Chris's questions

输入文件：

col01 20083d70-1bf0-40f4-9df9-6247d587c085 text text
col02 d233d76d-e60d-442e-9b44-f0e1dc9ebefb text text
col03 20083d70-1bf0-40f4-9df9-6247d587c085 text text
col04 20083d70-1bf0-40f4-9df9-6247d587c085 text text
col05 b26063fa-93e4-414d-82f2-7c2924c81cc6 text text
col06 dc3e2f9b-05bf-49b4-a63e-c4bc09524eb3 text text
col07 d9f5cf1d-a8f9-488d-b088-b7ad00b67ec6 text text
col08 b26063fa-93e4-414d-82f2-7c2924c81cc6 text text
col09 d9a558ff-494a-4ada-8ec2-e9b2f9178599 text text
col10 b26063fa-93e4-414d-82f2-7c2924c81cc6 text text
col11 b26063fa-93e4-414d-82f2-7c2924c81cc6 text text
col12 4d78f804-18f6-4cc8-98a3-c4aedc60aebf text text
col13 0fe487fe-2ff2-4ca4-adf8-de3c19115dde text text
col14 581207a4-f4a5-4373-806b-0e1f7c298bb1 text text

参考文件：

0fe487fe-2ff2-4ca4-adf8-de3c19115dde group1
20083d70-1bf0-40f4-9df9-6247d587c085 group2
4d78f804-18f6-4cc8-98a3-c4aedc60aebf group3
581207a4-f4a5-4373-806b-0e1f7c298bb1 group4
b26063fa-93e4-414d-82f2-7c2924c81cc6 group5
d233d76d-e60d-442e-9b44-f0e1dc9ebefb group6
d9a558ff-494a-4ada-8ec2-e9b2f9178599 group7
d9f5cf1d-a8f9-488d-b088-b7ad00b67ec6 group8
dc3e2f9b-05bf-49b4-a63e-c4bc09524eb3 group9

预期输出：

col01 20083d70-1bf0-40f4-9df9-6247d587c085(group2) text text
col02 d233d76d-e60d-442e-9b44-f0e1dc9ebefb(group6) text text
col03 20083d70-1bf0-40f4-9df9-6247d587c085(group2) text text
col04 20083d70-1bf0-40f4-9df9-6247d587c085(group2) text text
col05 b26063fa-93e4-414d-82f2-7c2924c81cc6(group5) text text
col06 dc3e2f9b-05bf-49b4-a63e-c4bc09524eb3(group9) text text
col07 d9f5cf1d-a8f9-488d-b088-b7ad00b67ec6(group8) text text
col08 b26063fa-93e4-414d-82f2-7c2924c81cc6(group5) text text
col09 d9a558ff-494a-4ada-8ec2-e9b2f9178599(group7) text text
col10 b26063fa-93e4-414d-82f2-7c2924c81cc6(group5) text text
col11 b26063fa-93e4-414d-82f2-7c2924c81cc6(group5) text text
col12 4d78f804-18f6-4cc8-98a3-c4aedc60aebf(group3) text text
col13 0fe487fe-2ff2-4ca4-adf8-de3c19115dde(group1) text text
col14 581207a4-f4a5-4373-806b-0e1f7c298bb1(group4) text text

现在我可以通过循环实现这sed -i一点while：

$ cat reference|while read a b;do sed -i s/"$a"/"$a($b)"/ input;done

是否可以以更简单的方式完成，仅使用 awk 而不使用 while 循环，在单个 awk 命令处理中使用输入和参考文件仅更改第二列中的值？

输入json文件：

[
  {
    "id": "1e9b2215-9efd-67cc-9113-1e21c74f09d3",
    "login": "u0159",
    "name": "Pat ABC",
    "role": "User",
    "groups": null,
    "disabled": false,
    "lLoginTime": 1567158950468,
    "lLoginFrom": "192.168.100.11"
  },
  {
    "id": "1e9b2215-9efd-67cc-9113-2b21c74f09d3",
    "login": "u0154",
    "name": "Roman ABC",
    "role": "User",
    "groups": [
      {
        "id": "1e8dc82a-596f-623c-8bc5-100000000000",
        "name": "Default ABC"
      },
      {
        "id": "1e99833f-9876-65d0-9d24-237228b9d9e0",
        "name": "AGE"
      }
    ],
    "disabled": false,
    "lLoginTime": 1567158950468,
    "lLoginFrom": "192.168.100.11"
  },
  {
    "id": "1e9b2215-9c2b-64f4-a0f1-47abdab7ae5f",
    "login": "u0155",
    "name": "Tomas CDE",
    "role": "User",
    "groups": [
      {
        "id": "1e8dc82a-596f-623c-8bc5-100000000000",
        "name": "ZZZ"
      },
      {
        "id": "1e8dc82a-596f-623c-8bc5-100000000000",
        "name": "Default ABC"
      },
      {
        "id": "1e99234f-9876-65d0-9d24-534528b9d9e0",
        "name": "NTE"
      }
    ],
    "disabled": false,
    "lLoginTime": 1566302653093,
    "lLoginFrom": "192.168.100.16"
  }
]

我想扩展以下过滤器：

$ jq -r '.[]|[.id, .login, ( if .groups == null then "n/a" else ([.groups[].name]|join("|")) end )]|@tsv' /tmp/test__
1e9b2215-9efd-67cc-9113-1e21c74f09d3    u0159   n/a
1e9b2215-9efd-67cc-9113-2b21c74f09d3    u0154   Default ABC|AGE
1e9b2215-9c2b-64f4-a0f1-47abdab7ae5f    u0155   ZZZ|Default ABC|NTE

从由 | 分隔的组列表中删除所有“默认 ABC” 输出时

我尝试了额外的 if 但失败并出现错误：

$ jq -r '.[]|[.id, .login, ( if .groups == null then "n/a" else ( if .groups[].name != "Default ABC" then [.groups[].name]|join("|")) end ) end )]|@tsv' /tmp/test__
jq: error: syntax error, unexpected ')' (Unix shell quoting issues?) at <top-level>, line 1:

感谢任何提示...

示例 json 的预期输出为：

1e9b2215-9efd-67cc-9113-1e21c74f09d3    u0159   n/a
1e9b2215-9efd-67cc-9113-2b21c74f09d3    u0154   AGE
1e9b2215-9c2b-64f4-a0f1-47abdab7ae5f    u0155   ZZZ|NTE

更新： oliv 提供的解决方案：

$ jq -r '.[]|[.id,.login,if .groups then (.groups|map(select(.name != "Default ABC").name)|join("|")) else "n/a" end]|@tsv' /tmp/test__
1e9b2215-9efd-67cc-9113-1e21c74f09d3    u0159   n/a
1e9b2215-9efd-67cc-9113-2b21c74f09d3    u0154   AGE
1e9b2215-9c2b-64f4-a0f1-47abdab7ae5f    u0155   ZZZ|NTE

输入json：

[
  {
    "id": "1e9b2215-9efd-67cc-9113-2b21c74f09d3",
    "login": "u0154",
    "name": "Roman ABC",
    "role": "User",
    "groups": [
      {
        "id": "1e99833f-9876-65d0-9d24-237228b9d9e0",
        "name": "AGE"
      }
    ],
    "disabled": false,
    "lLoginTime": 1567158950468,
    "lLoginFrom": "192.168.100.11"
  },
  {
    "id": "1e9b2215-9c2b-64f4-a0f1-47abdab7ae5f",
    "login": "u0155",
    "name": "Tomas CDE",
    "role": "User",
    "groups": [
      {
        "id": "1e8dc82a-596f-623c-8bc5-100000000000",
        "name": "ZZZ"
      },
      {
        "id": "1e99234f-9876-65d0-9d24-534528b9d9e0",
        "name": "NTE"
      }
    ],
    "disabled": false,
    "lLoginTime": 1566302653093,
    "lLoginFrom": "192.168.100.16"
  }
]

我需要打印制表符分隔的 .id、.login 和由 | 分隔的所有用户组 （管道）或逗号

我现在拥有的：

$ jq -r '.[]|[.id, .login, ([.groups[].name]|@tsv)]|@tsv' test.json
1e9b2215-9efd-67cc-9113-2b21c74f09d3    u0154   AGE
1e9b2215-9c2b-64f4-a0f1-47abdab7ae5f    u0155   ZZZ\tNTE

我需要修改过滤器，以便最后一列上的组用空格分隔或|代替当前\t

任何人都可以解释为什么“sort -nu”在下面的示例中只返回 1 个唯一行？

$ cat /tmp/test_
1e971713-31e4-62d0-a3d1-37c0610c10e5
1e97d59f-bc32-6126-b7b1-0b52ddd35273
1e971713-31e4-62d0-a3d1-37c0610c10e5
$ sort -nu /tmp/test_
1e971713-31e4-62d0-a3d1-37c0610c10e5
$

像“sort -u”这样的输出不应该有两条独特的行吗？

$ sort -u /tmp/test_
1e971713-31e4-62d0-a3d1-37c0610c10e5
1e97d59f-bc32-6126-b7b1-0b52ddd35273
$

${PIPESTATUS[0]在当前 shell 中设置变量时，如何读取命令？有没有办法以某种方式将它传递给当前的 shell？

我正在设置变量MAIL=$(ldapsearch.... | find_for_mail)，并想确定ldapsearch在我确定 LDAP 中确实不存在邮件属性之前是否没有失败。

$ true | false
$ echo ${PIPESTATUS[0]} ${PIPESTATUS[1]}
0 1
$ Z=$(true | false)
$ echo ${PIPESTATUS[0]}
1                        <--- why 1?
$ echo ${PIPESTATUS[1]}

$ Z=$(true | false;echo ${PIPESTATUS[0]} ${PIPESTATUS[1]})
$ echo $Z
0 1
$ Z=$(true | false;C="${PIPESTATUS[0]} ${PIPESTATUS[1]}")
$ echo $C

$ Z=$(true | false;export C="${PIPESTATUS[0]} ${PIPESTATUS[1]}")
$ echo $C

输入json：

[
  {
    "name": "cust1",
    "grp": [
      {
        "id": "46",
        "name": "BA2"
      },
      {
        "id": "36",
        "name": "GA1"
      },
      {
        "id": "47",
        "name": "NA1"
      },
      {
        "id": "37",
        "name": "TR3"
      },
      {
        "id": "38",
        "name": "TS1"
      }
    ]
  }
]

如何在@tsv 输出中获取用空格分隔的 .grp[] 名称？

$ jq -r '.[]|[.name, (.grp//[{"name":"-"}]|.[]|" ",.name)]|@tsv' test_inp
cust1           BA2             GA1             NA1             TR3             TS1

上述样本的预期输出：

cust[TAB]BA2[SPACE]GA1[SPACE]NA1[SPACE]TR3[SPACE]TS1

输入json：

{
  "id": "3885",
  "login": "050111",
  "lastLoginTime": 1529730115000,
  "lastLoginFrom": "192.168.66.230"
}
{
  "id": "3898",
  "login": "050112",
  "lastLoginTime": null,
  "lastLoginFrom": null
}

当 lastlogintime 存在时过滤器起作用：

$ jq -jr --arg ARG1 050111 'select(.login==$ARG1)|"user:", " ",.login,"\n","lastlogintime:", " ",(.lastLoginTime / 1000 | strftime("%Y-%m-%d %H:%M UTC")),"\n","lastloginfrom:", " ",.lastLoginFrom // "-","\n"' test_json3
user: 050111
lastlogintime: 2018-06-23 05:01 UTC
lastloginfrom: 192.168.66.230

但是当 lastlogintime 不存在时出现问题（不打印“-”）：

$ jq -jr --arg ARG1 050112 'select(.login==$ARG1)|"user:", " ",.login,"\n","lastlogintime:", " ",(.lastLoginTime / 1000 | strftime("%Y-%m-%d %H:%M UTC")),"\n","lastloginfrom:", " ",.lastLoginFrom // "-","\n"' test_json3
user: 050112
jq: error (at test_json3:12): null (null) and number (1000) cannot be divided

如何跳过空lastlogintimes的除法操作？

json输入：

[
  {
    "name": "cust1",
    "grp": [
      {
        "id": "46",
        "name": "BA2"
      },
      {
        "id": "36",
        "name": "GA1"
      },
      {
        "id": "47",
        "name": "NA1"
      },
      {
        "id": "37",
        "name": "TR3"
      },
      {
        "id": "38",
        "name": "TS1"
      }
    ]
  }
]

预期，输出是两行：

name: cust1
groups: BA2 GA1 NA1 TR3 TS1

我试图构建过滤器但没有成功..

$ jq -r '.[]|"name:", .name, "groups:", (.grp[]|[.name]|@tsv)' test_json
name:
cust1
groups:
BA2
GA1
NA1
TR3
TS1

更新：下面提供的解决方案工作正常，但我没有预测不存在组的情况：

[
  {
    "name": "cust1",
    "grp": null
  }
]

在这种情况下，提供的解决方案会返回错误：

$ jq -jr '.[]|"name:", " ",.name, "\n","groups:", (.grp[]|" ",.name),"\n"' test_json2
name: cust1
jq: error (at test_json2:6): Cannot iterate over null (null)

任何解决方法表示赞赏。

输入json：

{
  "id": "3885",
  "login": "050111",
  "lastLoginTime": 1529730115000,
  "lastLoginFrom": "192.168.66.230"
}
{
  "id": "3898",
  "login": "050112",
  "lastLoginTime": null,
  "lastLoginFrom": null
}

我想以制表符分隔格式获取登录、lastLoginTime 和 lastLoginFrom 的输出：

050111  1529730115000   192.168.66.230
050112              -                -

使用下面的 jq 过滤器，我得到的输出没有“空”值，我可以用“-”替换

$ jq -r '.|[.login, .lastLoginTime, .lastLoginFrom]|@tsv' test_json
050111  1529730115000   192.168.66.230
050112

有没有其他方法可以为这些空值打印“-”？

输入json：

[
  {
    "id": "89",
    "hostname": "abcd"
  },
  {
    "id": "89",
    "hostname": "babcd"
  }
]

如何修改以下过滤器以仅输出以“abcd”开头的主机名？

$ jq -r '.[]|select(.hostname | contains("abcd"))' jjjj
{
  "id": "89",
  "hostname": "abcd"
}
{
  "id": "89",
  "hostname": "babcd"
}
$ jq -r '.[]|select(.hostname | contains("^abcd"))' jjjj
$

有没有办法更改密码已过期时在密码更改过程中显示的默认提示？在 centos 和 rhel 上？I'd like to change (current) UNIX password:to Your previous password:orEnter expired password: 原来的提示误导了很多用户，他们在这个步骤中询问他们应该输入什么密码（他们在登录时已经输入了旧的过期密码，就在显示这个过期警告和提示之前，它再次要求他们输入老在(current) UNIX password:）

Changing password for user.
(current) UNIX password:
New password:
Retype new password: