当子查询中的所有条件都与 AND 一起否定时，solr_sub 查询会中断

Question

考虑以下查询

"-a:d +b:2 +(-c:[* TO *])"

此查询未返回任何结果。

"-a:d +b:2 -c:[* TO *]"

而此查询返回结果。

对于#1，

它先评估 c，然后评估 a，然后评估 b。

对于#2

它先评估a，然后评估b，然后评估c。

理想情况下是相同的。这有什么区别？

请帮忙。提前致谢。它是标准查询解析器。

Answer 1

将 Solr 中查询的每个部分视为代表一组文档。文档要么是集合的一部分，要么不是集合的一部分，然后您可以在这些集合之间执行操作作为查询的一部分。

AND

表示“应包含在这两个集合中的文档”（交集），

OR

表示“应包含在这两个集合中的文档”（并集），

NOT

或

 -

的意思是“从另一组中减去该组”。

给出您的第一个查询，它被解析为：

-a:d  # subtract the set of documents matched by "a:d" from the existing set
+b:2  # AND includes the set of documents matched by `b:2` (+ => needs to be present)
+(-c:[* TO *])"  # AND include the set of documents matched by:
                 #   the empty set minus the set of documents matching `c:[* TO *]`

如果我们忽略 Solr 有时会在您的查询中添加所有匹配文档的集合的前缀，以便您以表示“所有文档”而不是“无文档”的集合开始，并假设我们始终以空集合开始，我们可以看到为什么我们没有收到文件：

current: empty_set, statement: -a:d
  subtract the set matched by `a:d`
  # result is still an empty set
  
current: empty_set, statement +b:2
  add the set matched by `b:2`
  # result is a set that contains documents with `b:2`

current: (b:2), statement +(-c:[* TO *])
  INTERSECT it (since its given with +) with the set represented by:
    empty_set
      subtract the set matched by `c:[* TO *]` (documents with a value in `c`) 
      # result is still an empty set
  # result is an empty set, since the internal part is an empty set and we require a match through `+`

如果我们将最后一步更改为首先包含所有文档 (

*:*

):

current: (b:2), statement +(*:* -c:[* TO *])
  INTERSECT (since its given with +) with the set represented by:
    all_documents
      subtract the set matched by `c:[* TO *]` (documents with a value in `c`) 
      # result is all_documents except those with a value in `c`)

  # resulting set: documents with `b:2` and without a value for `c`.
  #                                     ^- not "either set", but those that are in both sets - so an intersection