JavaScript-来自Ajax的对象迭代

发布于 2025-02-14 01:55:46 字数 593 浏览 4 评论 0原文

我正在学习JavaScript，今天我学习了Ajax。我正在使用香草JS和简单的Ajax。我的任务是通过用户输入-ID从URL获取用户的对象。我尝试了.data，但仍然无法使用。任何帮助都将被应用！

async function display() {
  try {
    let id = +userID.value;
    const url = `https://jsonplaceholder.typicode.com/users/${id}`;
    const response = await fetch(url);
    const rootObject = await response.json();
    const users = rootObject.data;
    getUser(users);
  } catch (err) {
    alert(err.message);
  }
}

display()

我有另一个功能 - 当前是空的getusers（用户），因为它不起作用。如何仅从URL访问对象？

原文

I'm learning Javascript and today I learned AJAX.
I am using Vanilla JS and simple AJAX.
My task is to get the object of a user from the url by user input - ID.
I've tried the .data and still won't work.
Any help would be appriciated!

async function display() {
  try {
    let id = +userID.value;
    const url = `https://jsonplaceholder.typicode.com/users/${id}`;
    const response = await fetch(url);
    const rootObject = await response.json();
    const users = rootObject.data;
    getUser(users);
  } catch (err) {
    alert(err.message);
  }
}

display()

I have another function - getusers(users) that currently is empty because it isn't working.
How can I access the object only from the url?

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装纯掩盖桑 2025-02-21 01:55:46

返回的JSON中没有数据属性。只需解析JSON，然后将该数据作为参数GetUser。

function getUser(data) {
  for (const prop in data) {
    console.log(`${prop}: ${data[prop]}`);
  }
}

async function display(id) {
  try {
    const url = `https://jsonplaceholder.typicode.com/users/${id}`;
    const response = await fetch(url);
    const data = await response.json();
    getUser(data);
  } catch (err) {
    console.log(err.message);
  }
}

display(3);

There is no data property in the returned JSON. Just parse the JSON and then have that data as an argument to getUser.

function getUser(data) {
  for (const prop in data) {
    console.log(`${prop}: ${data[prop]}`);
  }
}

async function display(id) {
  try {
    const url = `https://jsonplaceholder.typicode.com/users/${id}`;
    const response = await fetch(url);
    const data = await response.json();
    getUser(data);
  } catch (err) {
    console.log(err.message);
  }
}

display(3);

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