16

我有一个这种格式的 json 数组:

[
    {
        id : "001",
        name : "apple",
        category : "fruit",
        color : "red"
    },
    {
        id : "002",
        name : "melon",
        category : "fruit",
        color : "green"
    },
    {
        id : "003",
        name : "banana",
        category : "fruit",
        color : "yellow"
    }
]

现在,我想用 Javascript 或 jQuery 以表格格式解析和显示它。该表有四列,每列表示该数组中每个元素的每个属性。该表的第一行是这四个键的名称。其他行是这些键的值。

我不知道如何编写 JavaScript 代码来实现这个功能。你能帮我解决这个问题吗?

4

4 回答 4

26

演示

var obj=[
        {
            id : "001",
            name : "apple",
            category : "fruit",
            color : "red"
        },
        {
            id : "002",
            name : "melon",
            category : "fruit",
            color : "green"
        },
        {
            id : "003",
            name : "banana",
            category : "fruit",
            color : "yellow"
        }
    ]
    var tbl=$("<table/>").attr("id","mytable");
    $("#div1").append(tbl);
    for(var i=0;i<obj.length;i++)
    {
        var tr="<tr>";
        var td1="<td>"+obj[i]["id"]+"</td>";
        var td2="<td>"+obj[i]["name"]+"</td>";
        var td3="<td>"+obj[i]["color"]+"</td></tr>";

       $("#mytable").append(tr+td1+td2+td3); 

    }   
于 2013-08-23T06:28:58.053 回答
11

使用jquery $.each您可以访问所有数据,也可以像这样在表格中设置

<table style="width: 100%">
     <thead>
          <tr>
               <th>Id</th>
               <th>Name</th>
               <th>Category</th>
               <th>Color</th>
           </tr>
     </thead>
     <tbody id="tbody">
     </tbody>
</table>

$.each(data, function (index, item) {
     var eachrow = "<tr>"
                 + "<td>" + item[1] + "</td>"
                 + "<td>" + item[2] + "</td>"
                 + "<td>" + item[3] + "</td>"
                 + "<td>" + item[4] + "</td>"
                 + "</tr>";
     $('#tbody').append(eachrow);
});
于 2013-08-23T06:21:37.483 回答
3
var data = [
    {
        id : "001",
        name : "apple",
        category : "fruit",
        color : "red"
    },
    {
        id : "002",
        name : "melon",
        category : "fruit",
        color : "green"
    },
    {
        id : "003",
        name : "banana",
        category : "fruit",
        color : "yellow"
    }
];

for(var i = 0, len = data.length; i < length; i++) {
    var temp = '<tr><td>' + data[i].id + '</td>';
    temp+= '<td>' + data[i].name+ '</td>';
    temp+= '<td>' + data[i].category + '</td>';
    temp+= '<td>' + data[i].color + '</td></tr>';
    $('table tbody').append(temp));
}
于 2013-08-23T06:20:53.823 回答
3
var jArr = [
{
    id : "001",
    name : "apple",
    category : "fruit",
    color : "red"
},
{
    id : "002",
    name : "melon",
    category : "fruit",
    color : "green"
},
{
    id : "003",
    name : "banana",
    category : "fruit",
    color : "yellow"
}
]

var tableData = '<table><tr><td>Id</td><td>Name</td><td>Category</td><td>Color</td></tr>';
$.each(jArr, function(index, data) {
 tableData += '<tr><td>'+data.id+'</td><td>'+data.name+'</td><td>'+data.category+'</td><td>'+data.color+'</td></tr>';
});

$('div').html(tableData);
于 2013-08-23T06:21:14.827 回答