在 $text 搜索后返回按列过滤的文档

Question

我正在对我的集合进行 $text 搜索，其中我制作了复合文本索引。我正在做的是一种解决方法（我认为）我无法以其他方式实现。我最初想要的是在用户给出的字段（列）上进行 $text 搜索。但我想这是不可能的，因为它们大约有 4 个文本字段，我猜 mongodb 不能那样工作。 所以现在，我要做的是搜索所有字段，然后过滤掉用户指定列中搜索关键字匹配的文档。

我已经编码直到全文搜索，但我不知道如何做文档过滤器的事情。有可能吗？

我的指数：Details.collection.createIndex({ name: 'text', desc: 'text', bio: 'text'})

收集文件：

{ "name" : "kat saphire",    "desc" : "I guess that's how it works",  "bio": "Creating a life, I love"}
{ "name" : "john doe",    "desc" : "I practice what I post", "bio": "Simplicity is the key to happiness"}
{ "name" : "kat lisa",    "desc" : "to be or not to be",   "bio": "In a world of worriers, be a warrior"}
{ "name" : "david beckham",  "desc" : "Live as if you were to die tomorrow",   "bio": "I practice what I post"}
{ "name" : "aura leo",    "desc" : "That which does not kill us makes us stronger", "bio": "She turned her can't into can and her dreams into plans"}

我的关键字搜索：practice what I post

全文搜索结果：

{ "name" : "john doe",    "desc" : "I practice what I post", "bio": "Simplicity is the key to happiness"}
{ "name" : "david beckham",  "desc" : "Live as if you were to die tomorrow",   "bio": "I practice what I post"}

现在我只想要一列的结果，例如“desc”

所以我想要的输出应该有：

{ "name" : "john doe",    "desc" : "I practice what I post", "bio": "Simplicity is the key to happiness"}

到目前为止，我使用聚合来获得我的结果。

Answer 1

你可以这样做：

const regex = { $regex: `${searchWord}`, $options: 'i' }
await db.collectionName.aggregate([{$match: { desc: regex }}])[0]