删除Awk命令中的列

Question

我想删除列，因为现在我有一个命令，其中包括一个命令中的分配，标头和拖车。我想在一个命令中进行分配，标头，拖车和删除第一个3列。

我的原始输出是

  Country   Gender    Plate       Name        Account Number        Car Name
    X        F          A          Fara        XXXXXXXXXX            Ixora
    X        F          b          Jiha        XXXXXXXXXX            Saga
    X        M          c          Jiji        XXXXXXXXXX            Proton

我的拆分文件：

第一个文件

06032017

  Country   Gender    Plate       Name    Account Number     Car Name        
    X        F          A     Fara        XXXXXXXXXX         Ixora

eof 1

第二文件
06032017

  Country   Gender    Plate       Name        Account Number        Car Name
    X        F          b          Jiha        XXXXXXXXXX            Saga

eof1

我的Desitre输出：

06032017

    Name    Account Number     Car Name        
    Fara        XXXXXXXXXX         Ixora

eof 1

06032017

  Name        Account Number        Car Name
  Jiha        XXXXXXXXXX            Saga

eof1

这是我分裂的命令：

awk -v date="$(date +"%d%m%Y")" -F| 'NR==1 {h=$0; next} 
{file="CAR_V1_"$1"_"$2"_"date".csv"; print (a[file]++?"": "DETAIL "date"" ORS h ORS) $0 > file}
END{for(file in a) print "EOF " a[file] > file}' HIRE_PURCHASE_testing.csv

Answer 1

有一个想法如何跳过某些字段：

$ echo "Field1 Field2 Field3 Field4 Field5 Field6"  |awk '{print $0}'
Field1 Field2 Field3 Field4 Field5 Field6  #No skipping here. Printed just for comparison
$ echo "Field1 Field2 Field3 Field4 Field5 Field6"  |awk '{print substr($0, index($0,$4))}'
Field4 Field5 Field6
$ echo "Field1 Field2 Field3 Field4 Field5 Field6"  |awk '{$1=$2=$3="";print}'
   Field4 Field5 Field6
#Mind the gaps in the beginning. You can use this if you need to "delete" particular columns.
#Actually you are not really delete columns but replace their contents with a blank value.
$ echo "Field1 Field2 Field3 Field4 Field5 Field6"  |awk '{gsub($1FS$2FS$3FS$4,$4);print}'
Field4 Field5 Field6   #Columns 1-2-3-4 have been replaced by column4. No gaps here.
$ echo "Field1 Field2 Field3 Field4 Field5 Field6"  |awk '{print $4,$5,$6}' 
Field4 Field5 Field6
# If you have a fixed number of fields (i.e 6 fields) you can just do not print the first three fields and print the last three fields

适应您存在的代码，行

print (a[file]++?"": "DETAIL "date"" ORS h ORS) $0 > file

如果更改为

print (a[file]++?"": "DETAIL "date"" ORS h ORS) substr($0, index($0,$4)) > file

或

print (a[file]++?"": "DETAIL "date"" ORS h ORS) $4,$5,$6 > file #if you have a fixed number of columns in the file)

应该足够了。

下次您需要帮助时，我会建议您简化问题以更轻松地获得帮助。因为其余的用户很难完全跟随您。

还可以在此处查看所有这些功能的工作方式：https：//www.gnu.org/software/gawk/manual/html_node/string-functions.html