我对所有没有返回200个响应(在特定时间间隔内)的API都很感兴趣。
我基本上需要这个:
select url from api_log
except/minus
select url from api_log where status='200'
我正在尝试翻译成ES:
select url, status, count(*) from api_log
group by url, status
ES样本数据
{
"_index": "api_log",
"_type": "_doc",
"_id": "1",
"_version": 1,
"_score": 1,
"_source": {
"in_time": "2019-05-13T17:20:51.108945",
"out_time": "2019-05-13T17:20:51.145549",
"duration": 36.6041660308838,
"status": "200",
"url": "/api/myFirstAPI"
}
}
,
{
"_index": "api_log",
"_type": "_doc",
"_id": "2",
"_version": 1,
"_score": 1,
"_source": {
"in_time": "2019-05-13T17:20:57.915694",
"out_time": "2019-05-13T17:20:57.941989",
"duration": 26.2949466705322,
"status": "403",
"url": "/api/mySecondAPI"
}
},
{
"_index": "api_log",
"_type": "_doc",
"_id": "3",
"_version": 1,
"_score": 1,
"_source": {
"in_time": "2019-05-13T17:22:35.274372",
"out_time": "2019-05-13T17:22:35.288944",
"duration": 14.5719051361084,
"status": "400",
"url": "/api/myFirstAPI"
}
}
对于上述数据,我希望结果网址为{'/ api / mySecondAPI'}。
仅用AGG进行请求/响应
POST /api_log/_search
{
"size": 0,
"aggs": {
"url": {
"terms": {
"field": "url.keyword"
},
"aggregations": {
"status": {
"terms": {
"field": "status.keyword"
}
}
}
}
}
}
对以上请求的答复
{
"took" : 880,
"timed_out" : false,
"_shards" : {
"total" : 1,
"successful" : 1,
"skipped" : 0,
"failed" : 0
},
"hits" : {
"total" : {
"value" : 10000,
"relation" : "gte"
},
"max_score" : null,
"hits" : [ ]
},
"aggregations" : {
"url" : {
"doc_count_error_upper_bound" : 0,
"sum_other_doc_count" : 394668,
"buckets" : [
{
"key" : "/api/myFirstRequest",
"doc_count" : 1352845,
"status" : {
"doc_count_error_upper_bound" : 0,
"sum_other_doc_count" : 0,
"buckets" : [
{
"key" : "200",
"doc_count" : 1187611
},
{
"key" : "302",
"doc_count" : 139932
},
{
"key" : "401",
"doc_count" : 22615
},
{
"key" : "500",
"doc_count" : 2250
},
{
"key" : "403",
"doc_count" : 437
}
]
}
},
...
...
...
从上面我需要过滤掉所有不包含状态为“ 200”的子存储桶的存储桶(URL)
我走了这么远。看起来很近,但是很远。...似乎无法弄清楚类型字段中应该包含什么。
带过滤器的请求
POST /api_log/_search
{
"size": 0,
"aggs": {
"page_name": {
"terms": {
"field": "url.keyword"
},
"aggregations": {
"status": {
"terms": {
"field": "status.keyword"
}
}
}
}
},
"post_filter": {
"bool": {
"must_not": [
{
"has_child" : {
"type" : "?????",
"query" : {
"term" : {"status" : "200"}
}
}
}
]
}
}
}
示例输入(来自apache日志):
t1 /api/FirstAPI 200 <-- Eliminate First API completely
t2 /api/FirstAPI 400
t3 /api/FirstAPI 403
t4 /api/SecondAPI 403
t5 /api/SecondAPI 400
t6 /api/ThirdAPI 500
t7 /api/ThirdAPI 500
t8 /api/SecondAPI 200 <---Eliminate Second API completely
t9 /api/ThirdAPI 500
t10 /api/ThirdAPI 403
鉴于上述输入,我只希望在时间范围t1-t10中从未给出200响应的页面。
预期结果
因此,输出应为 / api / ThirdAPI
如果我先过滤掉200个,然后再应用Agg,我将获得全部三个API。那不是我想要的。
答案 0 :(得分:0)
如果我理解正确,您只想从聚合中排除200。我没有在这里使用vowels = ['a', 'e', 'i', 'o', 'u']
res = ""
for letter in word:
if letter in vowels:
continue
res += letter
return res
的理由。您可以使用术语汇总。
Exclude or filter the status value in aggregations。这将计算所有post_filter
响应并添加到200
字段中,但会在聚合响应中排除存储桶,并且不会显示doc_count
200
替代:
根据您的输入,您似乎希望将POST /api_log/_search
{
"size": 0,
"aggs": {
"url": {
"terms": {
"field": "url.keyword"
},
"aggregations": {
"status": {
"terms": {
"field": "status.keyword",
"exclude": "200"
}
}
}
}
}
}
作为结果集的一部分(因为您正在使用post_filter),但是如果没有,那么如果不是这种情况,这是另一种方法。汇总是在查询响应上完成的;因此,如果您使用bool query从结果集中排除200个桶,则不会有状态为200的存储桶。
200