我有干草堆的问题 - 我不知道如何搜索所有外键符合条件的模型A 。
我的简化模型如下：
Group:
id
Meeting:
group = models.ForeignKey(Group)
day_of_week = models.IntegerField()
hour = models.IntegerField()
length = models.IntegerField()
基本上，一个小组可以召开很多会议，用户应该能够搜索所有会议都在给定时间范围内的小组。例如：
Group(1)
Meeting(day_of_week=Monday, hour=9, length=2)
Group(2)
Meeting(day_of_week=Monday, hour=10, length=1)
Meeting(day_of_week=Tuesday, hour=8, length=2)
Group(3)
Meeting(day_of_week=Monday, hour=10, length=1)
Meeting(day_of_week=Wednesday, hour=12, length=1)
并搜索：＆＃34;星期一从8到11＆＃34;，＆＃34;星期二，从12到14 （下午2点）＆＃34;，＆＃34 ;星期三，6点到17点（下午5点）＆＃34;应该返回第1组和第3组，因为来自这些组的所有会议都包含在用户指定的范围内，并且不返回第2组，因为第二次会议不在给定范围内（第一次会议是在第一次会议中）。
如果我要编写SQL，我可能会选择＆＃34;选择匹配会议的数量和所有会议的数量（如果这些数字相等） - ＆gt;然后所有会议都会举行：
SELECT g.id,
count(m2.id)
FROM groups g
JOIN meetings m2 ON m2.group_id = g.id
AND ((m2.day_of_week = 0 -- monday
AND m2.hour >= 8
AND m2.length<=3)
OR (m2.day_of_week=1 -- tuesday
AND m2.hour >= 12
AND m2.length<=2)
OR (m2.day_of_week=2 -- wednesday
AND m2.hour >= 6
AND m2.length<=11))
GROUP BY g.id
HAVING count(m2.id) =
(SELECT count(*)
FROM meetings
WHERE meetings.group_id=g.id);
但是我们使用haystack + elastic搜索索引，我完全不知道如何压缩模型来索引和编写查询。任何人都可以帮我吗？
答案 0 :(得分：1)
您可能需要压缩文档，使所有文档都必须是包含组信息的会议。
** ES 5的解决方案**
您的文档的映射将是：
PUT /meetings
{
"mappings": {
"meeting": {
"properties": {
"groupId": {
"type": "integer"
},
"dayOfWeek": {
"type": "integer"
},
"hourRange": {
"type": "integer_range"
}
}
}
}
}
然后你的五个文件看起来像这样：
POST /meetings/meeting/_bulk
{"index": {}}
{"groupId": 1, "dayOfWeek": 0, "hourRange": {"gte": 9, "lte": 11}}
{"index": {}}
{"groupId": 2, "dayOfWeek": 0, "hourRange": {"gte": 10, "lte": 11}}
{"index": {}}
{"groupId": 2, "dayOfWeek": 1, "hourRange": {"gte": 8, "lte": 10}}
{"index": {}}
{"groupId": 3, "dayOfWeek": 0, "hourRange": {"gte": 10, "lte": 11}}
{"index": {}}
{"groupId": 3, "dayOfWeek": 2, "hourRange": {"gte": 12, "lte": 13}}
最后，查询将如下所示：
POST /meetings/meeting/_search
{
"query": {
"bool": {
"should": [
{
"bool": {
"must": [
{
"term": {
"dayOfWeek": 0
}
},
{
"range": {
"hourRange": {
"gte": "8",
"lte": "11",
"relation": "within"
}
}
}
]
}
},
{
"bool": {
"must": [
{
"term": {
"dayOfWeek": 1
}
},
{
"range": {
"hourRange": {
"gte": "12",
"lte": "14",
"relation": "within"
}
}
}
]
}
},
{
"bool": {
"must": [
{
"term": {
"dayOfWeek": 2
}
},
{
"range": {
"hourRange": {
"gte": "6",
"lte": "17",
"relation": "within"
}
}
}
]
}
}
]
}
}
}
** ES的解决方案<5 **
PUT /meetings
{
"mappings": {
"meeting": {
"properties": {
"groupId": {
"type": "integer"
},
"dayOfWeek": {
"type": "integer"
},
"start": {
"type": "integer"
},
"end": {
"type": "integer"
}
}
}
}
}
然后你的五个文件看起来像这样：
POST /meetings/meeting/_bulk
{"index": {}}
{"groupId": 1, "dayOfWeek": 0, "start": 9, "end": 11}
{"index": {}}
{"groupId": 2, "dayOfWeek": 0, "start": 10, "end": 11}
{"index": {}}
{"groupId": 2, "dayOfWeek": 1, "start": 8, "end": 10}
{"index": {}}
{"groupId": 3, "dayOfWeek": 0, "start": 10, "end": 11}
{"index": {}}
{"groupId": 3, "dayOfWeek": 2, "start": 12, "end": 13}
最后，查询将如下所示：
POST /meetings/meeting/_search
{
"query": {
"bool": {
"should": [
{
"bool": {
"must": [
{
"term": {
"dayOfWeek": 0
}
},
{
"range": {
"start": {
"gte": "8"
}
}
},
{
"range": {
"end": {
"lte": "11"
}
}
}
]
}
},
{
"bool": {
"must": [
{
"term": {
"dayOfWeek": 1
}
},
{
"range": {
"start": {
"gte": "12"
}
}
},
{
"range": {
"end": {
"lte": "14"
}
}
}
]
}
},
{
"bool": {
"must": [
{
"term": {
"dayOfWeek": 2
}
},
{
"range": {
"start": {
"gte": "6"
}
}
},
{
"range": {
"end": {
"lte": "17"
}
}
}
]
}
}
]
}
}
}
答案 1 :(得分：1)
解决方案的关键是名为nested objects的ElasticSearch功能。幸运的是，所有ES版本都有此功能。嵌套对象在这里是关键，因为会议中的数据是严格关联的。
PUT /myindex
{
"mappings": {
"groups": {
"properties": {
"meetings": {
"type": "nested",
"properties": {
"dayOfWeek": { "type": "integer"},
"start": {"type": "integer"},
"end": {"type": "integer"}
}
},
"groupId": {"type":"integer"}
}
}
}
}
POST /myindex/groups/_bulk
{"index": {}}
{"groupId": 1, "meetings": [{"dayOfWeek": 0, "start": 9, "end": 11}]}
{"index": {}}
{"groupId": 2, "meetings": [{"dayOfWeek": 0, "start": 10, "end": 11}, { "dayOfWeek": 1, "start": 8, "end": 10}]}
{"index": {}}
{"groupId": 3, "meetings": [{"dayOfWeek": 0, "start": 10, "end": 11}, {"dayOfWeek": 2, "start": 12, "end": 13}]}
此时可以清楚地看到会议属于群组，我们将分组搜索。
无法直接编写查询以获取所有嵌套对象满足条件的所有组，但是......可以轻松地将其反转为：获取所有组 none 会议中包含错误时间。
GET /myindex/_search
{
"query": {
"bool": {
"must_not" : {
"nested": {
"path": "meetings",
"filter": {
"bool": {
"must_not": {
"bool": {
"should": [
{
"bool": {
"must": [
{"term" : { "dayOfWeek" : 0 }},
{"range": {"start": {"from":8, "to":11}}},
{"range": {"end": {"from":8, "to":11}}}
]
}
},
{
"bool": {
"must": [
{"term" : { "dayOfWeek" : 1 }},
{"range": {"start": {"from":12, "to":14}}},
{"range": {"end": {"from":12, "to":14}}}
]
}
},
{
"bool": {
"must": [
{"term" : { "dayOfWeek" : 2 }},
{"range": {"start": {"from":6, "to":17}}},
{"range": {"end": {"from":6, "to":17}}}
]
}
}
]
}
}
}
}
}
}
}
}
}
这将返回第1组和第3组。由于其中一次会议与错误的日期时间重叠，因此不会返回第2组。
第二个问题是与Django Haystack的集成，因为默认情况下它不支持引擎特定的功能，如ES中的嵌套字段。幸运的是，我不是唯一一个在django app和someone has already resolved it中需要它的人。