Javascript 无法填充json响应的数据

我有一个API调用，其响应如下

{
    "name": "Cocktails & Dreams, Speakeasy",
    "url": "https://www.zomato.com/ncr/cocktails-dreams-speakeasy-sector-15-gurgaon?utm_source=api_basic_user&utm_medium=api&utm_campaign=v2.1",
    "location": {
        "address": "SCO 23, Sector 15, Gurgaon",
        "locality": "Sector 15",
        "city": "Gurgaon",
        "city_id": 1,
        "latitude": "28.4579754130",
        "longitude": "77.0446266979",
        "zipcode": "0",
        "country_id": 1,
        "locality_verbose": "Sector 15, Gurgaon"
    },
    "price": 4,
    "thumbnail": "https://b.zmtcdn.com/data/res_imagery/300716_RESTAURANT_b026340288ddc8287349d90b0b712940.jpg?fit=around%7C200%3A200&crop=200%3A200%3B%2A%2C%2A",
    "rating": "4.8"
}

Ans这是静态文件夹中的my index.js，用于链接前端和后端

searchForm.addEventListener('submit', async e => {
  e.preventDefault();
  resultArea.innerHTML = '';
  const query = e.target.querySelector('#restaurant-name').value;
  if (query === '') {
    return
  }
  e.target.querySelector('#restaurant-name').value = '';
  const res = await fetch(`${hostname}/locations/${query}`, {
    headers: {
      "Content-Type": "application/x-www-form-urlencoded",
    },
    method: 'POST', 
  })
  const json = await res.json();
  populateData(json);
});

function populateData(results) {
  results(result => {
    const newResult = rTemp.content.cloneNode(true);
    newResult.querySelector('.result-title').innerText = result.name;
    newResult.querySelector('.result-neighborhood').innerText = result.location.locality;
    newResult.querySelector('.result-address').innerText = result.location.address;
    newResult.querySelector('.result-price').innerText = '$'.repeat(result.price);
    newResult.querySelector('.result-thumbnail').src = result.thumbnail;
    newResult.querySelector('.result-website').href = result.url;
    resultArea.appendChild(newResult);
  });
}

rTemp

=document.querySelector（'template'）；在前端

在这个例子中，我得到了以下错误

index.js:27 Uncaught (in promise) TypeError: results is not a function

正确的方法是什么？

而不是

function populateData(results) {
  results(result => {
    const newResult = rTemp.content.cloneNode(true);
    newResult.querySelector('.result-title').innerText = result.name;
    newResult.querySelector('.result-neighborhood').innerText = result.location.locality;
    newResult.querySelector('.result-address').innerText = result.location.address;
    newResult.querySelector('.result-price').innerText = '$'.repeat(result.price);
    newResult.querySelector('.result-thumbnail').src = result.thumbnail;
    newResult.querySelector('.result-website').href = result.url;
    resultArea.appendChild(newResult);
  });
}

使用

而不是

function populateData(results) {
  results(result => {
    const newResult = rTemp.content.cloneNode(true);
    newResult.querySelector('.result-title').innerText = result.name;
    newResult.querySelector('.result-neighborhood').innerText = result.location.locality;
    newResult.querySelector('.result-address').innerText = result.location.address;
    newResult.querySelector('.result-price').innerText = '$'.repeat(result.price);
    newResult.querySelector('.result-thumbnail').src = result.thumbnail;
    newResult.querySelector('.result-website').href = result.url;
    resultArea.appendChild(newResult);
  });
}

使用

---

你能在

fetch

方法中提供完整的url吗？我已经将

hostname

定义为

this.location.origin

你能在

fetch

方法中提供完整的url吗？我已经将

hostname

定义为

this.location.origin