shell中有没有方法来忽略转义字符？

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问题描述

在C＃中，有一个逐字字符串，以便

  string c =hello \t world; // hello world 
 string d = @hello \t world; // hello \t world

我是shell脚本的新手，shell中是否有类似的方法？

由于我有很多文件夹，名称如服装与配饰>服装>活动服装，我想知道是否有一个简单的方法来处理逃生没有写这么多的字符。

  test.sh 
 
 director =Apparel& Accessories>服装> Activewear
＃任何逃避空间的动作&> ??? 
 hadoop fs -ls $ director

解决方案

定义您的示例中的特定字符串， Apparel&配件>服装>活动服，双引号或单引号将工作;

在 shell （任何POSIX-兼容的外壳），您如何将引用到变量与您定义一样重要。

要安全地将引用到以前定义的变量中，而没有副作用，请将其包含在双引号中，例如$ directory 。

要定义 literal（逐字）字符串：

（相反，要使用嵌入式变量引用或定义变量>嵌入式命令替换或嵌入式算术表达式，请使用双重引号（）。 p>

如果您的字符串包含 NO 单个引号：
- 使用单引号字符串，例如：
```
 目录='Appare l&配件>服装> Activewear'
  
```

一个单引号的字符串不受shell 的任何解释，因此通常是定义文字的最安全的选项。 >。请注意，字符串可以跨越多行;例如：

  multiline ='line 1 
 line 2'
  / pre> 
 
  
  如果您的字符串 DOES包含单引号（例如，我在这里。），你想要一个解决方案，在所有 POSIX兼容的shell  p> 
 
 
 
  将字符串分割成多个（单引号）部分，单引号字符 strong>：
 
 
 
 
 
 
 
 注意：可惜的是，单引号字符串不能包含单引号，甚至没有转义。
  directory ='我'在这里。 
  
字符串分为单引号 I ，然后是文字'（作为 unquoted 字符串转义为 \' ），然后在这里单引号 m。。由于零件之间没有空格，所以结果是一个单个字符串，包含 I 之后的文字单引号。
 
 
 替代方法：如果您不介意使用多行语句，则可以使用这里引用的文档，如底部所述。 
 
 
  
  如果您的字符串 DOES包含单引号（例如我在这里。），你想要一个解决方案，在 bash ， ksh 和 zsh  ：
 
 
  
  使用 ANSI-C引号：
  directory = $'I\'m 
  
 
 
 
 
 
 
 
 
 注意：如您所见，ANSI-C引用允许将单引号转换为 \' ，但请注意附加含义：其他  \< char> 序列需要解释，太;例如， \\\
 被解释为换行符 - 请参阅 
 
 
 
 
 
  @chepner的帽子提示，他指出，将 POSIX兼容的方式直接包含在一个字符串中使用单引号即可逐字使用 是将 读取-r 与使用引号打开分隔符（ -r 选项确保字符串中的 \ 字符被视为文字）。
 ＃*任何*形式的引用，而不仅仅是单引号，在开放EOF将工作。 
＃注意，$ HOME将不会被扩展。 
＃（如果你没有引用开头的EOF，它会。）
读-r目录<'EOF'
我在这里$ HOME 
 EOF 
  
 
 
 
 
 请注意，这里的文档创建 stdin 在这种情况下，输入（读取）。因此，您不能使用此技术直接传递生成的字符串作为参数。
 
 
 
In C#, there is a verbatim string so that,
string c = "hello \t world";               // hello     world
string d = @"hello \t world";              // hello \t world
I am new to shell script, is there a similar method in shell?
Because I have many folders with the name like "Apparel & Accessories > Clothing > Activewear", I want to know if there is a easy way to process the escape characters without write so many .
test.sh

director="Apparel & Accessories > Clothing > Activewear"
# any action to escape spaces, &, >   ???
hadoop fs -ls $director
 解决方案 
For definining the specific string in your example, Apparel & Accessories > Clothing > Activewear, either double quotes or single quotes will work; referring to it later is a different story, however:
In the shell (any POSIX-compatible shell), how you refer to a variable is just as important as how you define it.
To safely refer to a previously defined variable without side-effects, enclose it in double quotes, e.g., "$directory".
To define [a variable as] a literal (verbatim) string:
(By contrast, to define a variable with embedded variable references or embedded command substitutions or embedded arithmetic expressions, use double quotes (").)
If your string contains NO single quotes: 
Use a single-quoted string, e.g.:
directory='Apparel & Accessories > Clothing > Activewear'
A single-quoted string is not subject to any interpretation by the shell, so it's generally the safest option for defining a literal. Note that the string may span multiple lines; e.g.:
        multiline='line 1
        line 2'
If your string  DOES contain single quotes (e.g., I'm here.) and you want a solution that works in all POSIX-compatible shells:
Break the string into multiple (single-quoted) parts and splice in single-quote characters:
Note: Sadly, single-quoted strings cannot contain single quotes, not even with escaping.
        directory='I'\''m here.'
The string is broken into into single-quoted I, followed by literal ' (escaped as an unquoted string as \'), followed by single-quoted m here.. By virtue of having NO spaces between the parts, the result is a single string containing a literal single quote after I.
Alternative: if you don't mind using a multiline statement, you can use a quoted here document, as described at the bottom.
If your string  DOES contain single quotes (e.g., I'm here.) and you want a solution that works in bash, ksh, and zsh:
Use ANSI-C quoting:
directory=$'I\'m here.'
Note: As you can see, ANSI-C quoting allows for escaping single quotes as \', but note the additional implications: other \<char> sequences are subject to interpretation, too; e.g., \n is interpreted as a newline character - see http://www.gnu.org/software/bash/manual/bash.html#ANSI_002dC-Quoting
Tip of the hat to @chepner, who points out that the POSIX-compatible way of directly including a single quote in a string to be used verbatim is to use read -r with a here document using a quoted opening delimiter (the -r option ensures that \ characters in the string are treated as literals).
# *Any* form of quoting, not just single quotes, on the opening EOF will work.
# Note that $HOME will by design NOT be expanded.
# (If you didn't quote the opening EOF, it would.)
read -r directory <<'EOF'
I'm here at $HOME
EOF
Note that here documents create stdin input (which read reads in this case). Therefore, you cannot use this technique to directly pass the resulting string as an argument.
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