blocks|key|962232|text|尝试以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|962233|grep+-v+-e+'%5E$'+foo.txt|code-block|syntax|javascript|962234|-e选项允许匹配正则表达式模式。|offset|length|style|CODE|962235|%5E$前后的单引号使其适用于Cshell。其他shell会喜欢使用单引号或双引号。|962236|更新:这适用于包含空白行或“全空白”的文件(例如带有"\r\n“样式行结尾的windows行)，而上面只删除具有空白行和unix样式行结尾的文件：|962237|grep+-v+-e+'%5E[[:space:]]*$'+foo.txt|962238|entityMap^0|0|0|0|2|0|0|2|0|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]]|9|@]|A|$]]|$1|M|3|N|5|6|7|Z|8|@$I|10|J|11|K|L]]|9|@]|A|$]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|T|$]]

Try the following:

<pre><code>grep -v -e '^$' foo.txt
</code></pre>

The <code>-e</code> option allows regex patterns for matching.

The single quotes around <code>^$</code> makes it work for Cshell. Other shells will be happy with either single or double quotes.

UPDATE: This works for me for a file with blank lines or "all white space" (such as windows lines with "\r\n" style line endings), whereas the above only removes files with blank lines and unix style line endings:

<pre><code>grep -v -e '^[[:space:]]*$' foo.txt
</code></pre>

blocks|key|1176924|text|保持简单。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1176925|grep+.+filename.txt|code-block|syntax|javascript|1176926|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|$]]

Keep it simple.

<pre><code>grep . filename.txt
</code></pre>

blocks|key|1176600|text|使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1176601|$+dos2unix+file
$+grep+-v+"%5E$"+file|code-block|syntax|javascript|1176602|或者只是简单的awk：|1176603|awk+'NF'+file|1176604|如果您没有dos2unix，那么可以使用诸如tr之类的工具|offset|length|1176605|tr+-d+'\r'+<+"$file"+>+t+;+mv+t+"$file"|1176606|entityMap|0|LINK|mutability|MUTABLE|url|https://linux.die.net/man/1/tr^0|0|0|0|0|M|2|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@$M|13|N|14|1|15]]|A|$]]|$1|O|3|P|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

Use:

<pre><code>$ dos2unix file
$ grep -v "^$" file
</code></pre>

Or just simply awk:

<pre><code>awk 'NF' file
</code></pre>

If you don't have dos2unix, then you can use tools like <a href="https://linux.die.net/man/1/tr" rel="noreferrer">tr</a>:

<pre><code>tr -d '\r' &lt; "$file" &gt; t ; mv t "$file"
</code></pre>

blocks|key|962384|text|grep+-v+"%5E[[:space:]]*$"

The+-v+makes+it+print+lines+that+do+not+completely+match

===Each+part+explained===
%5E+++++++++++++match+start+of+line
[[:space:]]+++match+whitespace-+spaces,+tabs,+carriage+returns,+etc.
*+++++++++++++previous+match+(whitespace)+may+exist+from+0+to+infinite+times
$+++++++++++++match+end+of+line|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|962385|运行代码-|unstyled|962386|$+echo+"
>+hello
>+++++++
>+ok"+%7C
>+grep+-v+"%5E[[:space:]]*$"
hello
ok|962387|要更多地了解它是如何工作的，我建议阅读正则表达式。http://www.regular-expressions.info/tutorial.html|offset|length|962388|entityMap|0|LINK|mutability|MUTABLE|url|http://www.regular-expressions.info/tutorial.html^0|0|0|0|P|1D|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|V|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|X|8|@]|9|@$K|Y|L|Z|1|10]]|A|$]]|$1|M|3|-4|5|F|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

<pre><code>grep -v "^[[:space:]]*$"

The -v makes it print lines that do not completely match

===Each part explained===
^ match start of line
[[:space:]] match whitespace- spaces, tabs, carriage returns, etc.
* previous match (whitespace) may exist from 0 to infinite times
$ match end of line
</code></pre>

Running the code-

<pre><code>$ echo "
&gt; hello
&gt; 
&gt; ok" |
&gt; grep -v "^[[:space:]]*$"
hello
ok
</code></pre>

To understand more about how/why this works, I recommend reading up on regular expressions. <a href="http://www.regular-expressions.info/tutorial.html" rel="noreferrer">http://www.regular-expressions.info/tutorial.html</a>

blocks|key|1176909|text|与前面的答案相同：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1176910|grep+-v+-e+'%5E$'+foo.txt|code-block|syntax|javascript|1176911|这里，grep+-e指的是grep的扩展版本。“%5E$”表示%5E(行首)和$(行尾)之间没有任何字符。'%5E‘和'$’是正则表达式字符。|offset|length|style|CODE|1176912|因此，命令grep+-v将打印与此模式不匹配的所有行(%5E和$之间没有字符)。|1176913|这样，空白行就被消除了。|1176914|entityMap|0|LINK|mutability|MUTABLE|url|http://linux.die.net/man/1/grep^0|0|0|3|7|D|4|0|0|5|7|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$I|11|J|12|K|L]]|9|@$I|13|J|14|1|15]]|A|$]]|$1|M|3|N|5|6|7|16|8|@$I|17|J|18|K|L]]|9|@]|A|$]]|$1|O|3|P|5|6|7|19|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

The same as the previous answers:

<pre><code>grep -v -e '^$' foo.txt
</code></pre>

Here, <code>grep -e</code> means the extended version of <a href="http://linux.die.net/man/1/grep" rel="nofollow noreferrer">grep</a>. '^$' means that there isn't any character between ^(Start of line) and $(end of line). '^' and '$' are regex characters.

So the command <code>grep -v</code> will print all the lines that do not match this pattern (No characters between ^ and $).

This way, empty blank lines are eliminated.

blocks|key|962735|text|如果在一行中有多个空行的序列，并且希望每个序列只有一个空行，请尝试|type|unstyled|depth|inlineStyleRanges|entityRanges|data|962736|grep+-v+"unwantedThing"+foo.txt+%7C+cat+-s|code-block|syntax|javascript|962737|cat+-s禁止重复的空输出行。|offset|length|style|CODE|962738|您的输出将从|962739|match1



match2|962740|至|962741|match1

match2|962742|原始输出中的三个空行将被压缩或“压缩”成一个空行。|962743|entityMap^0|0|0|0|6|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$I|11|J|12|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|15|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|17|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

If you have sequences of multiple blank lines in a row, and would like only one blank line per sequence, try

<pre><code>grep -v "unwantedThing" foo.txt | cat -s
</code></pre>

<code>cat -s</code> suppresses repeated empty output lines.

Your output would go from

<pre><code>match1



match2
</code></pre>

to

<pre><code>match1

match2
</code></pre>

The three blank lines in the original output would be compressed or "squeezed" into one blank line.

blocks|key|1176519|text|我更喜欢使用egrep，尽管在我使用包含空白行的真正文件的测试中，您的方法工作得很好(尽管在我的测试中没有引号)。这也行得通：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1176520|egrep+-v+"%5E(\r?\n)?$"+filename.txt|code-block|syntax|javascript|1176521|entityMap^0|6|5|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|$]]

I prefer using <code>egrep</code>, though in my test with a genuine file with blank line your approach worked fine (though without quotation marks in my test). This worked too:

<pre><code>egrep -v "^(\r?\n)?$" filename.txt
</code></pre>

blocks|key|962848|text|文件中的行是否包含空格字符？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|962849|如果是这样的话|962850|grep+"\S"+file.txt|offset|length|style|CODE|962851|否则|962852|grep+.+file.txt|962853|答案来自：https://serverfault.com/a/688789|962854|entityMap|0|LINK|mutability|MUTABLE|url|https://serverfault.com/a/688789^0|0|0|0|I|0|0|0|F|0|5|W|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Y|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Z|8|@$F|10|G|11|H|I]]|9|@]|A|$]]|$1|J|3|K|5|6|7|12|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|13|8|@$F|14|G|15|H|I]]|9|@]|A|$]]|$1|N|3|O|5|6|7|16|8|@]|9|@$F|17|G|18|1|19]]|A|$]]|$1|P|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|Q|$R|$5|S|T|U|A|$V|W]]]]

Do lines in the file have whitespace characters?
If so then
<code>grep &quot;\S&quot; file.txt</code>
Otherwise
<code>grep . file.txt</code>
Answer obtained from:
<a href="https://serverfault.com/a/688789">https://serverfault.com/a/688789</a>

blocks|key|1176794|text|awk+'NF'+file-with-blank-lines+>+file-with-no-blank-lines|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1176795|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>awk 'NF' file-with-blank-lines &gt; file-with-no-blank-lines
</code></pre>

blocks|key|1176562|text|我很努力地尝试了，但这似乎起作用了(假设\r在这里咬你)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1176563|printf+"\r"+%7C+egrep+-xv+"[[:space:]]*"|code-block|syntax|javascript|1176564|entityMap^0|K|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|$]]

I tried hard, but this seems to work (assuming <code>\r</code> is biting you here):

<pre><code>printf "\r" | egrep -xv "[[:space:]]*"
</code></pre>

blocks|key|1177193|text|此代码删除空行和以"#“开头的行|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1177194|+grep+-v+"%5E#"+file.txt+%7C+grep+-v+%5E[[:space:]]*$|code-block|syntax|javascript|1177195|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|$]]

This code removes blank lines and lines that start with &quot;#&quot;
<pre><code> grep -v &quot;^#&quot; file.txt | grep -v ^[[:space:]]*$
</code></pre>

blocks|key|962408|text|使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|962409|grep+pattern+filename.txt+%7C+uniq|code-block|syntax|javascript|962410|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|$]]

Use:

<pre><code>grep pattern filename.txt | uniq
</code></pre>

blocks|key|962571|text|使用Perl：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|962572|perl+-ne+'print+if+/\S/'|code-block|syntax|javascript|962573|\S表示匹配非空白字符。|offset|length|style|CODE|962574|entityMap^0|0|0|0|2|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]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Using Perl:

<pre><code>perl -ne 'print if /\S/'
</code></pre>

<code>\S</code> means match non-blank characters.

blocks|key|962617|text|下面是删除白色线条和以#符号开头的线条的另一种方法。我认为这对于读取配置文件非常有用。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|962618|[root@localhost+~]#+cat+/etc/sudoers+%7C+egrep+-v+'%5E(#%7C$)'
Defaults++++requiretty
Defaults+++!visiblepw
Defaults++++always_set_home
Defaults++++env_reset
Defaults++++env_keep+=++"COLORS+DISPLAY+HOSTNAME+HISTSIZE+INPUTRC+KDEDIR
LS_COLORS"
root++++ALL=(ALL)+++++++ALL
%25wheel++ALL=(ALL)+++++++ALL
stack+ALL=(ALL)+NOPASSWD:+ALL|code-block|syntax|javascript|962619|entityMap^0|B|1|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|$]]

Here is another way of removing the white lines and lines starting with the <code>#</code> sign. I think this is quite useful to read configuration files.

<pre><code>[root@localhost ~]# cat /etc/sudoers | egrep -v '^(#|$)'
Defaults requiretty
Defaults !visiblepw
Defaults always_set_home
Defaults env_reset
Defaults env_keep = "COLORS DISPLAY HOSTNAME HISTSIZE INPUTRC KDEDIR
LS_COLORS"
root ALL=(ALL) ALL
%wheel ALL=(ALL) ALL
stack ALL=(ALL) NOPASSWD: ALL
</code></pre>

blocks|key|962788|text|%5E\s\s%2B+-v“egrep”|type|unstyled|depth|inlineStyleRanges|entityRanges|data|962789|egrep已经执行正则表达式，\s是空格。|962790|%2B复制当前模式。|962791|%5E表示开始|962792|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

egrep -v "^\s\s+"

egrep already do regex, and the \s is white space. 

The + duplicates current pattern.

The ^ is for the start

I tried <code>grep -v '^$'</code> in Linux and that didn't work. This file came from a Windows file system.

Remove blank lines with grep

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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我在Linux中尝试了grep -v '^$'，但没有起作用。此文件来自Windows文件系统。

问使用grep删除空行
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