我有以下数据：

Question

。 我正在尝试将

$univerved

$ usevers的计数总结为：

$user

user1 unreserved =1
user1 reserved =4
user2 unreserved =0
user2 reserved =1
user3 unreserved =1
user3 reserved =0

等。 我还需要获得$未保留和$保留的总和一起计数，在这种情况下为110. 我尝试了下面的尝试；

grep handle | awk -F' ' '$1!=p{ if (NR>1) print p, s; p=$1; s=0} {s+=$12} END{print p, s}' | sort | uniq -c | sort -n

pureawk解决方案：

awk '/handle/ {
  u = $3;             # user in field 3
  n12 = $4;           # numbers 1 and 2 in field 4
  split(n12 ,n, "/"); # split numbers separated by slash into array
  un[u] += n[1];      # add to unreserved for user
  re[u] += n[2];      # add to reserved for user
  s += n[1]+n[2];     # add to total sum
}
END {
  for(i in re)        # loop over all associative array indices
  {
    # both arrays have the same index values
    printf "%s unreserved =%d\n", i, un[i];
    printf "%s reserved =%d\n", i, re[i];
  }
  printf "total =%d\n",s;
}'

（当然，您可以省略所有评论并将所有评论放入一行。）

您可以将输入馈送到

stdin

Answer 1

在您的示例输入中，我将获得此输出：

user1 unreserved =1
user1 reserved =4
user2 unreserved =0
user2 reserved =1
user3 unreserved =1
user3 reserved =0
user4 unreserved =1
user4 reserved =100
user5 unreserved =1
user5 reserved =1
total =110