问题描述
问题:基本上我想从不同领域的不同数组制作一个数组。
在下面的文档中,我想制作一个skus的id数组。
所以我想提取的数组是“optionCombinations.skus”和“supplements.skuInfo.skus”。 (我让数据看起来很简单)
数据:
{
"_id" : ObjectId("60d843f623b7ad1304bcf852"),"optionCombinations" : [
{
"skus" : [
{
"id" : "210625A00048","skuCode" : "210625A00048","salePrice" : 0,"supplyPrice" : 0,"stockQuantity" : 10,"brand" : "10000","manufacturer" : "100000"
}
]
}
],"supplements" : [
{
"skuInfo" : [
{
"skus" : [
{
"id" : "210625A00049","skuCode" : "210625A00049","stockQuantity" : 1,"manufacturer" : "100000"
}
]
},{
"skus" : [
{
"id" : "210625A00050","skuCode" : "210625A00050","manufacturer" : "100000"
}
]
}
]
}
]
}
]
}
查询:
db.shop_product.aggregate([
{$match : {_id : ObjectId('60d5836ed9549c2689304fd8')}},{$project : {'optionSkus' : '$composition.optionInfo.optionCombinations.skus','supplmentSkus' : {$reduce : {
input : '$composition.supplementInfo.supplements.skuInfo.skus',initialValue : [],in : {$concatArrays : ['$$value','$$this']}
}}}},{$project : {'skus' : {$reduce : {
input : ['$optionSkus','$supplmentSkus'],{$project : {'skus' : {$reduce : {
input : '$skus',{$project : {'skus' : '$skus.id'}}
])
我尝试过的:
Aggregationoperation reduce = new Aggregationoperation() {
@Override
public Document todocument(AggregationoperationContext context) {
ArrayOperators.Reduce reduce = ArrayOperators.Reduce.arrayOf("$composition.optionInfo.optionCombinations.skus")
.withInitialValue(new ArrayList<>())
.reduce(ArrayOperators.ConcatArrays.arrayOf(ArrayOperators.Reduce.Variable.VALUE.getTarget())
.concat(ArrayOperators.Reduce.Variable.THIS.getTarget()));
Document document = new Document("$reduce",new Document("optionSkus",reduce));
return document;
}
};
结果:
提前致谢
解决方法
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