访问js对象问题

时间:2019-06-24 13:19:32

标签: javascript javascript-objects

我创建了一个available_categories({类别:外套和夹克})对象和另一个对象category_names_for_display({coatsand-jackets:外套和夹克})对象。

目的是使用available_categories循环中的“键”(外套和夹克),从第二个对象获取category_name_for_display(外套和夹克)。

// build array of available categories
var obj = available_categories.find(o=>o.category === cat);
if (!(obj) && cat && cat.length > 0) {
    available_categories.push({
        category: cat
    });
}
//console.log('ac=' + available_categories);

// build array of category_names_for_display  
var obj = category_names_for_display.find(o=>o.cat === category_name_for_display);
if (!obj && cat && cat.length > 0 && category_name_for_display) {
    console.log('cat = ' + cat + 'cnfd= ' + category_name_for_display);

var arr = {};
arr[cat] = category_name_for_display;
category_names_for_display.push(arr);

new_var = category_names_for_display['cat'];
console.log('this  ' + new_var);
}
if (available_categories.length > 0) {

    //loop through the array to output more data
    $.each(available_categories, function(k , v) 
    {

        $.each(this, function(k , v) 
        {
            var checked_status = '';
            var selected_option_number = '';
            var undo_label = '';
            console.log('k=   ' + k + 'V=   ' + v);

            // here is where I have something wrong :(
            var catego = category_names_for_display['v'];

        }

    }

}

我只是让控制台显示未定义的值。我需要category_name_for_display是外套和夹克。

2 个答案:

答案 0 :(得分:0)

只需尝试创建一个对象并进行推送。

SELECT DDM_ZIP,
       MAX( TT_PAD_ZIP ) AS TT_PAD_ZIP
FROM   table1 t1
       INNER JOIN
       table2 t2
       ON t1.ddm_zip LIKE t2.tt_pad_zip || '%'
GROUP BY DDM_ZIP

答案 1 :(得分:0)

<form id="sample-form" action="url" method="post"> <input type="hidden" name="params"/> <button type="submit">Submit</button> </form> category_names_for_display,错误是您试图以array的身份访问category_names_for_display

您可以如下所示修改代码,以使object存储核心内容

var catego