blocks|key|406629|text|如果您使用jq并尝试从结果中删除引号，其他答案将会起作用，但有一种更好的方法。通过使用-r选项，您可以输出不带引号的结果。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|406630|$+echo+'{"foo":+"bar"}'+%7C+jq+'.foo'
"bar"

$+echo+'{"foo":+"bar"}'+%7C+jq+-r+'.foo'
bar|code-block|syntax|javascript|406631|entityMap^0|17|2|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

If you're using jq and trying to remove the quotes from the result, the other answers will work, but there's a better way. By using the <code>-r</code> option, you can output the result with no quotes.

<pre><code>$ echo '{"foo": "bar"}' | jq '.foo'
"bar"

$ echo '{"foo": "bar"}' | jq -r '.foo'
bar
</code></pre>

blocks|key|621937|text|只需调用一次sed即可完成此操作|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|621938|$+echo+"\"html\\test\\\""+%7C+sed+'s/%5E"$.*$"$/\1/'
html\test\|code-block|syntax|javascript|621939|entityMap^0|6|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

You can do it with only one call to <code>sed</code>:

<pre><code>$ echo "\"html\\test\\\"" | sed 's/^"$.*$"$/\1/'
html\test\
</code></pre>

blocks|key|622163|text|Bash中最简单的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|622164|$+s='"abc"'
$+echo+$s
"abc"
$+echo+"${s:1:-1}"
abc|code-block|syntax|javascript|622165|这称为子串扩展(参见Gnu+Bash+Manual和搜索${parameter:offset:length})。在本例中，它从位置1开始并在倒数第二个位置结束的s中获取子字符串。这是因为如果length是一个负值，它会被解释为从parameter末尾向后运行的偏移量。|offset|length|style|CODE|622166|entityMap|0|LINK|mutability|MUTABLE|url|https://www.gnu.org/savannah-checkouts/gnu/bash/manual/bash.html^0|0|0|S|Q|28|1|2N|6|36|9|A|F|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]|$I|13|J|14|K|L]]|9|@$I|15|J|16|1|17]]|A|$]]|$1|M|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

The easiest solution in Bash:

<pre><code>$ s='"abc"'
$ echo $s
"abc"
$ echo "${s:1:-1}"
abc
</code></pre>

This is called substring expansion (see <a href="https://www.gnu.org/savannah-checkouts/gnu/bash/manual/bash.html" rel="noreferrer">Gnu Bash Manual</a> and search for <code>${parameter:offset:length}</code>). In this example it takes the substring from <code>s</code> starting at position 1 and ending at the second last position. This is due to the fact that if <code>length</code> is a negative value it is interpreted as a backwards running offset from the end of <code>parameter</code>.

blocks|key|622395|text|如果你来这里是为了aws+cli+--query，可以试试--output+text。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|622396|entityMap^0|H|7|T|D|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

If you come here for aws cli <code>--query</code>, try <code>--output text</code>.

blocks|key|406862|text|Linux=`cat+/etc/os-release+%7C+grep+"ID"+%7C+head+-1+%7C+awk+-F=+'{+print+$2+}'`

echo+$Linux
Output:
"amzn"|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|406863|从变量中删除双引号的最简单方法是|unstyled|406864|Linux=`echo+"$Linux"+%7C+tr+-d+'"'`+|406865|Linux=$(eval+echo+$Linux)|406866|Linux=`echo+${Linux//\"/}`|406867|Linux=`echo+$Linux+%7C+xargs`|406868|All提供不带双引号的输出：|406869|echo+$Linux|406870|amzn|406871|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|X|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$B|C]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$B|C]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$B|C]]|$1|O|3|P|5|F|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|F|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|F|7|15|8|@]|9|@]|A|$]]]|V|$]]

<pre><code>Linux=`cat /etc/os-release | grep &quot;ID&quot; | head -1 | awk -F= '{ print $2 }'`

echo $Linux
Output:
&quot;amzn&quot;
</code></pre>
Simplest ways to remove double quotes from variables are
<pre><code>Linux=`echo &quot;$Linux&quot; | tr -d '&quot;'` 
</code></pre>
<pre><code>Linux=$(eval echo $Linux)
</code></pre>
<pre><code>Linux=`echo ${Linux//\&quot;/}`
</code></pre>
<pre><code>Linux=`echo $Linux | xargs`
</code></pre>
All provides the Output without double quotes:
echo $Linux
amzn

blocks|key|622329|text|我知道这是一个非常古老的问题，但这里有另一个sed变体，它可能对某些人有用。与其他一些不同的是，它只替换开头或结尾的双引号...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|622330|echo+"$opt"+%7C+sed+-r+'s/%5E"%7C"$//g'|code-block|syntax|javascript|622331|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I know this is a very old question, but here is another sed variation, which may be useful to someone. Unlike some of the others, it only replaces double quotes at the start or end...

<pre><code>echo "$opt" | sed -r 's/^"|"$//g'
</code></pre>

blocks|key|622091|text|我的版本|type|unstyled|depth|inlineStyleRanges|entityRanges|data|622092|strip_quotes()+{
++++while+[[+$#+-gt+0+]];+do
++++++++local+value=${!1}
++++++++local+len=${#value}
++++++++[[+${value:0:1}+==+\"+&&+${value:$len-1:1}+==+\"+]]+&&+declare+-g+$1="${value:1:$len-2}"
++++++++shift
++++done
}|code-block|syntax|javascript|622093|该函数接受变量名并在适当的位置去掉引号。它只去掉一对匹配的前导引号和尾随引号。它不会检查后面的引号是否转义(前面是\，它本身并不转义)。|offset|length|style|CODE|622094|根据我的经验，这样的通用字符串实用函数(我有一个这样的库)在直接操作字符串时效率最高，不使用任何模式匹配，特别是不创建任何子shell，或者调用任何外部工具，如sed、awk或grep。|622095|var1="\"test+\\+\"+end+\""
var2=test
var3=\"test
var4=test\"
echo+before:
for+i+in+var{1,2,3,4};+do
++++echo+$i="${!i}"
done
strip_quotes+var{1,2,3,4}
echo
echo+after:
for+i+in+var{1,2,3,4};+do
++++echo+$i="${!i}"
done|622096|entityMap^0|0|0|1L|1|0|28|3|2C|3|2G|4|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@$I|Y|J|Z|K|L]|$I|10|J|11|K|L]|$I|12|J|13|K|L]]|9|@]|A|$]]|$1|O|3|P|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|R|$]]

My version

<pre><code>strip_quotes() {
 while [[ $# -gt 0 ]]; do
 local value=${!1}
 local len=${#value}
 [[ ${value:0:1} == \" &amp;&amp; ${value:$len-1:1} == \" ]] &amp;&amp; declare -g $1="${value:1:$len-2}"
 shift
 done
}
</code></pre>

The function accepts variable name(s) and strips quotes in place. It only strips a matching pair of leading and trailing quotes. It doesn't check if the trailing quote is escaped (preceded by <code>\</code> which is not itself escaped). 

In my experience, general-purpose string utility functions like this (I have a library of them) are most efficient when manipulating the strings directly, not using any pattern matching and especially not creating any sub-shells, or calling any external tools such as <code>sed</code>, <code>awk</code> or <code>grep</code>.

<pre><code>var1="\"test \\ \" end \""
var2=test
var3=\"test
var4=test\"
echo before:
for i in var{1,2,3,4}; do
 echo $i="${!i}"
done
strip_quotes var{1,2,3,4}
echo
echo after:
for i in var{1,2,3,4}; do
 echo $i="${!i}"
done
</code></pre>

blocks|key|406556|text|有另一种方法可以做到这一点。像这样：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|406557|echo+${opt:1:-1}|code-block|syntax|javascript|406558|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

There is another way to do it. Like:

<pre><code>echo ${opt:1:-1}
</code></pre>

blocks|key|622384|text|如果您因为Makefile保留了引号而尝试删除引号，请尝试以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|622385|$(subst+$\",,$(YOUR_VARIABLE))|code-block|syntax|javascript|622386|基于另一个答案：https://stackoverflow.com/a/10430975/10452175|offset|length|622387|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/10430975/10452175^0|0|0|8|19|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

If you try to remove quotes because the Makefile keeps them, try this:
<pre class="lang-sh prettyprint-override"><code>$(subst $\&quot;,,$(YOUR_VARIABLE))
</code></pre>
Based on another answer: <a href="https://stackoverflow.com/a/10430975/10452175">https://stackoverflow.com/a/10430975/10452175</a>

Below is the snippet of a shell script from a larger script. It removes the quotes from the string that is held by a variable. I am doing it using sed, but is it efficient? If not, then what is the efficient way?

<pre><code>#!/bin/sh

opt="\"html\\test\\\""
temp=`echo $opt | sed 's/.$.*$/\1/' | sed 's/$.*$./\1/'`
echo $temp
</code></pre>

Shell script - remove first and last quote (") from a variable

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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下面是一个更大的脚本中的shell脚本的片段。它从变量所持有的字符串中删除引号。我正在使用sed来做这件事，但是它是否有效呢？如果不是，那么有效的方法是什么？#!/bin/shopt="\"html\\test\\\""temp=`echo $opt | sed 's/.$.*$/\1/' | sed 's/$.*$./\1/'`echo $temp

问Shell脚本-从变量中删除第一个和最后一个引号(")
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回答 9

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