如何计算由 2 个字段分隔符（“：” 和 “，”）分隔的行中单列的字段数？

您可以为此使用：split

awk -F: '$1 == "adm1Group" {print split($NF, a, /,/)}' file
3


awk -F: '$1 == "conda" {print split($NF, a, /,/)}' file
9

或者将它们全部打印在一起：

awk -F: '{print split($NF, a, /,/), "no of users in adm1Group:", $1}' file

2 no of users in adm1Group: adm
3 no of users in adm1Group: admins
1 no of users in adm1Group: ssl-cert
2 no of users in adm1Group: ala2
9 no of users in adm1Group: conda
3 no of users in adm1Group: adm1Group
1 no of users in adm1Group: docker

使用您展示的示例和尝试，请尝试以下代码。这将打印Input_file的每个组名称中存在的用户总数。awk

awk -F':' '
{
  num=0
  arr1[$1]=num=split($NF,arr2,",")
}
END{
  for(i in arr1){
    print "Group " i " has " arr1[i] " users."
  }
}
' Input_file

解释：为上述代码添加详细说明。

awk -F':' '                          ##Starting awk program where setting field separator as : here.
{
  num=0                              ##Setting num as 0 here.
  arr1[$1]=num=split($NF,arr2,",")   ##Creating arr1 array with index of $1 and has value of num, which contains total number of total elements in arr2 with delimiter of , here.
}
END{                                 ##Starting END block of this program from here.
  for(i in arr1){                    ##Traversing through arr1 here.
    print "Group " i " has " arr1[i] " users."  ##printing group name and its value(how many times users came for that group).
  }
}
' Input_file                         ##Mentioning Input_file name here.

如何计算每条线路的用户数？或单个 线？

我会用 GNU 来计算第 4 个字段内的数量，然后增加它，让内容为AWK,1file.txt

adm:x:4:syslog,adm1
admins:x:1006:adm2,adm12,manuel
ssl-cert:x:122:postgres
ala2:x:1009:aceto,salvemini
conda:x:1011:giovannelli,galise,aceto,caputo,haymele,salvemini,scala,adm2,adm12
adm1Group:x:1022:adm2,adm1,adm3
docker:x:998:manuel

然后

awk 'BEGIN{FS=":"}{printf "N of users in %s is %s\n", $1, gsub(/,/,"",$4)+1}' file.txt

给出输出

N of users in adm is 2
N of users in admins is 3
N of users in ssl-cert is 1
N of users in ala2 is 2
N of users in conda is 9
N of users in adm1Group is 3
N of users in docker is 1

解释：我告诉GNU字段分隔符（）是。对于我使用的每一行，它的作用类似于填充模板和打印，对于填充，我使用第 1 个字段 （） 和当命令使用空字符串 （） 替换时完成的更改次数 gsub 函数在第 4 个字段 （） 增加 1（因为姓氏没有尾随）。请注意，这确实会改变（删除字符），但对于此任务，所述副作用无关紧要。请注意，使用时需要隐式提供换行符 （），而不是 .AWKFS:printf$1,""$4,$4,printf\nprint

（在 Gawk 4.2.1 中测试）

使用 或 作为字段分隔符，然后打印字段数减去 3 个前导字段：:,

awk -F'[:,]' '{print $1, NF - 3}' file

awk -F'[:,]' -v group=conda '$1 == group {print NF - 3}' file

{m,g}awk '$!NF=sprintf("%20s\47s user(s) count = %\0478.f",$!_,NF-!_)' FS=':.+:|,'

          adm's  user(s) count =         2
       admins's  user(s) count =         3
     ssl-cert's  user(s) count =         1
         ala2's  user(s) count =         2
        conda's  user(s) count =         9
    adm1Group's  user(s) count =         3
       docker's  user(s) count =         1

稍加修改，现在完整的用户列表将在尾部提供。具体来说，小粗体项目 - 现在它正在覆盖而不是 ：：$1$0

{m,g}awk ' $!_ = sprintf("%15s\47s user(s) count = %\0476.f",$!_,NF-!_)' FS=':.+:|,'

        adm's user(s) count =      2 syslog adm1
     admins's user(s) count =      3 adm2 adm12 manuel
   ssl-cert's user(s) count =      1 postgres
       ala2's user(s) count =      2 aceto salvemini
      conda's user(s) count =      9 giovannelli galise aceto caputo haymele salvemini scala adm2 adm12
  adm1Group's user(s) count =      3 adm2 adm1 adm3
     docker's user(s) count =      1 manuel