js 获取url返回内容

var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function() {
    if (xhr.readyState === 4 && xhr.status === 200) {
        var response = xhr.responseText;
        console.log(response); // 打印返回的内容
    }
};
xhr.open('GET', 'https://api.example.com/data', true);
xhr.send();

fetch('https://api.example.com/data')
    .then(response => {
        if (!response.ok) {
            throw new Error('Network response was not ok');
        }
        return response.text(); // 或者 response.json() 如果你知道返回的是JSON格式
    })
    .then(data => {
        console.log(data); // 打印返回的内容
    })
    .catch(error => {
        console.error('There has been a problem with your fetch operation:', error);
    });