blocks|key|1189813|text|这是一个通用的答案！您必须将规则应用于特定的数据集|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1189814|ls+%7C+sort|code-block|syntax|javascript|1189815|示例：|1189816|ls+%7C+sort+-n+-t+_+-k+2|1189817|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

This is a generic answer! You have to apply rules to the specific set of data

<pre><code>ls | sort
</code></pre>

Example:

<pre><code>ls | sort -n -t _ -k 2
</code></pre>

blocks|key|1189876|text|ls+data_*+%7C+sort+-n+-t+_+-k+2|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1189877|-n:按数字排序|unstyled|1189878|-t:字段分隔符“_”|1189879|-k:对第二个字段进行排序，在本例中是第一个“_”之后的数字|1189880|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>ls data_* | sort -n -t _ -k 2
</code></pre>

-n: sorts numerically 
-t: field separator '_' 
-k: sort on second field, in your case the numbers after the first '_'

blocks|key|1189925|text|如果您的sort具有版本排序功能，请尝试：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1189926|ls+-1+%7C+sort+-V|code-block|syntax|javascript|1189927|(这是一个大写的V)。|1189928|entityMap^0|4|4|0|0|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]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

If your <code>sort</code> has version sort, try:

<pre><code>ls -1 | sort -V
</code></pre>

(that's a capital V).

blocks|key|1189969|text|对ls使用-v标志怎么样？该标志的目的是根据版本号对文件进行排序，但它在这里同样适用，并且不需要通过管道将结果发送到sort|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1189970|ls+-lv+data_*|code-block|syntax|javascript|1189971|entityMap^0|1|2|5|2|1M|4|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

How about using the <code>-v</code> flag to <code>ls</code>? The purpose of the flag is to sort files according to version number, but it works just as well here and eliminates the need to pipe the result to <code>sort</code>:

<pre><code>ls -lv data_*
</code></pre>

blocks|key|1190003|text|我能想到的一个建议是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1190004|for+i+in+`seq+1+5`
do++
+++cat+"data_${i}_box"
done|code-block|syntax|javascript|1190005|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|$]]

One suggestion I can think of is this :

<pre><code>for i in `seq 1 5`
do 
 cat "data_${i}_box"
done
</code></pre>

blocks|key|1190042|text|如果您的排序没有版本排序，下面是在bash中执行此操作的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1190043|cat+<your_former_ls_output_file>+%7C+awk+'+BEGIN+{+FS="_"+}+{+printf(+"%2503d\n",$2)+}'+%7C+sort+%7C+awk+'+{+printf(+"data_%25d_box\n",+$1)++}'|code-block|syntax|javascript|1190044|这一切都在一行之内。请记住，我还没有在您的特定数据上对此进行测试，因此它可能需要进行一些调整才能为您正确工作。然而，这勾勒出了一个很好的、健壮的和相对简单的解决方案。当然，您始终可以将开头的cat%2Bfilename与实际的ls交换，以动态创建文件数据。为了捕获实际的filename列，您可以选择正确的ls参数或通过cut或awk的管道。|1190045|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|$]]

Here's the way to do it in bash if your sort doesn't have version sort:

<pre><code>cat &lt;your_former_ls_output_file&gt; | awk ' BEGIN { FS="_" } { printf( "%03d\n",$2) }' | sort | awk ' { printf( "data_%d_box\n", $1) }'
</code></pre>

All in one line. Keep in mind, I haven't tested this on your specific data, so it might need a little tweaking to work correctly for you. This outlines a good, robust and relatively simple solution, though. Of course, you can always swap the cat+filename in the beginning with an the actual ls to create the file data on the fly. For capturing the actual filename column, you can choose between correct ls parameters or piping through either cut or awk.

blocks|key|1405359|text|也许你会喜欢SistemaNumeri.py+(“修复数字”)：它重命名你的|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1405360|data_1_box
data_2_box
...
data_10_box
...
data_99_box|code-block|syntax|javascript|1405361|在……里面|1405362|data_01_box
data_02_box
...
data_10_box
...
data_99_box|1405363|entityMap|0|LINK|mutability|MUTABLE|url|http://code.google.com/p/unusefulscripts/source/browse/trunk/SistemaNumeri.py^0|6|G|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

maybe you'll like <a href="http://code.google.com/p/unusefulscripts/source/browse/trunk/SistemaNumeri.py" rel="nofollow noreferrer">SistemaNumeri.py</a> ("fix numbers"): it renames your

<pre><code>data_1_box
data_2_box
...
data_10_box
...
data_99_box
</code></pre>

in

<pre><code>data_01_box
data_02_box
...
data_10_box
...
data_99_box
</code></pre>

blocks|key|1190127|text|我在文件夹中有文件，需要根据编号对它们进行排序。例如-|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1190128|abc_dr-1.txt
hg_io-5.txt
kls_er_we-3.txt
sd-4.txt
sl_rt_we_yh-2.txt|code-block|syntax|javascript|1190129|我需要根据数字对它们进行排序。所以我用这个来排序。|1190130|ls+-1+%7C+sort+-t+'-'+-nk2|1190131|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I have files in a folder and need to sort them based on the number. E.g. -
<pre><code>abc_dr-1.txt
hg_io-5.txt
kls_er_we-3.txt
sd-4.txt
sl_rt_we_yh-2.txt
</code></pre>
I need to sort them based on number.
So I used this to sort.
<pre><code>ls -1 | sort -t '-' -nk2
</code></pre>

with a previous bash script I created a list of files:

<pre><code>data_1_box
data_2_box
...
data_10_box
...
data_99_box
</code></pre>

the thing is that now I need to concatenate them, so I tried

ls -l data_*

but I get

<pre><code>.....
data_89_box
data_8_box
data_90_box
...
data_99_box
data_9_box
</code></pre>

but I need to get in the sucession 1, 2, 3, 4, .. 9, ..., 89, 90, 91, ..., 99

Can it be done in bash?

Bash and sort files in order

使用前面的bash脚本，我创建了一个文件列表：data_1_boxdata_2_box...data_10_box...data_99_box问题是，现在我需要将它们连接起来，所以我尝试ls -l data_*但我得到了.....data_89_boxdata_8_boxdata_90_box...data_99_bo...

问按顺序筛选和排序文件
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