Couchbase查詢「\ uefff」打破了一個條件鍵

我有這樣一個map（）函數在啤酒設計文檔：Couchbase查詢「 uefff」打破了一個條件鍵

function (doc, meta) { 

    if(doc.brewery_id) 
    emit([ doc.brewery_id, doc.abv], [doc.name, doc.abv, doc.type, doc.brewery_id, doc.style, doc.category]); 
}

我需要把所有的文檔與2個規則：

1. [brewery_id]開始與「21」

2 ABV] 3-4

之間

我的過濾器是：

startkey=["21st", 3] 
endkey=["21st\uefff", 4]

但結果不正確，規則1按預期工作，但規則2被忽略。請幫我看看有什麼不對。

謝謝！

聽的結果

{"total_rows":5891,"rows":[ 
{"id":"21st_amendment_brewery_cafe-bitter_american","key":["21st_amendment_brewery_cafe",3.6],"value":["Bitter American",3.6,"beer","21st_amendment_brewery_cafe","Special Bitter or Best Bitter","British Ale"]}, 
{"id":"21st_amendment_brewery_cafe-563_stout","key":["21st_amendment_brewery_cafe",5],"value":["563 Stout",5,"beer","21st_amendment_brewery_cafe","American-Style Stout","North American Ale"]}, 
{"id":"21st_amendment_brewery_cafe-south_park_blonde","key":["21st_amendment_brewery_cafe",5],"value":["South Park Blonde",5,"beer","21st_amendment_brewery_cafe","Golden or Blonde Ale","North American Ale"]}, 
{"id":"21st_amendment_brewery_cafe-amendment_pale_ale","key":["21st_amendment_brewery_cafe",5.2],"value":["Amendment Pale Ale",5.2,"beer","21st_amendment_brewery_cafe","American-Style Pale Ale","North American Ale"]}, 
{"id":"21st_amendment_brewery_cafe-potrero_esb","key":["21st_amendment_brewery_cafe",5.2],"value":["Potrero ESB",5.2,"beer","21st_amendment_brewery_cafe","Special Bitter or Best Bitter","British Ale"]}, 
{"id":"21st_amendment_brewery_cafe-general_pippo_s_porter","key":["21st_amendment_brewery_cafe",5.5],"value":["General Pippo's Porter",5.5,"beer","21st_amendment_brewery_cafe","Porter","Irish Ale"]}, 
{"id":"21st_amendment_brewery_cafe-watermelon_wheat","key":["21st_amendment_brewery_cafe",5.5],"value":["Watermelon Wheat",5.5,"beer","21st_amendment_brewery_cafe","Belgian-Style Fruit Lambic","Belgian and French Ale"]}, 
{"id":"21st_amendment_brewery_cafe-north_star_red","key":["21st_amendment_brewery_cafe",5.8],"value":["North Star Red",5.8,"beer","21st_amendment_brewery_cafe","American-Style Amber/Red Ale","North American Ale"]}, 
{"id":"21st_amendment_brewery_cafe-oyster_point_oyster_stout","key":["21st_amendment_brewery_cafe",5.9],"value":["Oyster Point Oyster Stout",5.9,"beer","21st_amendment_brewery_cafe","American-Style Stout","North American Ale"]}, 
{"id":"21st_amendment_brewery_cafe-21a_ipa","key":["21st_amendment_brewery_cafe",7.2],"value":["21A IPA",7.2,"beer","21st_amendment_brewery_cafe","American-Style India Pale Ale","North American Ale"]} 
] 
}

來源

2013-08-05 Phillip

可能的[couchbase中的複合視圖]的副本（http://stackoverflow.com/questions/17076267/composite-views-in-couchbase）。看到我的答案是Filipe Manana的解釋。簡短的回答：couchbase在視圖中只有一個索引，但你嘗試使用2. – m03geek

我不這麼認爲，因爲[link]（http://www.couchbase.com/docs/couchbase-manual-2.0/我們可以看到？startkey = [「carrot」，0]＆endkey = [「carrot」，20]，該過濾器基於2個字段 – Phillip

不可以。不瞭解我。在？startkey = [「carrot」，0]＆endkey = [「carrot」，20]中，第一個參數在兩個鍵中都是相等的。在你的例子中，你嘗試使用2個範圍，即？startkey = [「carrot-1」，0]和endkey = [「carrot-20」，20]。這是兩件不同的事情。我給了你一個同樣問題的鏈接。如果你仔細閱讀並按照Filipe的答案鏈接，你會明白爲什麼它是兩個不同的東西。 – m03geek

如果您需要2個不同範圍的篩選結果，你可以使用LINQ，但如果你有大量的文件也可能會很慢。所以，使其更快，你可以做兩件事情：

在memcached中或couchbase應用LinQ在「過濾器」緩存結果之後。
如果你的數據模型，可以爲範圍，也就是說，如果你可以從鍵移動你的範圍中的一個映射功能if像一個創建單獨的視圖：信息查看21sts：

map: function() { if (doc.subtype === "21sts") emit (doc.abv,null) }

其中有亞型==「21sts」的文檔，你可以從視圖中獲得與文檔：

map: function() { emit(doc.brewery_id, null) }

和startkey =「21」，endkey =「21st \ uefff」。

來源

2013-08-06 09:20:52 m03geek

Couchbase查詢「\ uefff」打破了一個條件鍵

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