blocks|key|1150645|text|尝试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1150646|echo+"$VAR"|code-block|syntax|javascript|1150647|而不是。|1150648|(对于echo，-e是一个有效的选项-这就是导致这种现象的原因)。|offset|length|style|CODE|1150649|entityMap^0|0|0|0|3|4|8|2|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]|$K|W|L|X|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|P|$]]

Try:

<pre><code>echo "$VAR"
</code></pre>

instead.

(<code>-e</code> is a valid option for <code>echo</code> - this is what causes this phenomenon).

blocks|key|1150654|text|变量VAR包含-e+xyz，如果通过$访问该变量，-e将被解释为echo的命令行选项。请注意，$VAR的内容不会自动包装到""中。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1150655|使用echo+"$VAR"解决您的问题。|1150656|entityMap^0|2|3|7|6|I|1|P|2|W|4|1B|4|1P|2|0|2|B|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|I|$]]

The variable <code>VAR</code> contains <code>-e xyz</code>, if you access the variable via <code>$</code> the <code>-e</code> is interpreted as a command-line option for <code>echo</code>. Note that the content of <code>$VAR</code> is not wrapped into <code>""</code> automatically. 

Use <code>echo "$VAR"</code> to fix your problem.

blocks|key|1150725|text|说把$VAR放在引号里的答案只会带来副作用。也就是说，当放在引号中时，echo(1)接收-e+xyz的单个参数，因为这不是一个有效的选项字符串，所以echo只打印出来。这是一个副作用，因为echo可以很容易地打印出关于格式错误的选项的错误。大多数程序都会这样做，但似乎GNU+echo+(来自coreutils)和内置到bash中的版本只是回显以连字符开头的字符串，但不是有效的参数字符串。此行为未记录在案，因此不应依赖它。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1150726|此外，如果$VAR包含有效的echo选项参数，则引用$VAR将无济于事：|1150727|$+VAR="-e"
$+echo+"$VAR"

$|code-block|syntax|javascript|1150728|大多数GNU程序将--作为参数，这意味着不再进行选项处理--+--之后的所有参数都将作为非选项参数处理。bash+echo不支持此功能，因此您无法使用它。即使有，它也不会是可移植的。echo还有其他的可移植性问题(-n与\c，没有-e)。|1150729|正确且可移植的解决方案是使用printf(1)。|1150730|printf+"%25s\n"+"$VAR"|1150731|entityMap^0|2|4|Z|7|18|6|22|4|2M|4|3U|4|42|9|4G|4|0|5|4|E|4|0|0|9|2|V|2|1G|9|2J|4|2Z|2|32|2|37|2|0|E|9|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1B|8|@$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|1G|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1H|8|@$9|1I|A|1J|B|C]|$9|1K|A|1L|B|C]|$9|1M|A|1N|B|C]|$9|1O|A|1P|B|C]|$9|1Q|A|1R|B|C]|$9|1S|A|1T|B|C]|$9|1U|A|1V|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|1W|8|@$9|1X|A|1Y|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|J|7|1Z|8|@]|D|@]|E|$K|L]]|$1|S|3|-4|5|6|7|20|8|@]|D|@]|E|$]]]|T|$]]

The answers that say to put <code>$VAR</code> in quotes are only correct by side effect. That is, when put in quotes, <code>echo(1)</code> receives a single argument of <code>-e xyz</code>, and since that is not a valid option string, <code>echo</code> just prints it out. It is a side effect as <code>echo</code> could just as easily print an error regarding malformed options. Most programs will do this, but it seems GNU <code>echo</code> (from <code>coreutils</code>) and the version built into <code>bash</code> simply echo strings that start with a hyphen but are not valid argument strings. This behaviour is not documented so it should not be relied upon.

Further, if <code>$VAR</code> contains a valid <code>echo</code> option argument, then quoting $VAR will not help:

<pre><code>$ VAR="-e"
$ echo "$VAR"

$
</code></pre>

Most GNU programs take <code>--</code> as an argument to mean no more option processing — all the arguments after <code>--</code> are to be processed as non-option arguments. <code>bash echo</code> does not support this so you cannot use it. Even if it did, it would not be portable. <code>echo</code> has other portability issues (<code>-n</code> vs <code>\c</code>, no <code>-e</code>).

The correct and portable solution is to use <code>printf(1)</code>.

<pre><code>printf "%s\n" "$VAR"
</code></pre>

blocks|key|936404|text|在zsh中，您可以在参数前使用单个破折号(-)。这样可以确保以下参数不会被解释为选项。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|936405|%25+VAR="-e+xyz"
%25+echo+-+$VAR
-e+xyz|code-block|syntax|javascript|936406|从zsh文档中：|936407|+++echo+[+-neE+]+[+arg+...+]
++++++++++...
++++++++++Note+that+for+standards+compliance+a+double+dash+does+not
++++++++++terminate+option+processing;+instead,+it+is+printed+directly.
++++++++++However,+a+single+dash+does+terminate+option+processing,+so+the
++++++++++first+dash,+possibly+following+options,+is+not+printed,+but
++++++++++everything+following+it+is+printed+as+an+argument.

+++++++++The+single+dash+behaviour+is+different+from+other+shells.|936408|请记住，此行为是特定于zsh的。|936409|entityMap^0|1|3|L|1|0|0|1|3|0|0|B|3|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|R|$]]

In <code>zsh</code>, you can use a single dash (<code>-</code>) before your arguments. This ensures that no following arguments are interpreted as options.
<pre><code>% VAR=&quot;-e xyz&quot;
% echo - $VAR
-e xyz
</code></pre>
From the <code>zsh</code> docs:
<pre><code> echo [ -neE ] [ arg ... ]
 ...
 Note that for standards compliance a double dash does not
 terminate option processing; instead, it is printed directly.
 However, a single dash does terminate option processing, so the
 first dash, possibly following options, is not printed, but
 everything following it is printed as an argument.

 The single dash behaviour is different from other shells.
</code></pre>
Keep in mind this behavior is specific to <code>zsh</code>.

<pre><code>VAR="-e xyz"
echo $VAR
</code></pre>

This prints <code>xyz</code>, for some reason. I don't seem to be able to find a way to get a string to start with <code>-e</code>.

What is going on here?

Bash: echo string that starts with "-"

VAR="-e xyz"echo $VAR出于某种原因，此命令将打印xyz。我似乎找不到一种方法来获得以-e开头的字符串。这里发生什么事情？

问Bash:以"-“开头的回显字符串
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