Unix - 用终结符逗号删除内部双引号

Question

输入文件：

"1","2col",""3col " "
"2","2col"," "3c,ol     " "
"3","2col"," 3co,l"     
"4","2col","3co,l"
"5","2col",""3co,l      ""              "
"6","2col",""3c,ol ""3c,ol"""

输出文件：

"1","2col","3col    "
"2","2col"," 3c,ol       "
"3","2col"," 3co,l"     
"4","2col","3co,l"
"5","2col","3co,l                       "
"6","2col","3c,ol 3c,ol"

请帮我用Unix命令获取上面的输出。请注意输出中的第3列已修改，所有内部双引号均已删除。

逗号是终结者。当双引号之间存在逗号时，它不被视为终止符。看到第6行和第2个逗号之后，逗号作为双引号之间的文本存在，这很好。

到目前为止我尝试过的：

sed 's/""|/|/g'
sed -e "s/\"\"//g"
perl -pe 's/(?<!^)(?<!\,)"(?!\,)(?!$)/""/g'

Answer 1

假设（第一列和第二列是“干净的”，例如它们不包含,）

输入：

"1","2col",""3col " "
"2","2col"," "3c,ol     " "
"3","2col"," 3co,l"     
"4","2col","3co,l"
"5","2col",""3co,l      ""              "
"6","2col",""3c,ol ""3c,ol"""

命令：

tr -d '"' < input | awk -F',' -v OFS=',' '{$1="\""$1"\"";$2="\""$2"\"";printf $1 OFS $2 OFS "\"";for(u=3;u<=NF;u++){if(u!=NF)printf $u OFS;else printf $u};printf "\"" RS}'

输出：

"1","2col","3col  "
"2","2col"," 3c,ol      "
"3","2col"," 3co,l     "
"4","2col","3co,l"
"5","2col","3co,l                    "
"6","2col","3c,ol 3c,ol"

说明：

• tr -d '"' < input将删除所有"
• | awk输出到awk
• -F',' -v OFS=','输入/输出字段分隔符定义为逗号
• 你使用"用$1="\""$1"\"";$2="\""$2"\"";围绕前两列你打印它们printf $1 OFS $2 OFS "\"";
• for(u=3;u<=NF;u++){if(u!=NF)printf $u OFS;else printf $u};printf "\"" RS}对于列的其余部分，您只需将它们重新附加在一起，然后在行尾添加最后一个"。

为了便于阅读：

'{
  $1="\""$1"\""
  $2="\""$2"\""
  printf $1 OFS $2 OFS "\""
  for(u=3;u<=NF;u++)
  {
    if(u!=NF)printf $u OFS
    else printf $u
  }
  printf "\"" RS
}'

Answer 2

使用引号查找前两个字段并连接其他字段。

awk -F '"' '
   BEGIN {q="\""}
   {printf "%s", q$2q$3q$4q$5q; for (i=6;i<=NF;i++) printf "%s", $i; print q}
   ' inputfile

编辑：另一种选择

paste -d, <( cut -d"," -f1,2 < inputfile) \
          <( cut -d"," -f3-  < inputfile | sed 's/"//g;s/.*/"&"/')

编辑：另一种选择

sed 's/old/new/g：将替换应用于regexp seds / old / new / number`的所有匹配：仅替换正则表达式的numberth匹配。当您在GNU sed中混合g和数字修饰符时，将忽略第一个匹配项，然后匹配并替换所有匹配项。 在这种情况下：

sed -r 's/"//g6;s/$/"/' inputfile