blocks|key|4538864|text|尝试使用awk，eks：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4538865|ls+-tlrh+/+%7C+awk+'NR==1;+/tmp/'|code-block|syntax|javascript|4538866|这将打印行号1和带有tmp的行|4538867|NR==1;打印行号1|4538868|/tmp/打印包含tmp的所有行|4538869|entityMap^0|4|3|0|0|6|1|A|3|0|0|6|A|1|0|0|5|9|3|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|R|$]]

Try use <code>awk</code>, eks:

<pre><code>ls -tlrh / | awk 'NR==1; /tmp/'
</code></pre>

This will print line number <code>1</code> and line with <code>tmp</code>

<code>NR==1;</code> print line number <code>1</code> 
<code>/tmp/</code> print all lines that contains <code>tmp</code>

blocks|key|755341|text|@Jotne的答案更好，但有时你可以使用grep+-E，如果你知道第一行中的某些东西，那么你可以搜索它或其他你想要的东西，比如用竖线符号来表示交替：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|755342|ps+-ef+%7C+grep+-E+"UID%7Cbash"|code-block|syntax|javascript|755343|输出|BOLD|755344|UID+++PID++PPID+++C+STIME+++TTY+++++++++++TIME+CMD
502+++510+++509+++0++8:01am+ttys000++++0:00.08+-bash
502+48806+++510+++0+10:18am+ttys000++++0:00.00+grep+-E+UID%7Cbash|755345|entityMap^0|K|7|0|0|0|2|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|U|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|V|8|@$9|W|A|X|B|M]]|D|@]|E|$]]|$1|N|3|O|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|P|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|Q|$]]

@Jotne's answer is better, but sometimes you can use <code>grep -E</code> if you know something in the first line, then you can search for that OR the other thing you want like this with the pipe symbol to express the alternation:

<pre><code>ps -ef | grep -E "UID|bash"
</code></pre>

Output

<pre><code>UID PID PPID C STIME TTY TIME CMD
502 510 509 0 8:01am ttys000 0:00.08 -bash
502 48806 510 0 10:18am ttys000 0:00.00 grep -E UID|bash
</code></pre>

blocks|key|755371|text|这不起作用的原因是，两个进程中的第一个进程(头-n1)读取的内容多于其输出的内容。它会消耗ls的输出，不会给grep进程留下任何东西；ps会逐行创建其输出。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|755372|解决这个问题的正确方法是为每个需要它的进程复制STDOUT，如下所述|755373|redirect+COPY+of+stdout+to+log+file+from+within+bash+script+itself|offset|length|755374|但是，在这里，只需逐行提供读取脚本就足够了，以避免任何缓冲问题：|755375|ls+-ltrh+/+%7C+while+{+read+a;+}+do+echo+$a;+done+%7C+{+head+-n+1;+grep+tmp;+}|code-block|syntax|javascript|755376|但是，这意味着grep看不到head消耗的行。|755377|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/3173131/redirect-copy-of-stdout-to-log-file-from-within-bash-script-itself^0|0|0|0|1U|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|10|8|@]|9|@$F|11|G|12|1|13]]|A|$]]|$1|H|3|I|5|6|7|14|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|15|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

The reason this does not work, is that the first of the two processes (head -n1) reads more than it outputs. It eats up the output of ls and leaves nothing for the grep process; ps creates its output linewise.

The correct way to solve this would be to duplicate STDOUT for every process that needs it, as described here

<a href="https://stackoverflow.com/questions/3173131/redirect-copy-of-stdout-to-log-file-from-within-bash-script-itself">redirect COPY of stdout to log file from within bash script itself</a>

However here it would suffice to simply feed the reading scripts line by line to avoid any buffering issues:

<pre><code>ls -ltrh / | while { read a; } do echo $a; done | { head -n 1; grep tmp; }
</code></pre>

However, this means that grep can not see the line(s), head has consumed.

blocks|key|3071327|text|好吧..。这工作to...but只是为了记录一个非常复杂的解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3071328|(ps+aux+%7C+tee+>(head+-n1+>&3+)+%7C+grep+bio+>&3+)+3>&1|offset|length|style|CODE|3071329|这并不像我希望的那样好，fd3的使用让它变得很奇怪|3071330|注意:理论上甚至有可能grep输出在头文件之前。|3071331|entityMap^0|0|0|1G|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@$D|P|E|Q|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|R|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|S|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|M|$]]

well.. this works to...but just for noting a very convoluted solution

<code>(ps aux | tee &gt;(head -n1 &gt;&amp;3 ) | grep bio &gt;&amp;3 ) 3&gt;&amp;1</code>

this is not as nice as i wanted it to be, fd3 usage makes it weird

note: it might even theorecally possible that grep output preceeds the header

blocks|key|755408|text|纯sed溶液;)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|755409|ps+aux%7Csed+'1p;/kwork/p;d'|offset|length|style|CODE|755410|entityMap^0|0|0|Q|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

pure sed solution ;)

<code>ps aux|sed '1p;/kwork/p;d'</code>

I am trying to understand this peculiar behavior. Basically, I'm trying to grep an output of a command while still keeping the first line/header. Thanks for the help in advance. 

<h2>Success Case</h2>

<pre><code>ps -ef | { head -1; grep bash; }
</code></pre>

Output: 

<pre><code>UID PID PPID C STIME TTY TIME CMD
username 1008 1 0 Jan21 tty1 00:00:00 -bash
username 1173 1008 0 Jan21 tty1 00:00:00 -bash
</code></pre>

<h2>Failed Case</h2>

<pre><code> ls -tlrh / | { head -1; grep tmp; }
</code></pre>

Output: 

<pre><code>total 100K
</code></pre>

(i.e.: it ignores the <code>/tmp</code> folder)

Grep-ing while keeping first line

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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我正在试着理解这种奇怪的行为。基本上，我试图grep一个命令的输出，同时仍然保留第一行/标题。提前感谢您的帮助。成功案例ps -ef | { head -1; grep bash; }输出：UID        PID  PPID  C STIME TTY          TIME CMDusername  1008...

问在保留第一行的同时执行Grep
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问在保留第一行的同时执行GrepEN