Question

我这里有一个csv文件有3个字段，在第三个字段我想将行分组不超过25个。
鉴于：
＆＃34; CW-03＆＃34 ;;（PE.502）; 24
＆＃34; CW-04＆＃34 ;;（PE.501）; 25
＆＃34; CW-05＆＃34 ;;（PE.518，PE.518-1）; 2
＆＃34; CW-06＆＃34 ;;（PE.517）; 1个
＆＃34; CW-07＆＃34 ;;（PE.516）; 3
＆＃34; CW-08＆＃34 ;;（PE.515）; 1个
＆＃34; CW-09＆＃34 ;;（PE.514）; 1个
＆＃34; CW-10＆＃34 ;;（PE.513，PE.513-1）; 23
＆＃34; CW-11＆＃34 ;;（PE.512）; 8
＆＃34; CW-12＆＃34 ;;（PE.511）; 1个
＆＃34; CW-13＆＃34 ;;（PE.510）; 3
＆＃34; CW-14＆＃34 ;;（PE.509）; 3
＆＃34; CW-15＆＃34 ;;（PE.508，PE.508-1）; 17
＆＃34; CW-16＆＃34 ;;（PE.507）; 1个
＆＃34; CW-17＆＃34 ;;（PE.506）; 1个
＆＃34; CW-18＆＃34 ;;（PE.505）; 1个
＆＃34; CW-19＆＃34 ;;（PE.521，PE.569）; 2
＆＃34; CW-20＆＃34 ;;（PE.520，PE.568）; 17
＆＃34; CW-21＆＃34 ;;（PE.519，PE.567）; 17
＆＃34; CW-22＆＃34 ;;（PE.526，PE.526-1，PE.574，PE.574-1）; 4
＆＃34; CW-23＆＃34 ;;（PE.525，PE.573）; 2
＆＃34; CW-24＆＃34 ;;（PE.524，PE.572）; 2
＆＃34; CW-25＆＃34 ;;（PE.523，PE.571）; 3
＆＃34; CW-26＆＃34 ;;（PE.522，PE.522-1，PE.570，PE.570-1）; 19个

期望的输出：
＆＃34; CW-03＆＃34 ;;（PE.502）; 24
＆＃34; CW-04＆＃34 ;;（PE.501）; 25
＆＃34; CW-05＆＃34 ;; （PE.518，PE.518-1，PE.517，PE.516，PE.515，PE.514）; 8
＆＃34; CW-10＆＃34 ;;（PE.513，PE.513-1）; 23
＆＃34; CW-11＆＃34 ;; （PE.509，PE.510，PE.511，PE.512）; 15

你能救我吗？

Answer 1

${interval}

使用以下命令

采用awk-script

$ cat awk-script
function print_line(str1,str2,sum)
{
  print str1 OFS "(" str2 ")" OFS sum  # print the format as your request
}
NR==1 {
  str1=$1                              
  str2=$3                              # Use the str2 to store the combined output for $3
  sum=$5                               # Use sum to store the sum of $5
  next
}
{
  sum+=$5
  if (sum>25){                         # If sum is greater than 25
    print_line(str1,str2,sum-$5)       # print the line in the desired format
    str1=$1                            # and reset the str1 to $1
    str2=""                            # str2 to ""
    sum=$5                             # sum to $5
  }
  str2=(str2?str2",":str2"")$3         # update the str2 based on if str2 is ""
}
END {
  print_line(str1,str2,sum)            # flush the final result
}

Answer 2

awk 解决方案：

function loginCheck($user_id, $pass){
   //here goes code which checks if user_id & pass exists
   //store in $RESULT if exists
    if(!empty($result)){

        session_set_cookie_params(30, '/');
        ini_set('session.gc_maxlifetime', 30);
        $_SESSION['username'] = $user_id;
        $_SESSION['logged_in'] = true;
    }
}

输出：

awk 'BEGIN{ FS=OFS=";" }
     f{ if(c+$3 > 25) { gsub(/\),\(/,",",r); print r,c; f=c=0 } else { c+=$3; r=r","$2 } }
     !f && c<25{ f=1; c=$3; r=$1 FS $2 }
     END{ if(c) gsub(/\),\(/,",",r); print r,c }' file

awk如何以40个为一组对行进行求和

2 个答案: