Firebase 动态 Link 创建 JavaScript

Question

var object={
"longDynamicLink": "https://[APP_NAME].page.link/?link=[LINK_HERE]",
    "suffix":{
        "option":"SHORT"
    }
}
$.ajax({
  url: 'https://firebasedynamiclinks.googleapis.com/v1/shortLinks?key=[KEY_HERE]',
  type: 'POST',
  dataType: "json",
  data: object,
  success: function(response, textStatus, jqXHR) {
    alert(response.shortLink);
  },
  error: function(jqXHR, textStatus, errorThrown){
    alert(textStatus, errorThrown);
  }
});

如果从请求中删除 "suffix"，则上述代码有效。这使得 "UNGUESSABLE" url，但我想要一个短 URL。正如 https://firebase.google.com/docs/dynamic-links/rest?authuser=0 的文档中所述，我添加了后缀选项参数，但结果是 400 响应。有什么想法吗？

Answer 1

我从来没有试过这个但是...

POST https://firebasedynamiclinks.googleapis.com/v1/shortLinks?key=api_key

var params = {
   "longDynamicLink": "https://example.page.link/?link=http://www.example.com/&apn=com.example.android&ibi=com.example.ios",
   "suffix": {
     "option": "SHORT"
   }
}

$.ajax({
    url: 'https://firebasedynamiclinks.googleapis.com/v1/shortLinks?key=[KEY_HERE]',
    type: 'POST',
    data: jQuery.param(params) ,
    contentType: "application/json",
    success: function (response) {
        alert(response.status);
    },
    error: function () {
        alert("error");
    }
});

Firebase 动态 Link 创建 JavaScript

Firebase Dynamic Link Creation With JavaScript

javascript

jquery

ajax

firebase

firebase-dynamic-links