blocks|key|3200926|text|awk+'{+s+=+"";+for+(i+=+9;+i+<=+NF;+i%2B%2B)+s+=+s+$i+"+";+print+s+}'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3200927|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 '{ s = ""; for (i = 9; i &lt;= NF; i++) s = s $i " "; print s }'
</code></pre>

blocks|key|276938|text|通常，perl会替换awk/sed/grep+et。而且更便携(同时也是一把更好的小刀)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|276939|perl+-lane+'print+"@F[8..$#F]"'|code-block|syntax|javascript|276940|当然，Timtowtdi也适用。|276941|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|$]]

Generally perl replaces awk/sed/grep et. al., and is much more portable (as well as just being a better penknife).

<pre><code>perl -lane 'print "@F[8..$#F]"'
</code></pre>

Timtowtdi applies of course.

blocks|key|278769|text|当您想要处理一系列字段时，awk实际上没有直接的方法来完成此任务。我会推荐cut：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|278770|cut+-d'+'+-f+9-+./infile|code-block|syntax|javascript|278771|编辑|278772|由于默认为制表符，添加了空格字段分隔符。感谢Glenn指出这一点|278773|entityMap^0|D|3|11|3|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|X|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

When you want to do a range of fields, <code>awk</code> doesn't really have a straight forward way to do this. I would recommend <code>cut</code> instead:

<pre><code>cut -d' ' -f 9- ./infile
</code></pre>

<h3>Edit</h3>

Added space field delimiter due to default being a tab. Thanks to Glenn for pointing this out

blocks|key|278775|text|ruby+-lane+'print+$F[3..-1].join("+")'+file|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|278776|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>ruby -lane 'print $F[3..-1].join(" ")' file
</code></pre>

blocks|key|278778|text|awk+'{print+substr($0,+index($0,$9))}'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|278779|编辑：请注意，如果第九个字段之前的任何字段包含与第九个字段相同的值，则此操作不起作用。|unstyled|offset|length|style|BOLD|278780|entityMap^0|0|0|2|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>awk '{print substr($0, index($0,$9))}'
</code></pre>

Edit:
Note, this doesn't work if any field before the ninth contains the same value as the ninth.

blocks|key|278785|text|awk+-v+m="\x01"+-v+N="3"+'{$N=m$N+;print+substr($0,+index($0,m)%2B1)}'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|278786|这将截断给定域nr.，N之前的内容，并打印该行的所有其余部分，包括域nr.N并保持原始间距(不会重新格式化)。字段的字符串是否也出现在行中的其他位置并不重要，这就是Ascherer的答案的问题。|unstyled|278787|定义函数：|278788|fromField+()+{+
awk+-v+m="\x01"+-v+N="$1"+'{$N=m$N;+print+substr($0,index($0,m)%2B1)}'
}|278789|并像这样使用它：|278790|$+echo+"++bat+++bi+++++++iru+++lau+bost+++"+%7C+fromField+3
iru+++lau+bost+++
$+echo+"++bat+++bi+++++++iru+++lau+bost+++"+%7C+fromField+2
bi+++++++iru+++lau+bost+++|278791|Output维护所有内容，包括N=0的尾随空格，它按原样返回整行，而n>NF则返回空字符串|278792|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|T|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$B|C]]|$1|K|3|L|5|F|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$B|C]]|$1|O|3|P|5|F|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|F|7|Z|8|@]|9|@]|A|$]]]|R|$]]

<pre><code>awk -v m="\x01" -v N="3" '{$N=m$N ;print substr($0, index($0,m)+1)}'
</code></pre>

This chops what is before the given field nr., N, and prints all the rest of the line, including field nr.N and maintaining the original spacing (it does not reformat). It doesn't mater if the string of the field appears also somewhere else in the line, which is the problem with Ascherer's answer.

Define a function:

<pre><code>fromField () { 
awk -v m="\x01" -v N="$1" '{$N=m$N; print substr($0,index($0,m)+1)}'
}
</code></pre>

And use it like this:

<pre><code>$ echo " bat bi iru lau bost " | fromField 3
iru lau bost 
$ echo " bat bi iru lau bost " | fromField 2
bi iru lau bost 
</code></pre>

Output maintains everything, including trailing spaces
For N=0 it returns the whole line, as is, and for n>NF the empty string

blocks|key|280604|text|要显示前3个字段并打印其余字段，您可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|280605|awk+'{s+=+"";+for+(i=4;+i<=+NF;+i%2B%2B)+s=+s+$i+:+";+print+$1+$2+$3+s}'+filename|code-block|syntax|javascript|280606|其中$1+$2+$3是前3个字段。|280607|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|$]]

To display the first 3 fields and print the remaining fields you can use:

<pre><code>awk '{s = ""; for (i=4; i&lt;= NF; i++) s= s $i : "; print $1 $2 $3 s}' filename
</code></pre>

where $1 $2 $3 are the first 3 fields.

blocks|key|3201199|text|sed+-re+'s,\s%2B,+,g'+%7C+cut+-d+'+'+-f+9-|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3201200|将所有空格替换为单个空格，而不是处理可变宽度的空格。然后对感兴趣的字段使用简单的cut。|unstyled|offset|length|style|CODE|3201201|它没有使用awk，所以没有关系，但考虑到其他答案/评论，它看起来很合适。|3201202|entityMap^0|0|14|3|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@$G|Q|H|R|I|J]]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|F|7|T|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>sed -re 's,\s+, ,g' | cut -d ' ' -f 9-
</code></pre>

Instead of dealing with variable width whitespace, replace all whitespace as single space. Then use simple <code>cut</code> with the fields of interest.

It doesn't use awk so isn't germane but seemed appropriate given the other answers/comments.

blocks|key|2987677|text|以下是ls+-l输出的示例：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2987678|-rwxr-----@+1+ricky.john++1493847943+++5610048+Apr+16+14:09+00-Welcome.mp4
-rwxr-----@+1+ricky.john++1493847943++27862521+Apr+16+14:09+01-Hello+World.mp4
-rwxr-----@+1+ricky.john++1493847943++21262056+Apr+16+14:09+02-Typical+Go+Directory+Structure.mp4
-rwxr-----@+1+ricky.john++1493847943++10627144+Apr+16+14:09+03-Where+to+Get+Help.mp4|code-block|syntax|javascript|2987679|我的解决方案是awk+'{print+substr($0,+61,+50)}'，用于打印$9后的任何内容|2987680|entityMap^0|3|5|0|0|7|W|18|2|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

Here is an example of <code>ls -l</code> output:

<pre><code>-rwxr-----@ 1 ricky.john 1493847943 5610048 Apr 16 14:09 00-Welcome.mp4
-rwxr-----@ 1 ricky.john 1493847943 27862521 Apr 16 14:09 01-Hello World.mp4
-rwxr-----@ 1 ricky.john 1493847943 21262056 Apr 16 14:09 02-Typical Go Directory Structure.mp4
-rwxr-----@ 1 ricky.john 1493847943 10627144 Apr 16 14:09 03-Where to Get Help.mp4
</code></pre>

My solution to print anything post <code>$9</code> is <code>awk '{print substr($0, 61, 50)}'</code>

blocks|key|278812|text|function+print_fields(field_num1,+field_num2){
++++input_line+=+$0

++++j+=+1;
++++for+(i=field_num1;+i+<=+field_num2;+i%2B%2B){
++++++++$(j%2B%2B)+=+$(i);

++++}
++++NF+=+field_num2+-+field_num1+%2B+1;
++++print+$0

++++$0+=+input_line
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|278813|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>function print_fields(field_num1, field_num2){
 input_line = $0

 j = 1;
 for (i=field_num1; i &lt;= field_num2; i++){
 $(j++) = $(i);

 }
 NF = field_num2 - field_num1 + 1;
 print $0

 $0 = input_line
}
</code></pre>

blocks|key|277109|text|使用cut而不是awk，并通过使用-c字符剪切命令克服了确定从哪一列开始的问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|277110|在这里，我要说的是，除了输出的前49个字符之外，给我所有的字符。|277111|+ls+-l+/some/path/*/*+%7C+cut+-c+50-|code-block|syntax|javascript|277112|ls命令末尾的/*/*/表示也显示子目录中的内容。|offset|length|style|CODE|277113|您还可以提取特定范围的字符ala+(从cut手册页)。例如，显示当前登录用户的姓名和登录时间：|277114|+++++++who+%7C+cut+-c+1-16,26-38|277115|entityMap^0|0|0|0|7|5|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|X|8|@$K|Y|L|Z|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|F|7|11|8|@]|9|@]|A|$G|H]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

Using cut instead of awk and overcoming issues with figuring out which column to start at by using the -c character cut command.

Here I am saying, give me all but the first 49 characters of the output. 

<pre><code> ls -l /some/path/*/* | cut -c 50-
</code></pre>

The <code>/*/*/</code> at the end of the ls command is saying show me what is in subdirectories too.

You can also pull out certain ranges of characters ala (from the cut man page). E.g., show the names and login times of the currently logged in users:

<pre><code> who | cut -c 1-16,26-38
</code></pre>

blocks|key|277125|text|通常需要原封不动地传递剩余的列。也就是说，不会折叠连续的空白区域。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|277126|想象一下处理ls+-l或ps+faux输出的情况(不推荐，仅给出最后一列可能包含空格序列的示例)。我们希望保留剩余列中任何连续的空格，这样名为my+++file.txt的文件就不会变成my+file.txt。|offset|length|style|CODE|277127|使用awk为行的其余部分保留空格是非常困难的。公认的基于awk的答案并非如此，即使经过建议的改进也是如此。|277128|sed或perl更适合这项任务。|277129|sed|BOLD|277130|echo+'1+2+3+4+5+6+7+8+9+++10'+%7C+sed+-E+'s/%5E([%5E+\t]*[+\t]*){8}//'|code-block|syntax|javascript|277131|结果：|277132|9+++10|277133|-E选项启用现代ERE正则表达式语法。这就省去了用反斜杠转义圆括号和大括号的麻烦。|277134|{8}是一个量词，表示恰好匹配前一项8次。|277135|sed+s命令将8次出现的空格分隔的单词替换为空字符串。这条线的其余部分保持不变。|277136|perl|277137|Perl+regex支持水平空格的\h转义。|277138|echo+'1+2+3+4+5+6+7+8+9+++10'+%7C+perl+-pe+'s/%5E(\H*\h*){8}//'|277139|277140|9+++10|277141|entityMap|0|LINK|mutability|MUTABLE|url|https://perldoc.perl.org/perlre^0|0|6|5|C|7|1Z|D|2K|B|0|2|3|0|0|3|4|4|0|0|3|0|0|0|0|0|2|0|0|3|0|4|1|0|0|4|0|H|2|0|A|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1L|8|@$D|1M|E|1N|F|G]|$D|1O|E|1P|F|G]|$D|1Q|E|1R|F|G]|$D|1S|E|1T|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1U|8|@$D|1V|E|1W|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|1X|8|@$D|1Y|E|1Z|F|G]|$D|20|E|21|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|22|8|@$D|23|E|24|F|N]]|9|@]|A|$]]|$1|O|3|P|5|Q|7|25|8|@]|9|@]|A|$R|S]]|$1|T|3|U|5|6|7|26|8|@]|9|@]|A|$]]|$1|V|3|W|5|Q|7|27|8|@]|9|@]|A|$R|S]]|$1|X|3|Y|5|6|7|28|8|@$D|29|E|2A|F|G]]|9|@]|A|$]]|$1|Z|3|10|5|6|7|2B|8|@$D|2C|E|2D|F|G]]|9|@]|A|$]]|$1|11|3|12|5|6|7|2E|8|@$D|2F|E|2G|F|G]]|9|@]|A|$]]|$1|13|3|14|5|6|7|2H|8|@$D|2I|E|2J|F|N]]|9|@]|A|$]]|$1|15|3|16|5|6|7|2K|8|@$D|2L|E|2M|F|G]]|9|@$D|2N|E|2O|1|2P]]|A|$]]|$1|17|3|18|5|Q|7|2Q|8|@]|9|@]|A|$R|S]]|$1|19|3|U|5|6|7|2R|8|@]|9|@]|A|$]]|$1|1A|3|1B|5|Q|7|2S|8|@]|9|@]|A|$R|S]]|$1|1C|3|-4|5|6|7|2T|8|@]|9|@]|A|$]]]|1D|$1E|$5|1F|1G|1H|A|$1I|1J]]]]

Usually it is desired to pass the remaining columns unmodified. That is, without collapsing contiguous white space.
Imagine the case of processing the output of <code>ls -l</code> or <code>ps faux</code> (not recommended, just giving examples where the last column may contain sequences of whitespace)). We'd want any contiguous white space in the remaning columns preserved so that a file named <code>my file.txt</code> doesn't become <code>my file.txt</code>.
Preserving white space for the remainder of the line is surprisingly difficult using <code>awk</code>. The accepted awk-based answer does not, even with the suggested refinements.
<code>sed</code> or <code>perl</code> are better suited to this task.
sed
<pre class="lang-sh prettyprint-override"><code>echo '1 2 3 4 5 6 7 8 9 10' | sed -E 's/^([^ \t]*[ \t]*){8}//'
</code></pre>
Result:
<pre><code>9 10
</code></pre>
The <code>-E</code> option enables modern ERE regex syntax. This saves me the trouble of backslash escaping the parentheses and braces.
The <code>{8}</code> is a quantifier indicating to match the previous item exactly 8 times.
The sed <code>s</code> command replaces 8 occurrences of white space delimited words by an empty string. The remainder of the line is left intact.
perl
<a href="https://perldoc.perl.org/perlre" rel="nofollow noreferrer">Perl regex</a> supports the <code>\h</code> escape for horizontal whitespace.
<pre><code>echo '1 2 3 4 5 6 7 8 9 10' | perl -pe 's/^(\H*\h*){8}//'
</code></pre>
Result:
<pre><code>9 10
</code></pre>

On shell, I pipe to awk when I need a particular column.

This prints column 9, for example:

<pre><code>... | awk '{print $9}'
</code></pre>

How can I tell awk to print all the columns including and after column 9, not just column 9?

How to print all the columns after a particular number using awk?

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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在shell上，当我需要一个特定的列时，我通过管道连接到awk。这将打印第9列，例如：... | awk '{print $9}'我怎样才能告诉awk打印所有的列，包括 9及之后的列，而不仅仅是第9列？

问如何使用awk打印特定数字后的所有列？
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问如何使用awk打印特定数字后的所有列？
EN