{
"employees": {
"year 2010": [
{
"fn": "john",
"ln": "doe",
}, {
"fn": "jane",
"ln": "doe",
}
],
"year 2012": [
{
"fn": "jack",
"ln": "frost",
}, {
"fn": "julie",
"ln": "frost",
}
],
"year 2014": [
{
"fn": "frank",
"ln": "green",
}, {
"fn": "steven",
"ln": "green",
}
]
}
}
如何使用jQuery each()函数遍历此JSON并将其输出到此结构中的页面。如果需要,JSON的格式可以不同,我只需要能够输出如下所示。
2010年
2012年
2014年
答案 0 :(得分:1)
jsFiddle上的完整代码。
var employees = json['employees'];
$.each(employees, function (key, data) {
console.log(key);
var cont = $(".container");
cont.append("<h3>" + key + "</h3>");
$.each(data, function (index, data) {
console.log(data);
cont.append("<p>" + data.fn + " " + data.ln + "</p>");
})
})
它有非常简单的标记,但您可以根据需要设置样式。
答案 1 :(得分:0)
var emps = {
"employees": {
"year 2010": [
{
"fn": "john",
"ln": "doe",
}, {
"fn": "jane",
"ln": "doe",
}
],
"year 2012": [
{
"fn": "jack",
"ln": "frost",
}, {
"fn": "julie",
"ln": "frost",
}
],
"year 2014": [
{
"fn": "frank",
"ln": "green",
}, {
"fn": "steven",
"ln": "green",
}
]
}
};
var arr = [];
$.each(emps.employees, function(i, v){
$("#employees").append('<ul>'+i+'</lu>');
$.each(v, function(j, e){
$("#employees").last().append('<li>'+e.fn+' '+ e.ln+'</li>');
});
});
&#13;
ul {
padding-left: 0px;
}
li {
padding-left: 30px;
}
&#13;
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<ul id="employees"></ul>
&#13;