Jquery 为什么我无法显示对象？

Question

$(function () {     
    setTimeout(function() {
            var subcategory1 = $("body").find("#navCatLvl2 a");
            var OdjSubcategory1 = {};

            $(subcategory1).each(function(subcategory1,elem1) {

                    var num1 = subcategory1 + 1;
                    $(this).trigger('click');                               


                var ArrOdjSubcategory1 = {
                    "SubCategoryId1" : num1,
                    "Name" : $(this).attr('data-text')
                };

                OdjSubcategory1[num1] = ArrOdjSubcategory1;

            });
        console.log('OdjSubcategory1 = ' + OdjSubcategory1);
    }, 2000);
    console.log(OdjSubcategory1);
});

我们得到错误:

ReferenceError: OdjSubcategory1 is not defined on line console.log(OdjSubcategory1);

为什么如果 console.log('OdjSubcategory1 = ' + OdjSubcategory1); 显示 OdjSubcategory1 = [object Object] 时出现错误？？

Answer 1

您的 console.log 存在于 setTimeout 范围之外。 OdjSubcategory1 不再存在(超出范围)。

请参阅下面代码中的注释:

$(function () {     
    setTimeout(function() {
            var subcategory1 = $("body").find("#navCatLvl2 a");
            var OdjSubcategory1 = {};

            $(subcategory1).each(function(subcategory1,elem1) {

                    var num1 = subcategory1 + 1;
                    $(this).trigger('click');                               


                var ArrOdjSubcategory1 = {
                    "SubCategoryId1" : num1,
                    "Name" : $(this).attr('data-text')
                };

                OdjSubcategory1[num1] = ArrOdjSubcategory1;

            });
        console.log('OdjSubcategory1 = ' + OdjSubcategory1);
    }, 2000); // The scope of everything declared within setTimeout ends here
    console.log(OdjSubcategory1);
});

在 setTimeout 中移动 console.log 将解决您的问题。