Question

从以下代码：

    BlogApp.query.blog.equalTo("objectId", url).find().then(function(blog) {
        var model = blog[0];
        model = model.toJSON();

        if (model == null {
                Materialize.toast('This object no long exists!', 4000, 'red');
            } else {

                BlogApp.fn.renderView({
                    View: BlogApp.Views.Blog,
                    data: {
                        model: model
                    }
                });

                BlogApp.blog = blog[0];
            }
        }
    });

我收到以下错误：

blog.js:1119 Uncaught TypeError: Cannot read property 'toJSON' of undefined
    at blog.js:1119
    at wrappedResolvedCallback (parse-1.2.19.js:3762)
    at parse-1.2.19.js:3713
    at Array.forEach (<anonymous>)
    at Object._.each._.forEach [as _arrayEach] (parse-1.2.19.js:95)
    at Parse.Promise.resolve (parse-1.2.19.js:3712)
    at Parse.Promise.<anonymous> (parse-1.2.19.js:3766)
    at Parse.Promise.wrappedResolvedCallback (parse-1.2.19.js:3762)
    at Parse.Promise.then (parse-1.2.19.js:3796)
    at wrappedResolvedCallback (parse-1.2.19.js:3765)

基本上，如果JSON为空或null，我想发出警告，但我无法做到。

我怎样才能发现这个错误？

Answer 1

您的Javascript无效，由于以下原因，该脚本甚至无法运行：if (model == null {

使用try-catch。

BlogApp.query.blog.equalTo("objectId", url).find().then(function(blog) {
  const model = blog[0];
  let jsonModel;
  try {
    jsonModel = model.toJSON();
  } catch (e) {
    Materialize.toast('This object no long exists!', 4000, 'red');
    return;
  }
  BlogApp.fn.renderView({
    View: BlogApp.Views.Blog,
    data: {
      model: jsonModel
    }
  });
  BlogApp.blog = blog[0];
});

或者，测试一下blog [0]是否存在，尽管它会减少错误：

BlogApp.query.blog.equalTo("objectId", url).find().then(function(blog) {
  if (!blog[0]) {
    Materialize.toast('This object no long exists!', 4000, 'red');
    return;
  }

Answer 2

BlogApp.query.blog.equalTo("objectId", url).find().then(function(blog) {
    var model = blog[0];

    if (!model) {
            Materialize.toast('This object no long exists!', 4000, 'red');
        } else {
            model = model.toJSON();

            BlogApp.fn.renderView({
                View: BlogApp.Views.Blog,
                data: {
                    model: model
                }
            });

            BlogApp.blog = blog[0];
        }
    }
});

如何在未定义的JSON上处理未捕获的TypeError？

2 个答案: