Christian Bongiorno's questions

我有这个脚本，它似乎printf '%s\0%s\0'以代码 1 退出，我不明白为什么。

我甚至强迫它使用 bash 版本 5.2 - 结果相同。

以下是示例脚本及其输出

#!/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

使用和输出

./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

我甚至尝试在 CLI 上简单运行它，并且它可以工作：

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

我有这些数据：

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
)

用文字来说，我需要：The jobs constituting the (top) 60% of errors

因此，在这种情况下，那将是（113）：

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

最终结果将是这样的，因为它们的总和 < 113：

我基本上需要一个过滤器来保持某种运行总和，然后一旦达到该值就丢弃所有内容。

我有这个查询，它不太有效，我不希望使用该with语句

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

我已经尝试过窗口总和：

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

这是我的数据样本

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
)

我想要的是按照首次计数最高的组对它们进行排序：

我甚至不知道如何用 SQL 来表达这个

Cresult在第一行中具有最高值，然后B具有第二高值A

假设我有这个 json：

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

我希望（一种通用形式）将它们合并为 1 个对象，其中每个键的值合并到相应值的数组中：

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

我正在尝试这个，但它根本没有产生任何输出。

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

关注此帖子

我有这个 bash 脚本，我期望它在点击时完全终止false，但它却没有。我意识到我做的事情与链接的帖子略有不同。

#!/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"

输出：

thing1 thing2 thing3
done

理想情况下，我想在 zsh 中执行此操作，但我也遇到了我不想要的行为：

#!/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"

输出：

  
done

如何实现这个功能？

我似乎没有找到有关此的任何文档（也许我不知道 Oracle 对此的解释），但我想创建一个接受函数的函数：比如，让它传递给我一个排序函数。

当我看到这样的代码时：

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

我感觉这正是这里发生的事情 ^^ 并且OVER子句是 lambda。

可以做到吗？我这样想对吗？如果有可行的示例就更好了。