如果逗号中不存在 csv 文件的第 6 列，如何在逗号后添加空格?

Question

Ubuntu 16.04
Bash 4.3.3

如果第 6 列中不存在逗号，我还需要一种方法在逗号后添加一个空格。我不得不对上面的行进行注释，因为它在 csv 文件中的所有逗号之后放置了一个空格。

错误："This is 6th column,Hey guys,Red White & Blue,I know it,Right On"

完美："This is 6th column, Hey guys, Red White & Blue, I know it, Right On"

我几乎可以看到awk打印出第 6 列，然后让sed做剩下的事情：

awk '{ print $6 }' "$feed " | sed -i 's/|/,/g; s/,/, /g; s/,s+/, /g'

这是我到目前为止所拥有的：

for feed in *; do
sed -r -i 's/([^,]{0,10})[^,]*/1/5' "$feed"
sed -i '
s/<b>//g; s/*//g;
s/([0-9])""/1inch/g;
#     s/|/,/g; s/,/, /g; s/,s+/, /g;
s/"one","drive"/"onetext","drive"/;
s/"comments"/"description"/;
s/"features"/"optiontext"/;
' "$feed"
done

s/|/,/g; s/,/, /g; s/,s+/, /g;有效，但是是全局的，不在列内。

Answer 1

听起来你所需要的只是这个(使用 GNU awk for FPAT(：

awk 'BEGIN{FPAT="[^,]*|"[^"]+""; OFS=","} {gsub(/, ?/,", ",$6)} 1'

例如：

$ cat file
1,2,3,4,5,"This is 6th column,Hey guys,Red White & Blue,I know it,Right On",7,8
$ awk 'BEGIN{FPAT="[^,]*|"[^"]+""; OFS=","} {gsub(/, ?/,", ",$6)} 1' file
1,2,3,4,5,"This is 6th column, Hey guys, Red White & Blue, I know it, Right On",7,8

实际上看起来你的整个 shell 脚本，包括对 GNU sed 的多次调用，只需一次调用 GNU awk 就可以更有效地完成，而不需要周围的 shell 循环，例如(未经测试(：

awk -i inplace '
BEGIN{FPAT="[^,]*|"[^"]+""; OFS=","} 
{
$0 = gensub(/([^,]{0,10})[^,]*/,"\1",5)
$0 = gensub(/([0-9])""/,"\1inch","g")
sub(/"one","drive"/,""onetext","drive"")
sub(/"comments"/,""description"")
sub(/"features"/,""optiontext"")
gsub(/, ?/,", ",$6)
}
' *

Answer 2

这可能对你有用(GNU sed(：

sed -r 's/[^,"]*("[^"]*")*/n&n/6;h;s/, ?/, /g;G;s/.*n(.*)n.*n(.*)n.*n/21/' file

用换行符包围第 6 个字段。复制该行。将所有逗号后跟一个可能的空格替换为一个逗号，后跟一个空格。附加原始行并使用模式匹配替换修改后的字段，丢弃改进行的其余部分。