在 csv 文件中，我有以下几列，我尝试用
更改第二列的值
awk -F ',' -v OFS=',' '$1 { $2=$2*2; print}' path/file.csv > output.csv.
但它返回零并删除双引号。
文件.csv
"sku","0.47","supplierName"
"sku","3.14","supplierName"
"sku","3.56","supplierName"
"sku","4.20","supplierName"
输出.csv
"sku",0,"supplierName"
"sku",0,"supplierName"
"sku",0,"supplierName"
"sku",0,"supplierName"
最佳答案
您可以在 FS 值中指定多个字符。
$ awk -v FS="\",\"" -v OFS="\",\"" '{$2=$2*2}1' file
"sku","0.94","supplierName"
"sku","6.28","supplierName"
"sku","7.12","supplierName"
"sku","8.4","supplierName"
如果你想四舍五入到小数点后两位，试试这个。
$ awk -v FS="\",\"" -v OFS="\",\"" '{$2=sprintf("%.2f",$2*2)}1' file
"sku","0.94","supplierName"
"sku","6.28","supplierName"
"sku","7.12","supplierName"
"sku","8.40","supplierName"
关于bash - 更改 CSV 文件中的列值，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/33517459/