Bash 用“替换引号”``&引用；及''&引用；

我有一个包含许多

“

标记的文档，但我想将其转换为TeX格式

TeX使用2'标记作为起始引号，2'标记作为结束引号

我只想在“

”

以偶数形式出现在一行时（例如，行上有2、4或6个

”

）对这些进行更改

"This line has 2 quotation marks."
--> ``This line has 2 quotation marks.''

"This line," said the spider, "Has 4 quotation marks."
--> ``This line,'' said the spider, ``Has 4 quotation marks.''

"This line," said the spider, must have a problem, because there are 3 quotation marks."
--> (unchanged)

我的句子从不跨行，所以不需要检查多行

很少有单引号的引号，所以我可以手动更改它们

---

如何转换这些内容？

这是我的一行代码，适用于我：

awk -F\" '{if((NF-1)%2==0){res=$0;for(i=1;i<NF;i++){to="``";if(i%2==0){to="'\'\''"}res=gensub("\"", to, 1, res)};print res}else{print}}' input.txt >output.txt

使用

awk

解释

awk '{
  n=gsub("\"","&") # set n to the number of quotes in the current line
}
!(n%2){ # if there are even number of quotes
  while(n--){ # as long as we have double-quotes
    n%2?Q=q:Q="`" # alternate Q between a backtick and single quote
    sub("\"",Q Q) # replace the next double quote with two of whatever Q is
  }
}1 # print out all other lines untouched' 
q=\' in # set the q variable to a single quote and pass the file 'in' as input

使用

sed

---

下面是我使用重复的

sed

的一行代码：

cat file.txt | sed -e 's/"\([^"]*\)"/`\1`/g' | sed '/"/s/`/\"/g' | sed -e 's/`\([^`]*\)`/``\1'\'''\''/g'

（注意：如果文件中已经有反勾号（`），它将无法正常工作，但如果没有，则应该这样做）

编辑： 通过简化消除了back tick bug，现在适用于所有情况：

cat file.txt | sed -e 's/"\([^"]*\)"/``\1'\'\''/g' | sed '/"/s/``/"/g' | sed '/"/s/'\'\''/"/g'

谨此陈辞：

{
    FS="\"" # set field separator to double quote
    if ((NF-1) % 2 == 0) { # if count of double quotes in line are even number
        res = $0 # save original line to res variable
        for (i = 1; i < NF; i++) { # for each double quote
            to = "``" # replace current occurency of double quote by ``
            if (i % 2 == 0) { # if its closes quote replace by ''
                to = "''"
            }
            # replace " by to in res and save result to res
            res = gensub("\"", to, 1, res)
        }
        print res # print resulted line
    } else {
        print # print original line when nothing to change
    }
}

cat file.txt                           # read file
| sed -e 's/"\([^"]*\)"/``\1'\'\''/g'  # initial replace
| sed '/"/s/``/"/g'                    # revert `` to " on lines with extra "
| sed '/"/s/'\'\''/"/g'                # revert '' to " on lines with extra "

---

这可能适合您：

sed 'h;s/"\([^"]*\)"/``\1''\'\''/g;/"/g' file

说明：

• 复制原始行

  h

• 替换成对的

  “

  的

  s/”\（[^“]*\）“/`\1'\\\'\'\''/g

• 检查是否存在奇数

  “

  ，如果发现，则恢复到原始行

  /”/g

---

它并不漂亮，但我找到了一个

sed

1行bash脚本何时漂亮？我一直在关注这个

awk

解决方案（以及无数的迭代：P）2个多小时以来，老实说，我不认为它能做得更好。大量使用

gsub

来识别引号的数量，很好地使用

模运算符

，以及很好地使用

三元运算符

+1我知道这一点。：/。

cat file.txt | sed -e 's/"\([^"]*\)"/``\1'\'\''/g' | sed '/"/s/``/"/g' | sed '/"/s/'\'\''/"/g'

cat file.txt                           # read file
| sed -e 's/"\([^"]*\)"/``\1'\'\''/g'  # initial replace
| sed '/"/s/``/"/g'                    # revert `` to " on lines with extra "
| sed '/"/s/'\'\''/"/g'                # revert '' to " on lines with extra "

sed 'h;s/"\([^"]*\)"/``\1''\'\''/g;/"/g' file