如何循环遍历JavaScript对象的深度嵌套属性?
我有一个JavaScript对象,有3个嵌套级别。我很难从第三层嵌套中获得值 我对SO做了一些研究,得到了基本的循环,但我似乎无法通过第一级 这是我的密码如何循环遍历JavaScript对象的深度嵌套属性?,javascript,Javascript,我有一个JavaScript对象,有3个嵌套级别。我很难从第三层嵌套中获得值 我对SO做了一些研究,得到了基本的循环,但我似乎无法通过第一级 这是我的密码 var customers = { "cluster": [{ "id": "cluster1.1", "color": "blue", "flights": "784", "profit": "524125", "clv": "2364", "segment":
var customers = {
"cluster": [{
"id": "cluster1.1",
"color": "blue",
"flights": "784",
"profit": "524125",
"clv": "2364",
"segment": [{
"id": "segment1.1",
"color": "green",
"flights": "82",
"profit": "22150",
"clv": "1564",
"node": [{
"id": "node1.1",
"color": "orange",
"xpos": "1",
"ypos": "1"
}, {
"id": "node1.2",
"color": "orange",
"xpos": "1",
"ypos": "2"
}, {
"id": "node1.3",
"color": "orange",
"xpos": "1",
"ypos": "3"
}, {
"id": "node1.4",
"color": "orange",
"xpos": "1",
"ypos": "4"
}]
}, {
"id": "segment1.2",
"color": "red",
"flights": "2",
"profit": "2150",
"clv": "1564",
"node": [{
"id": "node2.1",
"color": "tan",
"xpos": "2",
"ypos": "1"
}, {
"id": "node2.2",
"color": "tan",
"xpos": "2",
"ypos": "2"
}, {
"id": "node2.3",
"color": "tan",
"xpos": "2",
"ypos": "3"
}, {
"id": "node2.4",
"color": "tan",
"xpos": "2",
"ypos": "4"
}]
}]
}, {
"id": "cluster1.2",
"flights": "4",
"profit": "5245",
"clv": "2364",
"segment": [{
"id": "segment1.2",
"flights": "2",
"profit": "2150",
"clv": "1564",
"node": [{
"id": "node3.1",
"xpos": "3",
"ypos": "1"
}, {
"id": "node3.2",
"xpos": "3",
"ypos": "2"
}, {
"id": "node3.3",
"xpos": "3",
"ypos": "3"
}, {
"id": "node3.4",
"xpos": "3",
"ypos": "4"
}]
}]
}, {
"id": "cluster1.3",
"flights": "10",
"profit": "456978",
"clv": "548",
"segment": [{
"id": "segment1.3",
"flights": "2",
"profit": "2150",
"clv": "1564",
"node": [{
"id": "node4.1",
"xpos": "4",
"ypos": "1"
}, {
"id": "node4.2",
"xpos": "4",
"ypos": "2"
}, {
"id": "node4.3",
"xpos": "4",
"ypos": "3"
}, {
"id": "node4.4",
"xpos": "4",
"ypos": "4"
}]
}]
}]
};
如何在节点内循环并检索XPO和YPO?您有一个对象(客户
),其中一个数组存储在集群
,您可以使用它进行迭代
var i, cluster;
for (i = 0; i < customers.cluster.length; i++)
{
cluster = customers.cluster[i];
}
段
在节点上存储了一个数组,您可以使用该数组进行迭代:
var j, segment;
for (j = 0; j < cluster.segment.length; j++)
{
segment = cluster.segment[j];
}
var k, node;
for (k = 0; k < segment.node.length; k++)
{
node = segment.node[k];
}
vark,节点;
对于(k=0;k
您可以将所有这些结合起来,只需结合以下循环,即可迭代客户上每个集群的每个分段的每个节点:
var i, cluster, j, segment, k, node;
for (i = 0; i < customers.cluster.length; i++)
{
cluster = customers.cluster[i];
for (j = 0; j < cluster.segment.length; j++)
{
segment = cluster.segment[j];
for (k = 0; k < segment.node.length; k++)
{
node = segment.node[k];
//access node.xpos, node.ypos here
}
}
}
var i,簇,j,段,k,节点;
对于(i=0;i
customers.cluster[0]。segment[0]。node[0]。xpos
,0
值可以与其他数字交换。可能重复@zzzBov:lolz谢谢你帮我拼出来:)信不信由你,这不是家庭作业,我只是学这类东西的速度太慢了。在开始的时候我需要一点牵手,我很感激人们(像你一样!)花时间去了解细节。干杯。很好的解释,但是家庭作业的评论是高人一等的。对我来说,这里的收获不是答案中给出的努力,而是最后的失望。