如何在深度=1的对象树中返回给定节点(键名)的子节点,这意味着只返回第一个子节点。
具体问题:
下面是一个示例数据对象...
{
"1753": {
"1755": {
"1758": {
"1762": "1753_1755_1758_1762",
"1760": "1753_1755_1758_1760",
"1764": "1753_1755_1758_1764",
"1761": "1753_1755_1758_1761"
},
"1759": {
"1762": "1753_1755_1759_1762",
"1760": "1753_1755_1759_1760",
"1764": "1753_1755_1759_1764",
"1761": "1753_1755_1759_1761"
}
},
"1756": {
"1758": {
"1762": "1753_1756_1758_1762",
"1760": "1753_1756_1758_1760",
"1764": "1753_1756_1758_1764",
"1761": "1753_1756_1758_1761"
},
"1759": {
"1762": "1753_1756_1759_1762",
"1760": "1753_1756_1759_1760",
"1764": "1753_1756_1759_1764",
"1761": "1753_1756_1759_1761"
}
},
"1757": {
"1758": {
"1762": "1753_1757_1758_1762",
"1760": "1753_1757_1758_1760",
"1764": "1753_1757_1758_1764",
"1761": "1753_1757_1758_1761"
},
"1759": {
"1762": "1753_1757_1759_1762",
"1760": "1753_1757_1759_1760",
"1764": "1753_1757_1759_1764",
"1761": "1753_1757_1759_1761"
}
}
},
"1754": {
"1755": {
"1758": {
"1763": "1754_1755_1758_1763"
}
},
"1756": {
"1758": {
"1763": "1754_1756_1758_1763"
}
},
"1757": {
"1758": {
"1763": "1754_1757_1758_1763"
}
}
}
}; 对象的每一层代表页面上的一个下拉菜单。
我需要做的是,当有人从下拉菜单中选择某项内容时,我需要返回剩余菜单的所有可能性。
例如,假设我从dropdown-3中选择了值"1758“(因为它是对象中的第三级),请执行以下操作:
I would need to return info stating that because I selected "1758" from dropwdown-3...
-dropdown-1 can be 1753, 1754.
-dropdown-2 can be 1755,1756,1757
-dropdown-3 can remain unchanged
-dropdown-4 can be 1762,1760,1764,1761,1763到目前为止,我已经做到了这么多……当这种情况的某种组合发生时,问题出现了:
I have selected option "1758" from dropdown-3, AND THEN SELECTED "1754" from dropdown-1...
-dropdown-1 has the most recent selected value, it can remain unchanged
-dropdown-2 can be 1755,1756,1757
-dropdown-3 is set, but can still have its possible values modified & checked for accuracy
-dropdown-3 can be only 1758
-dropdown-4 can be only 1753这是一个样本对象,我的所有对象都会有不同的深度和复杂性…我已经在这方面做了相当多的尝试,希望有人能给我指路:)
谢谢
发布于 2012-04-16 11:41:03
想一想,你想基于选择(节点名)它的孩子,所以你不需要完整的对象树(虽然你可以使用它),你可以更容易地使用邻接表作为查找表(哈希表):
adj["1753"] = ["1755"];
adj["1755"] = ["1758"]
adj["1758"] = ["1762", "1760", "1764", "1761"]
adj["1762"] = "1753_1755_1758_1762"
...所以现在有人选择"1753",你查找adj"1753“,得到包含可能的chidlren的数组,和其他的一样。
但是如果你想使用你不知道深度的当前数据结构,那么你应该使用for (树中的var key )+递归来遍历树,但是这几乎没有意义,因为你需要一个给定的(父)节点名遍历整个树来“知道”父节点的位置(假设是唯一的键!)。
所以最好使用邻接表。
https://stackoverflow.com/questions/10168115
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