blocks|key|1193964|text|如果你想解释$replace，你不应该使用单引号，因为单引号会阻止变量替换。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1193965|尝试：|1193966|echo+$LINE+%7C+sed+-e+"s/12345678/${replace}/g"|code-block|syntax|javascript|1193967|文字记录：|1193968|pax>+export+replace=987654321
pax>+echo+X123456789X+%7C+sed+"s/123456789/${replace}/"
X987654321X
pax>+_|1193969|只需小心确保${replace}没有任何对sed有意义的字符(例如/+)，因为除非进行转义，否则它将导致混淆。但是，如果像您所说的那样，用一个数字替换另一个数字，这应该不是问题。|1193970|entityMap^0|6|8|0|0|0|0|0|6|A|L|3|X|1|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|Z|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|T|$]]

If you want to interpret <code>$replace</code>, you should not use single quotes since they prevent variable substitution.
Try:
<pre><code>echo $LINE | sed -e &quot;s/12345678/${replace}/g&quot;
</code></pre>
Transcript:
<pre><code>pax&gt; export replace=987654321
pax&gt; echo X123456789X | sed &quot;s/123456789/${replace}/&quot;
X987654321X
pax&gt; _
</code></pre>
Just be careful to ensure that <code>${replace}</code> doesn't have any characters of significance to <code>sed</code> (like <code>/</code> for instance) since it will cause confusion unless escaped. But if, as you say, you're replacing one number with another, that shouldn't be a problem.

blocks|key|1194052|text|您可以使用shell+(bash/ksh)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1194053|$+var="12345678abc"
$+replace="test"
$+echo+${var//12345678/$replace}
testabc|code-block|syntax|javascript|1194054|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|$]]

you can use the shell (bash/ksh). 

<pre><code>$ var="12345678abc"
$ replace="test"
$ echo ${var//12345678/$replace}
testabc
</code></pre>

blocks|key|1194272|text|不是专门针对这个问题的，但对于那些需要从前面的答案中清楚地扩展相同功能的人来说：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1194273|#+create+some+variables
str="someFileName.foo"
find=".foo"
replace=".bar"
#+notice+the+the+str+isn't+prefixed+with+$
#++++this+is+just+how+this+feature+works+:/
result=${str//$find/$replace}
echo+$result++++
#+result+is:+someFileName.bar

str="someFileName.sally"
find=".foo"
replace=".bar"
result=${str//$find/$replace}
echo+$result++++
#+result+is:+someFileName.sally+because+".foo"+was+not+found|code-block|syntax|javascript|1194274|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|$]]

Not specific to the question, but for folks who need the same kind of functionality expanded for clarity from previous answers:

<pre><code># create some variables
str="someFileName.foo"
find=".foo"
replace=".bar"
# notice the the str isn't prefixed with $
# this is just how this feature works :/
result=${str//$find/$replace}
echo $result 
# result is: someFileName.bar

str="someFileName.sally"
find=".foo"
replace=".bar"
result=${str//$find/$replace}
echo $result 
# result is: someFileName.sally because ".foo" was not found
</code></pre>

blocks|key|979990|text|找到graceful+solution。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|979991|echo+${LINE//12345678/$replace}|code-block|syntax|javascript|979992|entityMap|0|LINK|mutability|MUTABLE|url|https://www.shellscript.sh/tips/pattern-substitution/^0|2|H|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Found a <a href="https://www.shellscript.sh/tips/pattern-substitution/" rel="noreferrer">graceful solution</a>.

<pre><code>echo ${LINE//12345678/$replace}
</code></pre>

blocks|key|979699|text|单引号很强。一旦进入，您就无法调用变量替换，除非您离开。请使用双引号：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|979700|echo+$LINE+%7C+sed+-e+"s/12345678/$replace/g"|code-block|syntax|javascript|979701|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|$]]

Single quotes are very strong. Once inside, there's nothing you can do to invoke variable substitution, until you leave. Use double quotes instead:

<pre><code>echo $LINE | sed -e "s/12345678/$replace/g"
</code></pre>

blocks|key|979616|text|echo+$LINE+%7C+sed+-e+'s/12345678/'$replace'/g'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|979617|您仍然可以使用单引号，但是当您想要在正确的位置展开变量时，您必须“打开”它们。否则字符串将被“字面理解”(正如@paxdiablo所说的那样，他的答案也是正确的)|unstyled|979618|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>echo $LINE | sed -e 's/12345678/'$replace'/g'
</code></pre>

you can still use single quotes, but you have to "open" them when you want the variable expanded at the right place. otherwise the string is taken "literally" (as @paxdiablo correctly stated, his answer is correct as well)

blocks|key|979817|text|要让shell展开变量，需要使用双引号，如|type|unstyled|depth|inlineStyleRanges|entityRanges|data|979818|sed+-i+"s#12345678#$replace#g"+file.txt|code-block|syntax|javascript|979819|如果$replace包含特殊的sed字符(#、\)，这将中断。但您可以对$replace进行预处理以引用它们：|offset|length|style|CODE|979820|replace_quoted=$(printf+'%25s'+"$replace"+%7C+sed+'s/[#\]/\\\0/g')
sed+-i+"s#12345678#$replace_quoted#g"+file.txt|979821|entityMap^0|0|0|2|8|F|3|L|1|N|1|10|8|0|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]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$]]

To let your shell expand the variable, you need to use double-quotes like

<pre><code>sed -i "s#12345678#$replace#g" file.txt
</code></pre>

This will break if <code>$replace</code> contain special <code>sed</code> characters (<code>#</code>, <code>\</code>). But you can preprocess <code>$replace</code> to quote them:

<pre><code>replace_quoted=$(printf '%s' "$replace" | sed 's/[#\]/\\\0/g')
sed -i "s#12345678#$replace_quoted#g" file.txt
</code></pre>

blocks|key|1194412|text|我对此也有类似的要求，但我的替换变量包含一个“与”符号。像这样转义&符号就解决了我的问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1194413|replace="salt+&+pepper"
echo+"pass+the+salt"+%7C+sed+"s/salt/${replace/&/\&}/g"|code-block|syntax|javascript|1194414|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 had a similar requirement to this but my replace var contained an ampersand. Escaping the ampersand like this solved my problem:

<pre><code>replace="salt &amp; pepper"
echo "pass the salt" | sed "s/salt/${replace/&amp;/\&amp;}/g"
</code></pre>

blocks|key|979940|text|改用这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|979941|echo+$LINE+%7C+sed+-e+'s/12345678/$replace/g'|code-block|syntax|javascript|979942|这对我很有效，只需删除引号即可|979943|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Use this instead

<pre><code>echo $LINE | sed -e 's/12345678/$replace/g'
</code></pre>

this works for me just simply remove the quotes

blocks|key|1194181|text|我更喜欢使用双引号，因为单引号非常强大，因为如果不能更改其中的任何内容或者可以调用变量替换，单引号就像我们使用它们一样强大。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1194182|因此请使用已安装的双引号。|1194183|echo+$LINE+%7C+sed+-e+"s/12345678/$replace/g"|code-block|syntax|javascript|1194184|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I prefer to use double quotes , as single quptes are very powerful as we used them if dont able to change anything inside it or can invoke the variable substituion .

so use double quotes instaed.

<pre><code>echo $LINE | sed -e "s/12345678/$replace/g"
</code></pre>

I am using the below code for replacing a string
inside a shell script.

<pre><code>echo $LINE | sed -e 's/12345678/"$replace"/g'
</code></pre>

but it's getting replaced with <code>$replace</code> instead of the value of that variable.

Could anybody tell what went wrong?

Replace a string in shell script using a variable

我使用下面的代码替换shell脚本中的字符串。echo $LINE | sed -e 's/12345678/"$replace"/g'但它将被$replace代替，而不是该变量的值。有谁能说出哪里出了问题吗？

问使用变量替换shell脚本中的字符串
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