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使用awk删除列

[英]Remove column using awk

I want to remove column, 我想删除列,

this is my command: 这是我的命令:

awk -F"|" -v OFS="|" '{ $1=$2=$3"" ; $0=substr($0,37) } 1' PURCHASE_testing.csv > testing.csv

My input: 我的意见:

Account num|Car|Type|Name|Class|Price
xxxxxxxxxx |A  |Honda| Fara |A|rm20k
xxxxxxxxxx | B | Proton|Afiq |B|rm40k

I managed to remove first 3 column in row 1 but in row 2 and 3 the string not manage to remove. 我设法删除第1行中的前3列,但在第2行和第3行中,该字符串无法删除。

My output: 我的输出:

       Name      Class      Price              
0101   Fara       A          rm2
1234   Afiq       B          rm4

I want output be: 我想输出是:

Name Class Price
Fara   A   rm2
Afiq   B   rm4

To print all but first 3 columns use the following approach: 要打印除前3列以外的所有列,请使用以下方法:

For the input PURCHASE_testing.csv file contents: 对于输入PURCHASE_testing.csv文件内容:

Account num|Car|Type|Name|Class|Price 
0101 |A |Honda| Fara |A|rm20k 
1234 |B | Proton|Afiq |B|rm40k

awk -F"|" '{$1=$2=$3=""; print $0}' PURCHASE_testing.csv > testing.csv

Now, the testing.csv file contents should look like: 现在, testing.csv文件内容应如下所示:

Name Class Price 
 Fara  A rm20k 
Afiq  B rm40k

try this - 试试这个 -

awk 'BEGIN{FS="|"} {print $4,$5,substr($6,1,4)}' f
Name Class Pric
Fara  A rm20
Afiq  B rm40

Explanation : substr($6,1,4) fetching 1 to 4 character from column 6. 说明:substr($ 6,1,4)从第6列中获取1到4个字符。

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