如何将json值转换为prometheus？

Question

运行此命令后：

mtr -jnbz www.google.com |jq .report.hubs|jq -r 'keys_unsorted[] as $k | "\(.[$k])"'

我得到的结果是：

{"count":"1","host":"1.1.1.1","ASN":"AS???","Loss%":0,"Snt":10,"Last":36.28,"Avg":39.43,"Best":34.77,"Wrst":62.37,"StDev":8.15}
{"count":"2","host":"2.2.2.2","ASN":"AS???","Loss%":100,"Snt":10,"Last":0,"Avg":0,"Best":0,"Wrst":0,"StDev":0}

如何获得此结果(prometheus格式)：

mtr_loss{"count"="1","host"="1.1.1.1","ASN"="AS???"} 0
mtr_snt{"count"="1","host"="1.1.1.1","ASN"="AS???"} 10
mtr_last{"count"="1","host"="1.1.1.1","ASN"="AS???"} 36.28
mtr_avg{"count"="1","host"="1.1.1.1","ASN"="AS???"} 39.43
mtr_best{"count"="1","host"="1.1.1.1","ASN"="AS???"} 34.77
mtr_wrst{"count"="1","host"="1.1.1.1","ASN"="AS???"} 67.37
mtr_stdev{"count"="1","host"="1.1.1.1","ASN"="AS???"} 8.15

对于第二个字符串，依此类推

我将非常感谢您的任何提示和技巧

问候

Answer 1

下面的代码可能更容易遵循、维护和重用：

# produce the {"k"="value", ...} representation:
def kv:
  . as $in
  | reduce keys_unsorted[] as $k ([]; . + ["\"\($k)\"=\"\($in[$k])\""] )
  | join(",")
  | "{" + . + "}" ;

# downcase and remove %
def ht($s):
  keys_unsorted[] as $key
  | .[$key] as $value
  | "mtr_\($key|ascii_downcase|gsub("%";""))\($s) \($value)";

({count,host,ASN} | kv) as $s
| {"Loss%", Snt, Last, Avg, Best, Wrst, StDev}
| ht($s)

Answer 2

这很简单:)：

 mtr -jnbz  www.google.com | 
 tr -d "%" |
 jq -r '.report.hubs[] 
        | "mtr_loss{count=\"\(.count)\",host=\"\(.host)\",asn=\"\(.ASN)\"} \(.Loss)\nmtr_sent{count=\"\(.count)\",host=\"\(.host)\",asn=\"\(.ASN)\"} \(.Snt)\nmtr_last{count=\"\(.count)\",host=\"\(.host)\",asn=\"\(.ASN)\"} \(.Last)\nmtr_avg{count=\"\(.count)\",host=\"\(.host)\",asn=\"\(.ASN)\"} \(.Avg)\nmtr_best{count=\"\(.count)\",host=\"\(.host)\",asn=\"\(.ASN)\"} \(.Best)\nmtr_wrst{count=\"\(.count)\",host=\"\(.host)\",asn=\"\(.ASN)\"} \(.Wrst)\nmtr_stdev{count=\"\(.count)\",host=\"\(.host)\",asn=\"\(.ASN)\"} \(.StDev)" '