请查看此fiddle
如何在循环中使用slice
使其从特定索引开始返回结果?
JSON文件:
[
{
"title": "A",
"link": "google.com",
"image": "image.com",
"price": "$1295.00",
"brand": "ABC",
"color": "Black",
"material": "Rubber"
}
]
我希望它从brand
开始返回结果:
brand - ABC
color - Black
material - Rubber
我不知道将.slice(4)
放在循环中的位置。我使用
得到了未定义的错误
$.each(value.slice(4),function(key, value)
以下是代码:
JS:
$.ajax({
url: "https://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20json%20where%20url%20%3D%22http%3A%2F%2Fgoo.gl%2FaZgYDB%22&format=json&diagnostics=true&callback=",
success: function (data) {
var item_html="";
$(data.query.results.json).each(function(key, value) {
$.each(value,function(key, value){
item_html += '<h3>'+key+' - '+value+'</h3>';
});
});
$('#area').append(item_html);
}
});
答案 0 :(得分:2)
使用单独的属性名称数组,以便您可以对其进行切片并以保证的顺序获取名称。
var props = [
"title",
"link",
"image",
"price",
"brand",
"color",
"material"
];
$.ajax({
url: "https://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20json%20where%20url%20%3D%22http%3A%2F%2Fgoo.gl%2FaZgYDB%22&format=json&diagnostics=true&callback=",
dataType: 'json',
success: function (data) {
var item_html="";
var propslice = props.slice(4);
$.each(data.query.results.json, function(i, obj) {
$.each(propslice, function(i, key) {
value = obj[key];
item_html += '<h3>'+key+' - '+value+'</h3>';
});
});
$('#area').append(item_html);
}
});
如果您想跳过少量属性,可以在对象中列出它们,并针对该列表进行测试:
var excluded_props = {
title: true,
link: true,
image: true,
price: true
};
$.ajax({
url: "https://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20json%20where%20url%20%3D%22http%3A%2F%2Fgoo.gl%2FaZgYDB%22&format=json&diagnostics=true&callback=",
dataType: 'json',
success: function (data) {
var item_html="";
$.each(data.query.results.json, function(i, obj) {
$.each(obj, function(key, value) {
if (!excluded_props[key]) {
value = obj[key];
item_html += '<h3>'+key+' - '+value+'</h3>';
}
});
});
$('#area').append(item_html);
}
});
答案 1 :(得分:1)
对象无法满足您的要求。 Javascript中的对象不是有序的,只有数组。您有几个选择:
*示例:
[
{
"title": {
"value" : "A",
"order" : 4
},
"link": {
"value" : "google.com",
"order" : 5
...
}
]