我正在为我的集合使用模型树结构。作为引用,我使用父字段。我需要从当前对象及其所有父对象获取属性。路径中的最后一个元素有一个字段“target”。所以我从
var result = parent = Articles.findOne({target: this.params._id});
do {
parent = Articles.findOne({_id: parent.parent}).parent;
for (var attrname in parent) { result[attrname] = parent[attrname]; }
}
while (parent.parent === null);
这对我来说似乎效率很低。是否可以用一行来获取包含所有元素的对象?然后我就可以处理该对象。
示例文档
{
"_id" : "LD6h5ZcDuJjexfKfx",
"title" : "title",
"publisher" : "public",
"author" : "author"
}
{
"_id" : "KSiyh8zHRq8RZQ2E6",
"edition" : "edition",
"year" : "2020",
"parent" : "LD6h5ZcDuJjexfKfx"
}
{
"_id" : "5yCk4y25wrLBLZhyY",
"pageNumbers" : "1-10",
"target" : "9sjhzPhyTuQ5Kbh6v",
"parent" : "KSiyh8zHRq8RZQ2E6"
}
因此,从 "target": "9sjhzPhyTuQ5Kbh6v"
开始,我想获取两个父文档(在本例中)。
至少我需要数据集
"title" : "title",
"publisher" : "public",
"author" : "author",
"edition" : "edition",
"year" : "2020",
"pageNumbers" : "1-10"
最佳答案
如果您想在单个查询中执行此操作,那么您需要遵循 array of ancestors Mongodb 中的模式。否则,您需要像您所做的那样递归遍历叶节点上方的分支。对于像您这样的深度较低的层次结构,这并不是一个很大的惩罚。
有了祖先数组,您的文档树将如下所示:
{
"_id" : "LD6h5ZcDuJjexfKfx",
"title" : "title",
"publisher" : "public",
"author" : "author",
}
{
"_id" : "KSiyh8zHRq8RZQ2E6",
"edition" : "edition",
"year" : "2020",
"ancestors" : ["LD6h5ZcDuJjexfKfx"],
"parent" : "LD6h5ZcDuJjexfKfx"
}
{
"_id" : "5yCk4y25wrLBLZhyY",
"pageNumbers" : "1-10",
"target" : "9sjhzPhyTuQ5Kbh6v",
"ancestors" : ["LD6h5ZcDuJjexfKfx","KSiyh8zHRq8RZQ2E6"],
"parent" : "KSiyh8zHRq8RZQ2E6"
}
获取文档及其父级:
Articles.find({ $or: [ { target: target },
_id: { $in: Articles.findOne({ target: target }).ancestors }]});
关于javascript - 获取mongoDB模型树结构中的所有父文档,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32682744/