用awk替换一列CSV文件中的字符gsub

Question

我想使用awk将CSV文件转换为仅包含原始列的子集的新CSV文件。而且我也想用仅有一列的下划线替换空格。我试过像这样：

gawk -F "," ' 
{ 
    name=gsub(/ /,"_",$1); 
    label=$2; 
    print ","name","label"," 
}' ./in.csv >> ./out.csv 

但gsub（）返回匹配的次数，而不是替换字符串。所以我得到的是这样的：的

,1,label 

代替：

,name_nospace,label 

如何使用AWK GSUB这样的替换字符只有一个列？

Answer 1

别：换人的数量，而不是字符串

name=gsub() 

为gsub回报。只是

gsub() 

和打印你拨弄领域，即：在这种情况下，一个sed

gsub(/ /,"_",$1); 
label=$2; 
print "," $1 "," label "," # or whatever you were doing 

Answer 2

gawk -F "," ' 
{ 
    gsub(/ /,"_",$1); 
    # print only: ,NameValue,LabelValue, as output 
    # so 4 field with first and last empty as in OP 
    print "," $1 "," $2 "," 
}' ./in.csv >> ./out.csv 

也可

sed -e ':under' -e 's/^\([^[ ,]*\) /\1_/;t under' -e 's/^\([^,]*,[^,]*,\).*/,\1/' ./in.csv >> ./out.csv 

Answer 3

要修改 “名”，变：

name=gsub(/ /,"_",$1) 

到（呆子，只有较新的mawk）：

name=gensub(/ /,"_","g",$1) 

或（任何AWK）：

name=$1 
gsub(/ /,"_",name) 

你也应该设置而不是硬编码的逗号OFS，特别是如果你修改字段，因此你的脚本应该写成：

awk ' 
BEGIN { FS=OFS="," } 
{ 
    name=$1 
    gsub(/ /,"_",name) 
    label=$2 
    print "", name, label, "" 
}' ./in.csv 

假设有某种原因使用的变量，而不是直接修改字段。