问题描述
我知道 $in
运算符,它似乎在数组中搜索某个项目的存在,但我只想在该项目位于数组的第一个位置时找到匹配项.
I'm aware of the $in
operator, which appears to search for an item's presence in array, but I only want to find a match if the item is in the first position in an array.
例如:
{
"_id" : ObjectId("0"),
"imgs" : [
"http://foo.jpg",
"http://bar.jpg",
"http://moo.jpg",
]
},
{
"_id" : ObjectId("1"),
"imgs" : [
"http://bar.jpg",
"http://foo.jpg",
"http://moo.jpg",
]
}
我正在寻找类似于以下内容的查询:
I'm looking for a query akin to:
db.products.find({"imgs[0]": "http://foo.jpg"})
这将/应该返回 ObjectId("0")
而不是 ObjectId("1")
,因为它只检查数组中的第一个图像.
This would/should return the ObjectId("0")
but not ObjectId("1")
, as it's only checking against the first image in the array.
如何实现?我知道我可以创建一个单独的字段,其中包含 firstImg
的单个字符串,但这并不是我真正想要的.
How can this be achieved? I'm aware I could just create a separate field which contains a single string for firstImg
but that's not really what I'm after here.
推荐答案
我相信你想要 imgs.0
.例如,给定您的示例文档,您想说: db.products.find({"imgs.0": "http://foo.jpg"})
I believe you want imgs.0
. eg, given your example document, you want to say: db.products.find({"imgs.0": "http://foo.jpg"})
请注意,引用数组索引仅适用于第一级数组.Mongo 不支持更深入地搜索数组索引.
Be aware that referencing array indexes only works for the first-level array. Mongo doesn't support searching array indexes any deeper.
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