Question

如果URL不正确，

JavaScript 获取API 不会返回正确的错误文本。它总是在 catch 语句中返回无法获取错误。

window.onload = () => {
  fetch("https://www.w3schools.com/nodejs/incorrecturl")
    .then(res => res.json())
    .then(data => console.log(data))
    .catch(error => alert(error.toString())) // Here i need proper error message, instead of "failed to fetch".
}

Answer 1

尝试

var myRequest = new Request('https://www.w3schools.com/nodejs/incorrecturl');
fetch(myRequest).then(function(response) {
    console.log(response.status);
});

OR

fetch('https://www.w3schools.com/nodejs/incorrecturl').then(function(response) {
    console.log(response.status);
});

引用：https://developer.mozilla.org/en-US/docs/Web/API/Response/status

如果网址不正确，fetch API不会在catch中显示确切错误

1 个答案: