blocks|key|220391|text|echo+"%7C$COMMAND%7C"%7Ctr+'\n'+'+'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|220392|将换行符(在POSIX/Unix中不是回车符)替换为空格。|unstyled|220393|老实说，我会考虑从bash切换到更正常的东西。或者从一开始就避免生成这种畸形数据。|220394|嗯，这似乎也是一个可怕的安全漏洞，这取决于数据来自哪里。|220395|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|F|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>echo "|$COMMAND|"|tr '\n' ' '
</code></pre>

will replace the newline (in POSIX/Unix it's not a carriage return) with a space.

To be honest I would think about switching away from bash to something more sane though. Or avoiding generating this malformed data in the first place.

Hmmm, this seems like it could be a horrible security hole as well, depending on where the data is coming from.

blocks|key|220477|text|通过删除所有回车符来清除变量：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|220478|COMMAND=$(echo+$COMMAND%7Ctr+-d+'\n')|code-block|syntax|javascript|220479|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|$]]

Clean your variable by removing all the carriage returns:

<pre><code>COMMAND=$(echo $COMMAND|tr -d '\n')
</code></pre>

blocks|key|220515|text|如果在启用extglob选项的情况下使用bash，则可以通过以下方式仅删除尾随的空格：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|220516|shopt+-s+extglob
COMMAND=$'\nRE+BOOT\r+++\n'
echo+"%7C${COMMAND%25%25*([$'\t\r\n+'])}%7C"|code-block|syntax|javascript|220517|这将输出以下内容：|220518|%7C
RE+BOOT%7C|220519|或者将%25%25替换为##，以仅替换前导空格。|220520|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

If you are using bash with the extglob option enabled, you can remove just the trailing whitespace via:

<pre><code>shopt -s extglob
COMMAND=$'\nRE BOOT\r \n'
echo "|${COMMAND%%*([$'\t\r\n '])}|"
</code></pre>

This outputs:

<pre><code>|
RE BOOT|
</code></pre>

Or replace %% with ## to replace just the leading whitespace.

blocks|key|3759|text|对我来说起作用的是echo+$testVar+%7C+tr+"\n"+"+"|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3760|其中testVar包含我的变量/脚本输出|3761|entityMap^0|9|R|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|N|8|@]|D|@]|E|$]]]|I|$]]

What worked for me was <code>echo $testVar | tr "\n" " "</code>

Where testVar contained my variable/script-output

blocks|key|3767|text|您可以简单地使用echo+-n+"%7C$COMMAND%7C"。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3768|entityMap^0|8|K|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

You can simply use <code>echo -n "|$COMMAND|"</code>.

blocks|key|220595|text|为了解决实际问题的一个可能根源，您有可能正在获取crlf文件。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|220596|CRLF示例：|220597|.env+(crlf)|220598|VARIABLE_A="abc"
VARIABLE_B="def"|code-block|syntax|javascript|220599|run.sh|220600|#!/bin/bash
source+.env
echo+"$VARIABLE_A"
echo+"$VARIABLE_B"
echo+"$VARIABLE_A+$VARIABLE_B"|220601|返回：|220602|abc
def
+def|220603|但是，如果您将其转换为LF：|220604|.env+(lf)|220605|220606|220607|#!/bin/bash
source+.env
echo+"$VARIABLE_A"
echo+"$VARIABLE_B"
echo+"$VARIABLE_A+$VARIABLE_B"|220608|220609|abc
def
abc+def|220610|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|16|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|17|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|18|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|19|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|1A|8|@]|9|@]|A|$I|J]]|$1|O|3|P|5|6|7|1B|8|@]|9|@]|A|$]]|$1|Q|3|R|5|H|7|1C|8|@]|9|@]|A|$I|J]]|$1|S|3|T|5|6|7|1D|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1E|8|@]|9|@]|A|$]]|$1|W|3|G|5|H|7|1F|8|@]|9|@]|A|$I|J]]|$1|X|3|L|5|6|7|1G|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|H|7|1H|8|@]|9|@]|A|$I|J]]|$1|10|3|P|5|6|7|1I|8|@]|9|@]|A|$]]|$1|11|3|12|5|H|7|1J|8|@]|9|@]|A|$I|J]]|$1|13|3|-4|5|6|7|1K|8|@]|9|@]|A|$]]]|14|$]]

To address one possible root of the actual issue, there is a chance you are sourcing a crlf file.
CRLF Example:
.env (crlf)
<pre><code>VARIABLE_A=&quot;abc&quot;
VARIABLE_B=&quot;def&quot;
</code></pre>
run.sh
<pre><code>#!/bin/bash
source .env
echo &quot;$VARIABLE_A&quot;
echo &quot;$VARIABLE_B&quot;
echo &quot;$VARIABLE_A $VARIABLE_B&quot;
</code></pre>
Returns:
<pre><code>abc
def
 def
</code></pre>
If however you convert to LF:
.env (lf)
<pre><code>VARIABLE_A=&quot;abc&quot;
VARIABLE_B=&quot;def&quot;
</code></pre>
run.sh
<pre><code>#!/bin/bash
source .env
echo &quot;$VARIABLE_A&quot;
echo &quot;$VARIABLE_B&quot;
echo &quot;$VARIABLE_A $VARIABLE_B&quot;
</code></pre>
Returns:
<pre><code>abc
def
abc def
</code></pre>

I have the following variable.

<pre><code>echo "|$COMMAND|"
</code></pre>

which returns 

<pre><code>|
REBOOT|
</code></pre>

How can I remove that first newline?

How to remove a newline from a string in Bash

Unix

我有以下变量。echo "|$COMMAND|"它会返回|REBOOT|如何删除第一个换行符？

问如何在Bash中删除字符串中的换行符
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