weixin_33704234 2019-05-22 13:52 采纳率: 0%
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为什么忽略AJAX调用？

I don't know why the process ignores my AJAX call. It simply jumps from console.log("1"); to console.log("2");.

Can someone explain to me what is going wrong?

render: function() {
  let view = this;
  component.prototype.render.call(view);

  console.log("1");

  $.ajax = ({
    type: "GET",
    cache: false,
    url: "news.json",
    dataType: "json",
    success: function(json) {

      console.log("success");

      for (let i = 0; i < json.length; i++) {
        let news = modelNews;
        news.title = json[i].title;
        news.type = json[i].type;
        news.img = json[i].img;
        news.link = json[i].link;

        view.$('#newsfeed').append(news.getNewsFeedLook());
      }
    },
    error: function() {
      console.log("error");
    }
  });

  console.log("2");

}
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1条回答 默认 最新

  • 普通网友 2019-05-22 13:57
    关注

    Your code is not calling the ajax function of jQuery, it is reassigning it.

    $.ajax = ({
    type: "GET",
    cache: false,
    url: "news.json",
    dataType: "json",
    success: function(json) {

      console.log("success");

      for (let i = 0; i < json.length; i++) {
        let news = modelNews;
        news.title = json[i].title;
        news.type = json[i].type;
        news.img = json[i].img;
        news.link = json[i].link;

        view.$('#newsfeed').append(news.getNewsFeedLook());
      }
    },
    error: function() {
      console.log("error");
    }
  });

    This is the proper call, a function call that is. Pay closer attention to minor error like this!

    $.ajax({
    type: "GET",
    cache: false,
    url: "news.json",
    dataType: "json",
    success: function(json) {

      console.log("success");

      for (let i = 0; i < json.length; i++) {
        let news = modelNews;
        news.title = json[i].title;
        news.type = json[i].type;
        news.img = json[i].img;
        news.link = json[i].link;

        view.$('#newsfeed').append(news.getNewsFeedLook());
      }
    },
    error: function() {
      console.log("error");
    }
  });
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