我怎样才能逃避用作Bash的命令行参数的任意字符串？ [英] How can I escape an arbitrary string for use as a command line argument in Bash?
问题描述
我有一个字符串列表，我想通过这些字符串作为参数在一个单一的bash命令行调用。对于简单的字母数字串就足够了只是通过他们逐字：
＆GT; script.pl富酒吧巴兹是否
富
酒吧
巴兹
是
没有
我明白，如果一个参数包含空格或反斜线或双引号，我需要反斜杠转义双引号和反斜线，然后双引号的说法。
＆GT; script.pl富酒吧巴兹\\是\\\\\\\\无\\
富
酒吧
巴兹
是\\不
但是，当一个参数包含一个感叹号，发生这种情况：
＆GT; script.pl！富
-bash：未找到事件：富
双引号不工作：
＆GT; script.pl！富
-bash：未找到事件：富
也确实反斜杠逃逸（注意反斜线是如何在输出present）：
＆GT; script.pl\\！富
\\！富
我不知道很多关于猛砸，但是我知道，还有其他的特殊字符，做类似的事情。 什么是安全逃离任意字符串用作Bash的命令行参数的一般程序？让我们假设该字符串可以是任意长度，并包含特殊字符的任意组合。我想一个越狱（）子程序，我可以按照以下（Perl的例子）使用方法：
$ CMD =加盟，图{逃生（$ _）; } @args;
下面是应当由该功能可以安全地躲过一些例子字符串（我知道的一些这些看起来类似Windows，这是故意的）：
是
没有
你好，世界[字符串中有一个逗号和空格]
C：\\ Program Files文件\\ [用反斜杠路径中有空格]
[即一个双引号]
\\ [反斜线]
\\\\ [两个反斜杠]
\\\\\\【三反斜杠]
\\\\\\\\ [四个反斜杠]
\\\\\\\\\\ [五反斜线]
\\ [双引号，反斜线]
\\ T [双引号，反斜线，T]
\\\\ T [双引号，反斜线，反斜线，T]
！1
！一个
！\\ /[双引号，感叹号，反斜杠，正斜杠，单引号，双引号]
杰夫的！ [双引号，J，E，F，F，撇号，S，感叹号，双引号]
$ PATH
％路径％
＆安培;
＆所述;＆GT; |＆放大器; ^
* @ $$一个$ @＃？-_
编辑：
这会做的伎俩？使用反斜线每一个不同寻常的人物，并省略单或双引号。 （例如在Perl，但任何语言可以做到这一点）
子逃逸{
$ _ [0] =〜S /（[^ A-ZA-Z0-9 _]）/ \\\\ $ 1 /克;
返回$ _ [0];
}
如果你想安全地报价的任何的对击，就可以使用其内置的的printf％Q 格式：
猫strings.txt ：
是
没有
你好，世界
C：\\ Program Files文件\\
\\
\\\\
\\\\\\
\\\\\\\\
\\\\\\\\\\
\\
\\ t
\\\\ŧ
！1
！一个
！\\ /'
杰夫的！
$ PATH
％路径％
＆安培;
＆所述;＆GT; |＆放大器; ^
* @ $$一个$ @＃？-_
猫quote.sh ：
＃！/斌/庆典
而IFS =读-r -d $的'\\ n'
做
printf的％Q$ REPLY
printf的'\\ n'
完成＆LT; strings.txt
./ quote.sh ：
是
没有
你好，世界
C：\\\\程序\\文件\\\\
\\
\\\\
\\\\\\\\
\\\\\\\\\\\\
\\\\\\\\\\\\\\\\
\\\\\\\\\\\\\\\\\\\\
\\\\\\
\\\\\\Ť
\\\\\\\\\\Ť
\\！1
\\！一个
\\\\！\\\\ / \\'\\
\\杰夫\\的\\！\\
\\ $ PATH
％路径％
\\＆安培;
\\＆LT; \\＆GT; \\ | \\＆放大器; \\ ^
\\ * @ \\ $ \\ $ A \\ $ @＃\\？-_
这些字符串可以strings.txt被逐字复制到例如回声来输出原始的字符串。
I have a list of strings and I want to pass those strings as arguments in a single Bash command line call. For simple alphanumeric strings it suffices to just pass them verbatim:
> script.pl foo bar baz yes no
foo
bar
baz
yes
no
I understand that if an argument contains spaces or backslashes or double-quotes, I need to backslash-escape the double-quotes and backslashes, and then double-quote the argument.
> script.pl foo bar baz "\"yes\"\\\"no\""
foo
bar
baz
"yes"\"no"
But when an argument contains an exclamation mark, this happens:
> script.pl !foo
-bash: !foo: event not found
Double quoting doesn't work:
> script.pl "!foo"
-bash: !foo: event not found
Nor does backslash-escaping (notice how the literal backslash is present in the output):
> script.pl "\!foo"
\!foo
I don't know much about Bash yet but I know that there are other special characters which do similar things. What is the general procedure for safely escaping an arbitrary string for use as a command line argument in Bash? Let's assume the string can be of arbitrary length and contain arbitrary combinations of special characters. I would like an escape() subroutine that I can use as below (Perl example):
$cmd = join " ", map { escape($_); } @args;
Here are some more example strings which should be safely escaped by this function (I know some of these look Windows-like, that's deliberate):
yes
no
Hello, world [string with a comma and space in it]
C:\Program Files\ [path with backslashes and a space in it]
" [i.e. a double-quote]
\ [backslash]
\\ [two backslashes]
\\\ [three backslashes]
\\\\ [four backslashes]
\\\\\ [five backslashes]
"\ [double-quote, backslash]
"\T [double-quote, backslash, T]
"\\T [double-quote, backslash, backslash, T]
!1
!A
"!\/'" [double-quote, exclamation, backslash, forward slash, apostrophe, double quote]
"Jeff's!" [double-quote, J, e, f, f, apostrophe, s, exclamation, double quote]
$PATH
%PATH%
&
<>|&^
*@$$A$@#?-_
EDIT:
Would this do the trick? Escape every unusual character with a backslash, and omit single or double quotes. (Example is in Perl but any language can do this)
sub escape {
$_[0] =~ s/([^a-zA-Z0-9_])/\\$1/g;
return $_[0];
}
If you want to securely quote anything for Bash, you can use its built-in printf %q formatting:
cat strings.txt:
yes
no
Hello, world
C:\Program Files\
"
\
\\
\\\
\\\\
\\\\\
"\
"\T
"\\T
!1
!A
"!\/'"
"Jeff's!"
$PATH
%PATH%
&
<>|&^
*@$$A$@#?-_
cat quote.sh:
#!/bin/bash
while IFS= read -r -d $'\n'
do
printf %q "$REPLY"
printf '\n'
done < strings.txt
./quote.sh:
yes
no
Hello\,\ world
C:\\Program\ Files\\
\"
\\
\\\\
\\\\\\
\\\\\\\\
\\\\\\\\\\
\"\\
\"\\T
\"\\\\T
\!1
\!A
\"\!\\/\'\"
\"Jeff\'s\!\"
\$PATH
%PATH%
\&
\<\>\|\&\^
\*@\$\$A\$@#\?-_
These strings can be copied verbatim to for example echo to output the original strings in strings.txt.
这篇关于我怎样才能逃避用作Bash的命令行参数的任意字符串？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
