[英]Does ArangoDB have faceted search?
有誰知道ArangoDB是否支持分面搜索以及性能與其他支持它的產品(例如Solr,MarkLogic)或不支持它的產品(例如Mongo)相比如何?
在搜索網站,閱讀文檔以及搜索Google網上論壇后,我認為不會在任何地方進行討論。
謝謝
ArangoDB有一種查詢語言,支持分組查詢。 這允許您實現分面搜索。 為了確定我們對分面搜索有相同的理解,讓我解釋一下,我認為它是什么意思。 例如,您有一個產品清單。 每種產品都有一些屬性(例如名稱,型號)和一些類別(例如制造商)。 然后我可以搜索包含單詞的名稱或名稱。 這將列出所有產品以及指示在哪個類別中有多少產品。 這是你的意思嗎?
因此,舉例:假設您有三個屬性(name,attribute1,attribute2)和兩個類別(category1,category2)的文檔:
> for (i = 0; i < 10000; i++) db.products.save({category1: i % 5, category2: i % 7, attribute1: i % 13, attribute2: i % 17, name: "Lore Ipsum " + i, productId: i})
所以典型的文件是:
> db.products.any()
{
"_id" : "products/8788564659",
"_rev" : "8788564659",
"_key" : "8788564659",
"productId" : 9291,
"category1" : 1,
"category2" : 2,
"attribute1" : 9,
"attribute2" : 9,
"name" : "Lore Ipsum 9291"
}
如果要搜索屬性1在2和3(含)之間的所有文檔,可以使用
> db._query("FOR p IN products FILTER p.attribute1 >= 2 && p.attribute1 <= 3 SORT p.name LIMIT 3 RETURN p").toArray();
[
{
"_id" : "products/7159077555",
"_rev" : "7159077555",
"_key" : "7159077555",
"productId" : 1003,
"category1" : 3,
"category2" : 2,
"attribute1" : 2,
"attribute2" : 0,
"name" : "Lore Ipsum 1003"
},
{
"_id" : "products/7159274163",
"_rev" : "7159274163",
"_key" : "7159274163",
"productId" : 1004,
"category1" : 4,
"category2" : 3,
"attribute1" : 3,
"attribute2" : 1,
"name" : "Lore Ipsum 1004"
},
{
"_id" : "products/7161633459",
"_rev" : "7161633459",
"_key" : "7161633459",
"productId" : 1016,
"category1" : 1,
"category2" : 1,
"attribute1" : 2,
"attribute2" : 13,
"name" : "Lore Ipsum 1016"
}
]
或者如果您只對產品標識感興趣
> db._query("FOR p IN products FILTER p.attribute1 >= 2 && p.attribute1 <= 3 SORT p.name LIMIT 3 RETURN p.productId").toArray();
[
1003,
1004,
1016
]
現在為了獲得類別1的方面
> db._query("LET l = (FOR p IN products FILTER p.attribute1 >= 2 && p.attribute1 <= 3 SORT p.name RETURN p) return [ slice(l,@skip,@count), (FOR p in l collect c1 = p.category1 INTO g return { category1: c1, count: length(g[*].p)}) ]", { skip: 0, count: 3 }).toArray()
[
[
[
{
"_id" : "products/7159077555",
"_rev" : "7159077555",
"_key" : "7159077555",
"productId" : 1003,
"category1" : 3,
"category2" : 2,
"attribute1" : 2,
"attribute2" : 0,
"name" : "Lore Ipsum 1003"
},
{
"_id" : "products/7159274163",
"_rev" : "7159274163",
"_key" : "7159274163",
"productId" : 1004,
"category1" : 4,
"category2" : 3,
"attribute1" : 3,
"attribute2" : 1,
"name" : "Lore Ipsum 1004"
},
{
"_id" : "products/7161633459",
"_rev" : "7161633459",
"_key" : "7161633459",
"productId" : 1016,
"category1" : 1,
"category2" : 1,
"attribute1" : 2,
"attribute2" : 13,
"name" : "Lore Ipsum 1016"
}
],
[
{
"category1" : 0,
"count" : 307
},
{
"category1" : 1,
"count" : 308
},
{
"category1" : 2,
"count" : 308
},
{
"category1" : 3,
"count" : 308
},
{
"category1" : 4,
"count" : 308
}
]
]
]
要深入到category1並使用facet進行類別2:
> db._query("LET l = (FOR p IN products FILTER p.attribute1 >= 2 && p.attribute1 <= 3 && p.category1 == 1 SORT p.name RETURN p) return [ slice(l,@skip,@count), (FOR p in l collect c2 = p.category2 INTO g return { category2: c2, count: length(g[*].p)}) ]", { skip: 0, count: 3 }).toArray()
[
[
[
{
"_id" : "products/7161633459",
"_rev" : "7161633459",
"_key" : "7161633459",
"productId" : 1016,
"category1" : 1,
"category2" : 1,
"attribute1" : 2,
"attribute2" : 13,
"name" : "Lore Ipsum 1016"
},
{
"_id" : "products/7169497779",
"_rev" : "7169497779",
"_key" : "7169497779",
"productId" : 1056,
"category1" : 1,
"category2" : 6,
"attribute1" : 3,
"attribute2" : 2,
"name" : "Lore Ipsum 1056"
},
{
"_id" : "products/6982720179",
"_rev" : "6982720179",
"_key" : "6982720179",
"productId" : 106,
"category1" : 1,
"category2" : 1,
"attribute1" : 2,
"attribute2" : 4,
"name" : "Lore Ipsum 106"
}
],
[
{
"category2" : 0,
"count" : 44
},
{
"category2" : 1,
"count" : 44
},
{
"category2" : 2,
"count" : 44
},
{
"category2" : 3,
"count" : 44
},
{
"category2" : 4,
"count" : 44
},
{
"category2" : 5,
"count" : 44
},
{
"category2" : 6,
"count" : 44
}
]
]
]
為了使搜索字符串更加用戶友好,有必要在Javascript中編寫一些小幫助函數。 我認為支持小組https://groups.google.com/forum/#!forum/arangodb是討論您的要求的正確位置。
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.