我有一个具有3级嵌套的JavaScript对象。我很难从第3级嵌套中获取值。
我已经对SO进行了一些研究并得到了基本的循环,但我似乎无法超越第一级。
这是我的代码
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xpos和ypos?
答案 0 :(得分:21)
您有一个对象(customers
),其数组存储在cluster
,您可以使用
var i, cluster;
for (i = 0; i < customers.cluster.length; i++)
{
cluster = customers.cluster[i];
}
cluster
有一个存储在segment
的数组,您可以使用以下代码进行迭代:
var j, segment;
for (j = 0; j < cluster.segment.length; j++)
{
segment = cluster.segment[j];
}
segment
有一个存储在node
的数组,您可以使用以下代码进行迭代:
var k, node;
for (k = 0; k < segment.node.length; k++)
{
node = segment.node[k];
}
您可以将所有这些组合在一起,通过组合这些循环来遍历客户中每个群集的每个细分的每个节点:
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
}
}
}