[英]How do I loop through deeply nested properties of a JavaScript object?
我有一個具有3級嵌套的JavaScript對象。 我很難從第3級嵌套中獲取值。
我已經對SO進行了一些研究並得到了基本的循環,但我似乎無法超越第一級。
這是我的代碼
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"
}]
}]
}]
};
如何在節點內循環並檢索xpos和ypos?
您有一個對象( customers
),其中包含一個存儲在cluster
的數組,您可以使用該對象進行迭代
var i, cluster;
for (i = 0; i < customers.cluster.length; i++)
{
cluster = customers.cluster[i];
}
cluster
有一個存儲在segment
的數組,你可以迭代:
var j, segment;
for (j = 0; j < cluster.segment.length; j++)
{
segment = cluster.segment[j];
}
segment
有一個存儲在node
的數組,您可以通過以下方式迭代:
var k, node;
for (k = 0; k < segment.node.length; k++)
{
node = segment.node[k];
}
您可以將所有這些組合在一起,通過組合這些循環來迭代客戶中每個集群的每個段的每個節點:
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
}
}
}
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