Question

这是我的档案

$ cat head_datafile_pipe_deleimiter.csv
"Rec_Open_Date"|"MSISDN"|"IMEI"|"Data_Volume_Bytes"|"Device_Manufacturer"|"Device_Model"|"Product_Description"|"Data_Volume_MB"
"2016-07-17"|"686"|"630"|"618320"|"Apple Inc"|"Apple iPhone S A1530"|"PREPAY PLUS - $0 -"|"0.589676"
"2016-07-17"|"560"|"570"|"42841779"|"Motorola Mobility LLC, a Lenovo Company"|"Moto X 2nd Generation, X112360445"|"$39.95 Plan"|"40.8571"
"2016-07-17"|"811"|"340"|"2465082"|"Samsung Korea"|"Samsung SM-G900I"|"$69.95 Plan"|"2.35089"
"2016-07-17"|"785"|"610"|"41498628"|"Apple Inc"|"Apple iPhone 6S Plus A1687"|"$29.95 Carryover Plan 1GB"|"39.5762"
"2016-07-17"|"908"|"310"|"6497563"|"Samsung Korea"|"Samsung GT-I9195"|"PREPAY PLUS - $0 -"|"6.19656"
"2016-07-17"|"919"|"610"|"0"|"Samsung Korea"|"Samsung SM-G925I"|"$19 CO COMBO - NOT RECURRENT"|"0"
"2016-07-17"|"356"|"290"|"33189681"|"Apple Inc"|"Apple iPhone 6S A1688"|"$39.95 Plan"|"31.6521"
"2016-07-17"|"009"|"160"|"30340"|"Samsung Korea"|"Samsung SM-J500Y"|"PREPAY PLUS - $1 - #33"|"0.0289345"
"2016-07-17"|"574"|"400"|"549067"|"HUAWEI Technologies Co Ltd"|"HUAWEI Y6"|"PREPAY PLUS - $0 -"|"0.523631"

我想将此输出存储在数组

中

$ awk -F'|' 'NR>1{print $7}' head_datafile_pipe_deleimiter.csv | sort | uniq
"$19 CO COMBO - NOT RECURRENT"
"$29.95 Carryover Plan 1GB"
"$39.95 Plan"
"$69.95 Plan"
"PREPAY PLUS - $0 -"
"PREPAY PLUS - $1 - #33"

我这样做的方法是将其写入日志文件

$ awk -F'|' 'NR>1{print $7}' head_datafile_pipe_deleimiter.csv | sort | uniq > logfile
$ cat logfile
"$19 CO COMBO - NOT RECURRENT"
"$29.95 Carryover Plan 1GB"
"$39.95 Plan"
"$69.95 Plan"
"PREPAY PLUS - $0 -"
"PREPAY PLUS - $1 - #33"

然后将其存储在数组

中

$ u_vals=(`cat "logfile"`)

prning数组中的所有元素

$ echo "${u_vals[@]}"
"$19 CO COMBO - NOT RECURRENT" "$29.95 Carryover Plan 1GB" "$39.95 Plan" "$69.95 Plan" "PREPAY PLUS - $0 -" "PREPAY PLUS - $1 - #33"

打印第一个元素

$ echo "${u_vals[0]}"
"$19

获取数组的长度（零是第一个数组）

$ echo "${#u_vals[@]}"
25

打印最后一个元素

$ echo "${u_vals[24]}"
#33"

我真的有2个问题首先

我想要的是在一个命令中创建我的数组，如果可能的话，不必写入文件

$ u_vals=(`awk -F'|' 'NR>1{print $7}' head_datafile_pipe_deleimiter.csv | sort | uniq`)

，其次，更重要的是，我希望数组有6个元素，如下所示，但空格似乎是问题

$ cat -n logfile
     1  "$19 CO COMBO - NOT RECURRENT"
     2  "$29.95 Carryover Plan 1GB"
     3  "$39.95 Plan"
     4  "$69.95 Plan"
     5  "PREPAY PLUS - $0 -"
     6  "PREPAY PLUS - $1 - #33"

 ## this will loop through the array but sperrates the elements by spaces 
for elem in "${u_vals[@]}"; do  echo "$elem"; done

EDIT1

这回答了我的第一个问题，但不回答第二个问题

$ u_vals=($(awk -F'|' 'NR>1{print $7}' head_datafile_pipe_deleimiter.csv | sort | uniq))

$ echo "${u_vals[@]}"
"$19 CO COMBO - NOT RECURRENT" "$29.95 Carryover Plan 1GB" "$39.95 Plan" "$69.95 Plan" "PREPAY PLUS - $0 -" "PREPAY PLUS - $1 - #33"

$ echo "${u_vals[0]}"
"$19

EDIT2

根据下面的回答，这是我选择的方式，请注意我使用awk而不是下面发布的while。不确定哪个是最好的，但我只是喜欢并更好地理解awk。

$ mapfile -t u_vals <<<"$(awk -F'|' 'NR>1{print $7}' head_datafile_pipe_deleimiter.csv | sort | uniq)"


$ declare -p u_vals
declare -a u_vals='([0]="\"\$19 CO COMBO - NOT RECURRENT\"" [1]="\"\$29.95 Carryover Plan 1GB\"" [2]="\"\$39.95 Plan\"" [3]="\"\$69.95 Plan\"" [4]="\"PREPAY PLUS - \$0 -\"" [5]="\"PREPAY PLUS - \$1 - #33\"")'


$ for elem in "${u_vals[@]}"; do  echo "$elem"; done
"$19 CO COMBO - NOT RECURRENT"
"$29.95 Carryover Plan 1GB"
"$39.95 Plan"
"$69.95 Plan"
"PREPAY PLUS - $0 -"
"PREPAY PLUS - $1 - #33"


$ printf "%s\n" "${u_vals[@]}"
"$19 CO COMBO - NOT RECURRENT"
"$29.95 Carryover Plan 1GB"
"$39.95 Plan"
"$69.95 Plan"
"PREPAY PLUS - $0 -"
"PREPAY PLUS - $1 - #33"

Answer 1

使用BASH，您可以将mapfile与process substitution：

一起使用

mapfile -t u_vals < <(
   p=1; while IFS='|' read -ra arr; do (( p )) && p=0 || echo "${arr[6]}"; done < file.csv|
   sort -u)

测试输出：

printf "%s\n" "${u_vals[@]}"

"$19 CO COMBO - NOT RECURRENT"
"$29.95 Carryover Plan 1GB"
"$39.95 Plan"
"$69.95 Plan"
"PREPAY PLUS - $0 -"
"PREPAY PLUS - $1 - #33"

进程替换中的

while loop正在执行此操作：

丢弃第一个标题行
使用管道作为分隔符拆分每列
仅提取第7列
仅排序和使用唯一身份

Code Demo

从文件中的列创建bash数组

EDIT1

EDIT2

1 个答案: