替换“#”、“$”、“%”、“&”和“_”与“\#”、“\$”、“\%”、“\&”和“\_”

发布于 2024-12-28 07:56:26 字数 615 浏览 5 评论 0原文

我有一个纯文本文档，我想在 LaTeX 中编译它。但是，有时它包含字符“#”、“$”、“%”、“&”和“_”。为了在 LaTeX 中正确编译，我必须首先用“#”、“\$”、“\%”、“\&”和“_”替换这些字符。我在 sed 中使用了这一行：

sed -i 's/\#/\\\#/g' ./file.txt
sed -i 's/\$/\\\$/g' ./file.txt
sed -i 's/\%/\\\%/g' ./file.txt
sed -i 's/\&/\\\&/g' ./file.txt
sed -i 's/\_/\\\_/g' ./file.txt

这是正确的吗？

不幸的是，该文件太大，无法在任何 GUI 软件中打开，因此使用文本编辑器检查我的 sed 行是否正确非常困难。我尝试使用 grep 进行搜索，但搜索未按预期工作（例如，下面，我搜索了包含“$”的任何行）：

grep "\$" file.txt

将“\”放在这些前面的最佳方法是什么人物？
如何使用 grep 成功检查包含替换的行？

原文

I have a plain text document, which I want to compile inside LaTeX. However, sometimes it has the characters, "#", "$", "%", "&", and "_". To compile properly in LaTeX, I must first replace these characters with "#", "\$", "\%", "\&", and "_". I have used this line in sed:

sed -i 's/\#/\\\#/g' ./file.txt
sed -i 's/\$/\\\$/g' ./file.txt
sed -i 's/\%/\\\%/g' ./file.txt
sed -i 's/\&/\\\&/g' ./file.txt
sed -i 's/\_/\\\_/g' ./file.txt

Is this correct?

Unfortunately, the file is too large to open in any GUI software, so checking if my sed line is correct with a text editor is difficult. I tried searching with grep, but the search does not work as expected (e.g. below, I searched for any lines containing "$"):

grep "\$" file.txt

What is the best way to put "\" in front of these characters?
How can I use grep to successfully check the lines with the replacements?

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不必了 2025-01-04 07:56:26

您可以通过一次调用 sed 来完成替换：

sed -i -E 's/([#$%&_\])/\\&/g' file.txt

替换文本中的 & 将填充括号中的单个字符。请注意，由于 \ 是 LaTeX 转义字符，因此您也必须在原始文件中对其进行转义。

You can do the replacement with a single call to sed:

sed -i -E 's/([#$%&_\])/\\&/g' file.txt

The & in the replacement text fills in for whichever single character is enclosed in parentheses. Note that since \ is the LaTeX escape character, you'll have to escape it as well in the original file.

回复收藏 0 原文

巨坚强 2025-01-04 07:56:26

sed -i 's/\#/\\\#/g' ./file.txt
sed -i 's/\$/\\\$/g' ./file.txt
sed -i 's/\%/\\\%/g' ./file.txt
sed -i 's/\&/\\\&/g' ./file.txt
sed -i 's/\_/\\\_/g' ./file.txt

大多数情况下，您不需要第一个（搜索）字符串上的 \ ，只需 $ （它是一个特殊字符，表示行尾；其余的并不特别）。并且在替换时，只需要两个\\，而不是三个。另外，您可以使用多个 -e 语句来完成所有操作：

sed -i.bak -e 's/#/\\#/g'  \
           -e 's/\$/\\$/g' \
           -e 's/%/\\%/g'  \
           -e 's/&/\\&/g'  \
           -e 's/_/\\_/g' file.txt

您不需要对任何内容进行双重转义（\\ 除外），因为它们是单转义的引。在您的 grep 中，bash 正在解释 $ 上的转义符，因为它是一个特殊字符（特别是变量的符号），因此 grep 正在获取并搜索 $，这是一个特殊字符，表示行尾。您需要将其用单引号引起来，以防止 bash 解释 \ ('\$'，或者添加另一对 \\: "\\\$" 大概是您获取\` 的地方，但在 sed 中不需要它。 > 正如其所写。

sed -i 's/\#/\\\#/g' ./file.txt
sed -i 's/\$/\\\$/g' ./file.txt
sed -i 's/\%/\\\%/g' ./file.txt
sed -i 's/\&/\\\&/g' ./file.txt
sed -i 's/\_/\\\_/g' ./file.txt

You don't need the \ on the first (search) string on most of them, just $ (it's a special character, meaning the end of a line; the rest aren't special). And in the replacement, you only need two \\, not three. Also, you could do it all in one with several -e statements:

sed -i.bak -e 's/#/\\#/g'  \
           -e 's/\$/\\$/g' \
           -e 's/%/\\%/g'  \
           -e 's/&/\\&/g'  \
           -e 's/_/\\_/g' file.txt

You don't need to double-escape anything (except the \\) because these are single-quoted. In your grep, bash is interpreting the escape on the $ because it's a special character (specifically, a sigil for variables), so grep is getting and searching for just the $, which is a special character meaning the end of a line. You need to either single-quote it to prevent bash from interpreting the \ ('\$', or add another pair of \\: "\\\$". Presumably, that's where you're getting the\` from, but you don't need it in the sed as it's written.