.data.list.splice(index, 1);
    //return selected; // returns selected array
  }
  return conv.data.list; // returns remaining values of original array
}

listremover().then(function(res) {
  // removed `conv.add()` not sure where this is defined...
  //console.log(`Ok, your choice is ${selected}. The remaining options are: ${conv.data.list[0]}, ${conv.data.list[1]} and ${conv.data.list[2]}. Which one is next?`);
  console.log(res);
});

Array.splice() 结果值 'undefined'

Array.splice() results in the value 'undefined'

javascript

arrays

node.js

actions-on-google