Question

我有一个csv文件，格式如下。

event1,event2,event3,event4  
"Ozz","time","1234","test1"  
"Max","time1","5678","test2"

我需要替换第一列值而不更改标题。我需要将第二列值复制到第三列。所以预期的输出是：

event1,event2,event3,event4  
"Razor","time","time","test1"  
"Maxine","time1","time1","test2"

输出将是另一个csv文件我不知道该如何解决这个问题？

Answer 1

您可以使用类似

的内容

awk 'BEGIN{FS=OFS=","} $1~/^".*"$/{gsub("^\"Ozz\"$", "\"Razor\"",$1); gsub("^\"Max\"$","\"Maxine\"", $1); $3=$2;print $0} /event/' inputFile

输出

 event1,event2,event3,event4
"Razor","time","time","test1"
"Maxine","time1","time1","test2"

Answer 2

如果文件很小（少于100行），使用纯bash是一个不错的选择：

exec 3<inputfile
read -ru 3 a; echo $a
while IFS=, read -ru 3 v a b b; do
    v=${v/#\"/};v=${v/%\"/} # Trim surrounding "
    case $v in
    Ozz) v=Razor;;
    Max) v=Maxine;;
    *) echo "Wrong value '$v'">&2; v=Default;;
    esac
    echo "\"$v\",$a,$a,$b"
done

输出：

event1,event2,event3,event4
"Razor","time","time","test1"
"Maxine","time1","time1","test2"

替换可以从另一个文件中读取，并且可以使用散列进行替换。

awk shell脚本用双引号处理csv文件

2 个答案: