我尝试用Haskell(Aeson)向Elasticsearch发出Http请求。
Elasticsearch主体看起来如此:
{
"query": {
"function_score": {
"query": {
"bool": {
"should": [
{"term": {"word_n_gram": "str"}},
{"term": {"word_n_gram": "not"}}
]
}
},
"functions": [
{
"script_score": {
"script": {
"lang": "groovy",
"file": "test-score",
"params": {
"boostBy": {
"str": 1,
"not": 1
}
}
}
}
}
]
}
}
}
它运作正常。
所以,我做了#34;等同于#34;在哈斯克尔:
data QueryRequest = QueryRequest {
query :: Query
} deriving (Eq, Generic, Show)
instance ToJSON QueryRequest
data Query = Query {
function_score :: FunctionScore
} deriving (Eq, Generic, Show)
instance ToJSON Query
data FunctionScore = FunctionScore {
queryIn :: QueryIn
, functions :: [Functions]
} deriving (Eq, Generic, Show)
instance ToJSON FunctionScore
data QueryIn = QueryIn {
bool :: BoolQuery
} deriving (Eq, Generic, Show)
instance ToJSON QueryIn
data BoolQuery = BoolQuery {
should :: [ShouldQuery]
} deriving (Eq, Generic, Show)
依旧......
重点是,在haskell中,我不能有两次"查询"声明我写queryIn
的原因但由于我正在提出请求,而Elasticsearch正在等待query
两次,我收到此错误:
FailureResponse {responseStatus = Status {statusCode = 400,statusMessage =" Bad Request"},responseContentType = application / json; charset = UTF-8,responseBody =" {\"错误\":{\" ROOT_CAUSE \":[{\"类型\":\" parsing_exception \" \&#34 ; reason \":\"没有[query]注册[queryIn] \",\" line \":1,\" col \& #34; 39}],\"类型\":\" parsing_exception \" \"原因\":\"无[query]已注册[queryIn] \",\" line \":1,\" col \":39},\" status \ " 400}"}
这是一个逻辑错误。但我不知道如何解决它......
我制作" RequestQuery"这样:
toElasticSearchQuery :: T.Text -> RW.QueryRequest
toElasticSearchQuery word =
RW.QueryRequest {
RW.query = RW.Query {
RW.function_score = RW.FunctionScore {
RW.queryIn = RW.QueryIn {
RW.bool = RW.BoolQuery {
RW.should = toShouldQueryList (splitInNGrams word)
}
},
RW.functions = [
RW.Functions {
RW.scriptScore = RW.ScriptScore {
RW.script = RW.Script {
RW.lang = scriptLang,
RW.file = scriptFile,
RW.params = RW.Params {
RW.boostBy = fixGramConter (splitInNGrams word)
}
}
}
}
]
}
}
}
当然,我无法在RW.FunctionScore中写入RW.query。 我不知道如何解决它,因为响应不会有问题,但对于请求,这是一个问题。
也许有人曾经尝试过类似的东西。
答案 0 :(得分:2)
我修好了。我必须从toJSON
aeson
个实例
instance ToJSON FunctionScore where
toJSON (FunctionScore q f) = object ["query" .= q, "functions" .= f]
它运作良好。