string 如何在 bash printf 中转义一系列反斜杠？
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How do I escape a series of backslashes in a bash printf?
提问by Charney Kaye
The following script yielded an unexpected output:
以下脚本产生了意外的输出：
printf "escaped slash: \ \n"
printf "2 escaped slashes: \\ \n"
printf "3 escaped slashes: \\\ \n"
printf "4 escaped slashes: \\\\ \n"
Run as a bash script under Ubuntu 14, I see:
在 Ubuntu 14 下作为 bash 脚本运行，我看到：
escaped slash: \
2 escaped slashes: \
3 escaped slashes: \
4 escaped slashes: \
Err.. what?
呃……什么？
采纳答案by Eugeniu Rosca
Assuming that printfFORMATstring is surrounded by double quotes, printftakes one additional level of expansion, compared to e.g. echo(both being shell builtin commands).
假设printfFORMAT字符串被双引号包围printf，与 eg echo（两者都是 shell 内置命令）相比，需要一个额外的扩展级别。
What you expect from printfcan actually be achieved using single quotes:
printf使用单引号实际上可以实现您的期望：
printf '1 escaped slash: \ \n'
printf '2 escaped slashes: \\ \n'
printf '3 escaped slashes: \\\ \n'
printf '4 escaped slashes: \\\\ \n'
outputs:
输出：
1 escaped slash: \
2 escaped slashes: \
3 escaped slashes: \\
4 escaped slashes: \\
回答by Cyrus
printfis a bash builtin. Look at help printf:
printf是一个内置的 bash。看看help printf：
printf [-v var] format [arguments]
Formats and prints ARGUMENTS under control of the FORMAT.
You should pass the format and the argument. So add the format "%s\n"before the argument:
您应该传递格式和参数。所以"%s\n"在参数之前添加格式：
printf "%s\n" "escaped slash: \"
printf "%s\n" "2 escaped slashes: \\"
printf "%s\n" "3 escaped slashes: \\\"
printf "%s\n" "4 escaped slashes: \\\\"
Output:
输出：
escaped slash: \ 2 escaped slashes: \ 3 escaped slashes: \\ 4 escaped slashes: \\
