-3 
var g_Vehicle = {
        g_Vehicle1: [
            Year: "2011",
            Make: "abc",
            Model: "Avenger",
            SubModel: "def"
        ],
        g_Vehicle2: [
            Year: "2012",
            Make: "abc",
            Model: "200",
            SubModel: "deft"
        ],
        g_Vehicle3: [
            Year: "2011",
            Make: "dfg",
            Model: "300",
            SubModel: "sde"
        ],
}
4

4 回答 4

2

那不是 JSON，那是 JavaScript 对象字面量。无效的，您的语法不正确...

var g_Vehicle = {
        g_Vehicle1: [
        //  v------------------------here
            Year: "2011",
            Make: "abc",
            Model: "Avenger",
            SubModel: "def"
        ],
        ...

[...]是您定义数组的方式，但您放入其中的是您定义对象属性的方式。如果您将[and更改]为{and }，它将是有效的，但使用起来很尴尬。相反，只需将整个事物设为对象数组：

var g_Vehicles = [
        {
            Year: "2011",
            Make: "abc",
            Model: "Avenger",
            SubModel: "def"
        },
        {
            Year: "2012",
            Make: "abc",
            Model: "200",
            SubModel: "deft"
        },
        {
            Year: "2011",
            Make: "dfg",
            Model: "300",
            SubModel: "sde"
        }
];

然后你可以遍历它，生成你想要的任何类型的下拉菜单。例如，这是创建select下拉菜单的一种方法：

var markup = ['<select>'];
$.each(g_Vehicles, function(index, vehicle) {
    markup.push('<option value=' + index + '>' +
                vehicle.Make + '-' + vehicle.Model + '-' + vehicle.SubModel +
                '</option>');
});
markup.push('</select>');
$("selector_for_some_container").html(markup.join(""));
于 2012-11-01T08:26:21.570 回答
0

这是一个 JavaScript 对象文字而不是 JSON。

所以使用jQuery的$.each函数：

g_Vehicle.each(function(index) {
    $('select').append($('<option>' + g_Vehicle[index].Make + '</option>'));
});
于 2012-11-01T08:27:22.313 回答
0

这是一些可以帮助您入门的代码。根据您填写的下拉菜单对此进行扩展。

<html>
<head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
<script type="text/javascript">

var g_Vehicle = [
    {
        Year: "2011",
        Make: "Ram",
        Model: "Avenger",
        SubModel: "Dart",
        Vin: "1C1"
    },
    {
        Year: "2012",
        Make: "Dodge",
        Model: "200",
        SubModel: "Dart",
        Vin: "1C2"
    },
    {
        Year: "2011",
        Make: "Chrysler",
        Model: "300",
        SubModel: "Durango",
        Vin: "1C3"
    }
];

var year="2011";

$(document).ready(function(){
    for(var i in g_Vehicle){
       if(g_Vehicle[i].Year == year){
           $('#make').append($('<option>' + g_Vehicle[i].Make + '</option>'));
       }
    }
});


</script>
<body>
<select id="make"></select>
</body>
</html>
于 2012-11-01T08:27:30.997 回答
0

与jQuery一起做......

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
<script>
var g_Vehicle = {
  g_Vehicle1: {
      Year: "2011",
      Make: "abc",
      Model: "Avenger",
      SubModel: "def"
  },
  g_Vehicle2: {
      Year: "2012",
      Make: "abc",
      Model: "200",
      SubModel: "deft"
  },
  g_Vehicle3: {
      Year: "2011",
      Make: "dfg",
      Model: "300",
      SubModel: "sde"
  }
}

$(document).ready(function() {

    (function() {
        var l = '';
        for(var id in g_Vehicle) {
            l += '<option value="'+id+'">'+id+'</option>'           
        }
        $('#mymenu')
        .html(l)
        .change(function() {

            //Test here
            var r = g_Vehicle[$(this).val()];           
            alert('Year: ' + r.Year+'\nMake: ' + r.Make+'\nModel: ' + r.Model+'\nSubModel: ' + r.SubModel+'\n');

        });     
    })();
});
</script>
</head>
<body>
<form id="myform" name="myform">
<select id="mymenu" name="mymenu"></select>
</form>
</body>
</html>
于 2012-11-01T08:48:46.687 回答