Christian Bongiorno's questions

Tenho esse script e parece que ele printf '%s\0%s\0'sai com o código 1 e não consigo entender o porquê.

Eu até forcei a usar o bash versão 5.2 - mesmos resultados.

Aqui está um script de exemplo e sua saída

#!/usr/bin/env bash
echo $BASH_VERSION
set -x
set -Eeuo pipefail

function prompt_creds {
  local username password
  read -rp "username " username
  read -rsp "password " password 
  printf '%s\0%s\0' "${username}" "${password}"
}

function main() {
  IFS=$'\0' read -r username password < <(prompt_creds )
  echo "name name is ${username}" "my password is ${password}"
}

main

uso e saída

./sample.sh 
3.2.57(1)-release
+ set -Eeuo pipefail
+ main
+ IFS=
+ read -r username password
++ prompt_creds
++ local username password
++ read -rp 'username ' username
username user
++ read -rsp 'password ' password
password ++ printf '%s\0%s\0' user pass
bash-5.2$ echo $?
1

Eu até tentei executá-lo simplesmente na CLI e funcionou:

bash-5.2$ IFS=$'\0' read -r u p < <(printf '%s\0%s\0' user pass)
bash-5.2$ echo $u $p
userpass

Tenho estes dados:

select * from (
    select 'A' as JOB, 15 as errors from dual union all
    select 'B' as JOB, 17 as errors from dual union all
    select 'C' as JOB, 29 as errors from dual union all
    select 'D' as JOB, 27 as errors from dual union all
    select 'E' as JOB, 35 as errors from dual union all
    select 'F' as JOB, 32 as errors from dual union all
    select 'G' as JOB, 75 as errors from dual union all
    select 'H' as JOB, 31 as errors from dual union all
    select 'I' as JOB, 12 as errors from dual union all
    select 'J' as JOB, 10 as errors from dual
)

E, em palavras, eu preciso:The jobs constituting the (top) 60% of errors

Então, neste caso, seria (113):

select sum(errors) * .4 as cut_off from ...

Os resultados finais seriam estes, pois sua soma < 113:

Basicamente, preciso de um filtro que mantenha algum tipo de soma corrente e, depois, descarte tudo quando atingir esse valor.

Tenho essa consulta, que não funciona muito bem e eu preferiria não usar a withdeclaração

with data as (
    select 'A' as JOB, 15 as errors from dual union all
        select 'B' as JOB, 17 as errors from dual union all
        select 'C' as JOB, 29 as errors from dual union all
        select 'D' as JOB, 27 as errors from dual union all
        select 'E' as JOB, 35 as errors from dual union all
        select 'F' as JOB, 32 as errors from dual union all
        select 'G' as JOB, 75 as errors from dual union all
        select 'H' as JOB, 31 as errors from dual union all
        select 'I' as JOB, 12 as errors from dual union all
        select 'J' as JOB, 10 as errors from dual
)
select k.*
from (
    select t.*,
           errors + LAG(errors, 1, 0) OVER (order by errors desc ) previous
    from data t
) k where previous >= (select sum(errors) *.4 from data) order by errors desc

E eu tentei a soma em janela:

select k.*
from (
    select t.*,
           SUM(errors) OVER (
               partition by JOB
               order by errors desc
               RANGE BETWEEN UNBOUNDED PRECEDING
                AND CURRENT ROW
          ) as limit
    from (
        select 'A' as JOB, 15 as errors from dual union all
        select 'B' as JOB, 17 as errors from dual union all
        select 'C' as JOB, 29 as errors from dual union all
        select 'D' as JOB, 27 as errors from dual union all
        select 'E' as JOB, 35 as errors from dual union all
        select 'F' as JOB, 32 as errors from dual union all
        select 'G' as JOB, 75 as errors from dual union all
        select 'H' as JOB, 31 as errors from dual union all
        select 'I' as JOB, 12 as errors from dual union all
        select 'J' as JOB, 10 as errors from dual
    ) t
) k order by errors desc

Aqui está uma amostra dos meus dados

select * from (
    select 'A' as POD, 164 as result from dual union all
    select 'A' as POD, 3 as result from dual union all
    select 'A' as POD, 2 as result from dual union all
    select 'B' as POD, 409 as result from dual union all
    select 'B' as POD, 128 as result from dual union all
    select 'B' as POD, 5 as result from dual union all
    select 'C' as POD, 12391 as result from dual union all
    select 'C' as POD, 624 as result from dual union all
    select 'C' as POD, 405 as result from dual union all
    select 'C' as POD, 26 as result from dual union all
    select 'C' as POD, 3 as result from dual union all
    select 'C' as POD, 2 as result from dual
)

E o que eu quero é que eles sejam classificados por grupo com a maior primeira contagem:

Não tenho certeza de como expressar isso como SQL

Ctem o mais alto resultna primeira linha, depois Btem o próximo mais alto entãoA

Digamos que eu tenha este json:

[
  {
    "first": 12355,
    "second": "abc"
  },
  {
    "first": 89010,
    "second": "def"
  },
  {
    "first": 23423,
    "second": "hij"
  },
  {
    "first": 23456,
    "second": "klm"
  },
  {
    "first": 11111,
    "second": "nop"
  }
]

E eu gostaria (de uma forma genérica) que os mescle em 1 objeto onde o valor de cada chave seja mesclado em um array dos valores correspondentes:

{
    "first" : [12355,89010,23423,23456,11111],
    "second" : ["abc","def","hij","klm","nop"]
}

Eu estava tentando isso, mas não produz nenhuma saída.

reduce .[] as $final (
    {};
    ((. | keys) as $k |
        map(
            ( (.[$k] // []) += ($final[$k] // []))
        )
    )
)

Seguindo neste post

Eu tenho esse script bash que eu esperaria que terminasse completamente ao clicar no falsee, ainda assim, não termina. Eu reconheço que estou fazendo coisas um pouco diferentes do post vinculado.

#!/usr/bin/env bash
set -Eeuo pipefail

function thing() {
  false
  echo "thing1|thing2|thing3"
}
IFS='|' read -r one two three <<< "$(thing )"
echo "${one} ${two} ${three}"
echo "done"

Saída:

thing1 thing2 thing3
done

O ideal seria fazer isso no zsh, mas também recebo um comportamento que não desejo:

#!/usr/bin/env zsh
set -Eeuo pipefail

function thing() {
  false
  echo "thing1|thing2|thing3"
}
IFS='|' read -r one two three <<< "$(thing )"
echo "${one} ${two} ${three}"
echo "done"

saída:

  
done

Como fazer isso funcionar?

Parece que não estou encontrando nenhuma documentação sobre isso (e talvez não conheça o vernáculo do Oracle para isso), mas gostaria de criar uma função que receba uma função: Digamos, faça com que ela me passe uma função de classificação.

Quando olho para um código como este:

LAG(FOO) OVER (PARTITION BY XXX ORDER BY YYY) AS thing

Tenho a sensação de que é exatamente isso que está acontecendo aqui ^^ e a OVERcláusula é o lambda.

É possível fazer? Estou pensando nisso corretamente? Um exemplo prático seria ótimo.