如何将 mongodb 文档中的所有数组元素更改为某个值?
问题描述：
假设我有以下文件
{
_id: ObjectId("5234cc89687ea597eabee675"),
code: "xyz",
tags: [ "school", "book", "bag", "headphone", "appliance" ],
qty: [
{ size: "S", num: 10, color: "blue" },
{ size: "M", num: 45, color: "blue" },
{ size: "L", num: 100, color: "green" }
]
}
{
_id: ObjectId("5234cc8a687ea597eabee676"),
code: "abc",
tags: [ "appliance", "school", "book" ],
qty: [
{ size: "6", num: 100, color: "green" },
{ size: "6", num: 50, color: "blue" },
{ size: "8", num: 100, color: "brown" }
]
}
{
_id: ObjectId("5234ccb7687ea597eabee677"),
code: "efg",
tags: [ "school", "book" ],
qty: [
{ size: "S", num: 10, color: "blue" },
{ size: "M", num: 100, color: "blue" },
{ size: "L", num: 100, color: "green" }
]
}
我想将文档中代码为efg"的所有元素的数量更改为 0.我该怎么做?我应该使用带有位置运算符的循环吗?
I want to change the num of all the elements in the document having the code "efg" to 0. How do I do that ? Should I use a loop with the positional operator ?
答
最好的方法是匹配数组元素并单独更新位置 $ 操作符使用 Bulk() API .你真的不应该破坏你的 qty 数组.
The best way to do this is match the array element and update individually with the positional $ operator using the Bulk() API . You really shouldn't blow your qty array.
var bulk = db.mycollection.initializeOrderedBulkOp(),
count = 0;
db.mycollection.find({ "code" : "efg" }).forEach(function(doc){
var qty = doc["qty"];
for (var idx = 0; idx < qty.length; idx++){
bulk.find({
"_id": doc._id,
"qty": { "$elemMatch": { "num": qty[idx]["num"]}}
}).update({ "$set": { "qty.$.num": 0 }})
}
count++;
if (count % 200 == 0) {
// Execute per 200 operations and re-init.
bulk.execute();
bulk = db.mycollection.initializeOrderedBulkOp();
}
})
// Clean up queues
if (count % 200 != 0)
bulk.execute();