将cURL命令转换为ajax[重复]

Question

我正在尝试使用网站提供的Zoom.us API。他们给我cURL命令来创建一个新用户：

curl --data 'api_key=your_api_key&api_secret=your_api_secret&email=[e-mail]&type=1&first_name=John&last_name=Smith' https://api.zoom.us/v1/user/create

我翻译成Ajax：

$.ajax({
        url: 'https://api.zoom.us/v1/user/create',
        type: "POST",
        cache: true,
        async: false,
        headers: {
            'Content-Type': 'application/json'
        },
        data: JSON.stringify({ 'api_key': 'key', 'api_secret': 'secret', 'email': '[e-mail]', 'first_name': 'John', 'last_name': 'Smith' }),
        success: function (res) {
            console.log(res);
        },
        error: function (err) {
            console.error(err);
        }
    });

XMLHttpRequest cannot load https://api.zoom.us/v1/user/create. 
Response to preflight request doesn't pass access control check: No 'Access-Control-Allow-Origin' header is present on the requested resource. 
Origin 'http://mywebsite.com' is therefore not allowed access. The response had HTTP status code 403.

$.ajax({
    url: 'https://api.zoom.us/v1/user/create',
    cache: true,
    async: false,
    data: { 'api_key': 'key', 'api_secret': 'secret', 'email': e, 'first_name': 'john', 'last_name': 'smith' },
    success: function (res) {
        console.log(res);
    },
    error: function (err) {
        console.error(err);
    }
});

产生新的405错误：

XMLHttpRequest cannot load api.zoom.us/v1/user/create?api_key=key&api_secret =secret&email=test%40email.com&first_name=Juan&last_name=Gon‌​zalez. 
No 'Access-Control-Allow-Origin' header is present on the requested resource. Origin 'website.com'; is therefore not allowed access. 

Answer 1

尝试添加datatype:'jsonp'，就像bellow一样

    $.ajax({
    url: 'https://api.zoom.us/v1/user/create',
    cache: true,
    dataType: 'jsonp'
    async: false,
    data: { 'api_key': 'key', 'api_secret': 'secret', 'email': e, 'first_name': 'john', 'last_name': 'smith' },
    success: function (res) {
        console.log(res);
    },
    error: function (err) {
        console.error(err);
    }
});