blocks|key|1829985|text|如果你真的想使用ls，那么使用awk格式化它的输出：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1829986|ls+-R+/path+%7C+awk+'
/:$/&&f{s=$0;f=0}
/:$/&&!f{sub(/:$/,"");s=$0;f=1;next}
NF&&f{+print+s"/"$0+}'|code-block|syntax|javascript|1829987|entityMap^0|8|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|$]]

If you really want to use <code>ls</code>, then format its output using awk:

<pre><code>ls -R /path | awk '
/:$/&amp;&amp;f{s=$0;f=0}
/:$/&amp;&amp;!f{sub(/:$/,"");s=$0;f=1;next}
NF&amp;&amp;f{ print s"/"$0 }'
</code></pre>

blocks|key|1930709|text|使用find：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1930710|find+.
find+/home/dreftymac|code-block|syntax|javascript|1930711|如果您只想要文件(省略目录、设备等)：|1930712|find+.+-type+f
find+/home/dreftymac+-type+f|1930713|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|$]]

Use find:

<pre><code>find .
find /home/dreftymac
</code></pre>

If you want files only (omit directories, devices, etc):

<pre><code>find . -type f
find /home/dreftymac -type f
</code></pre>

blocks|key|1828005|text|ls+-ld+$(find+.)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1828006|如果要按修改时间对输出进行排序：|1828007|ls+-ltd+$(find+.)|1828008|entityMap^0|0|G|0|0|0|H|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|K|$]]

<code>ls -ld $(find .)</code>

if you want to sort your output by modification time:

<code>ls -ltd $(find .)</code>

blocks|key|1930848|text|尝试以下更简单的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1930849|find+"$PWD"|code-block|syntax|javascript|1930850|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|$]]

Try the following simpler way:

<pre><code>find "$PWD"
</code></pre>

blocks|key|528682|text|哦，答案真的很长。它帮助很大，最后，我创建了我自己的，这是我正在寻找的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|528683|要列出某个目录及其子目录中的所有文件：|528684|find+"$PWD"+-type+f|code-block|syntax|javascript|528685|要列出目录及其子目录中的所有目录，请执行以下操作：|528686|find+"$PWD"+-type+d|528687|要列出目录及其子目录中的所有目录和文件：|528688|find+"$PWD"|528689|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Oh, really a long list of answers. 
It helped a lot and finally, I created my own which I was looking for :

To List All the Files in a directory and its sub-directories:

<pre><code>find "$PWD" -type f
</code></pre>

To List All the Directories in a directory and its sub-directories:

<pre><code>find "$PWD" -type d
</code></pre>

To List All the Directories and Files in a directory and its sub-directories:

<pre><code>find "$PWD"
</code></pre>

blocks|key|523030|text|最佳命令是：tree+-fi|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|523031|-f打印每个文件的完整路径前缀|523032|-i不打印缩进|523033|例如：|523034|$+tree+-fi
.
./README.md
./node_modules
./package.json
./src
./src/datasources
./src/datasources/bookmarks.js
./src/example.json
./src/index.js
./src/resolvers.js
./src/schema.js|code-block|syntax|javascript|523035|为了使用文件而不是链接，您必须从输出中删除>：|523036|tree+-fi+%7Cgrep+-v+\>|523037|如果您想知道每个文件的性质(例如，只读ASCII文件)，请尝试while循环：|523038|tree+-fi+%7C
grep+-v+\>+%7C
while+read+-r+first+;+do+
++++file+"${first}"
done+%7C
grep+ASCII|523039|entityMap^0|6|8|0|0|0|0|0|L|1|0|0|V|5|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|13|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|14|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|15|8|@]|D|@]|E|$]]|$1|L|3|M|5|N|7|16|8|@]|D|@]|E|$O|P]]|$1|Q|3|R|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|S|3|T|5|N|7|1A|8|@]|D|@]|E|$O|P]]|$1|U|3|V|5|6|7|1B|8|@$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|W|3|X|5|N|7|1E|8|@]|D|@]|E|$O|P]]|$1|Y|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|Z|$]]

Best command is: <code>tree -fi</code>
-f print the full path prefix for each file 
-i don't print indentations
e.g.
<pre><code>$ tree -fi
.
./README.md
./node_modules
./package.json
./src
./src/datasources
./src/datasources/bookmarks.js
./src/example.json
./src/index.js
./src/resolvers.js
./src/schema.js
</code></pre>
In order to use the files but not the links, you have to remove <code>&gt;</code> from your output:
<pre><code>tree -fi |grep -v \&gt;
</code></pre>
If you want to know the nature of each file, (to read only ASCII files for example) try a <code>while</code> loop:
<pre><code>tree -fi |
grep -v \&gt; |
while read -r first ; do 
 file &quot;${first}&quot;
done |
grep ASCII
</code></pre>

blocks|key|1930645|text|我不知道完整路径，但您可以使用-R进行递归。或者，如果你对ls不感兴趣，你可以直接使用find+*。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1930646|entityMap^0|F|2|T|2|17|6|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

I don't know about the full path, but you can use <code>-R</code> for recursion. Alternatively, if you're not bent on <code>ls</code>, you can just do <code>find *</code>.

blocks|key|523248|text|du+-a|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|523249|对于一些有限的设备shell来说很方便，因为其中的find/locate不可用。|unstyled|523250|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>du -a
</code></pre>

Handy for some limited appliance shells where find/locate aren't available.

blocks|key|1930670|text|找到/将会做到这一点|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1930671|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

blocks|key|1828038|text|对于你们所有未来的人来说，最简单的方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1828039|du|code-block|syntax|javascript|1828040|但是，这也显示了每个文件夹中包含的内容的大小，您可以使用awk仅输出文件夹名：|1828041|du+%7C+awk+'{print+$2}'|1828042|编辑-对不起，是我的错。我以为它只是需要的文件夹。我把这个留在这里，以防将来有人需要它。|1828043|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|$]]

The easiest way for all you future people is simply:

<pre><code>du
</code></pre>

This however, also shows the size of whats contained in each folder
You can use awk to output only the folder name:

<pre><code>du | awk '{print $2}'
</code></pre>

Edit- Sorry sorry, my bad. I thought it was only folders that were needed.
Ill leave this here in case anyone in the future needs it anyways...

blocks|key|1828181|text|使用以下格式运行bash命令：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1828182|find+/path+-type+f+-exec+ls+-l+\{\}+\;|code-block|syntax|javascript|1828183|同样，要删除-l详细信息并仅返回绝对路径：|offset|length|style|CODE|1828184|find+/path+-type+f+-exec+ls+\{\}+\;|1828185|entityMap^0|0|0|6|2|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Run a bash command with the following format:
<pre><code>find /path -type f -exec ls -l \{\} \;
</code></pre>
Likewise, to trim away <code>-l</code> details and return only the absolute paths:
<pre><code>find /path -type f -exec ls \{\} \;
</code></pre>

blocks|key|1931315|text|不要让它变得复杂。我刚刚使用了这个，并得到了一个漂亮的输出：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1931316|ls+-lR+/path/I/need|code-block|syntax|javascript|1931317|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|$]]

Don't make it complicated. I just used this and got a beautiful output:

<pre><code>ls -lR /path/I/need
</code></pre>

blocks|key|1830041|text|可以自由使用所有可能的ls选项：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1830042|find+-type+f+%7C+xargs+ls+-1|offset|length|style|CODE|1830043|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|$]]

With having the freedom of using all possible ls options:

<code>find -type f | xargs ls -1</code>

blocks|key|523228|text|我认为对于扁平列表，最好的方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|523229|find+-D+tree+/fullpath/to-dir/+|code-block|syntax|javascript|523230|(或为了将其保存在txt文件中)|523231|find+-D+tree+/fullpath/to-dir/+>+file.txt|523232|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 think for a flat list the best way is:

<pre><code>find -D tree /fullpath/to-dir/ 
</code></pre>

(or in order to save it in a txt file)

<pre><code>find -D tree /fullpath/to-dir/ &gt; file.txt
</code></pre>

blocks|key|1930991|text|以下是显示目录名称的部分答案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1930992|ls+-mR+*+%7C+sed+-n+'s/://p'|code-block|syntax|javascript|1930993|解释：|1930994|ls+-mR+*列出以':‘结尾的完整目录名，然后单独列出该目录中的文件|offset|length|style|CODE|1930995|sed+-n+'s/://p'查找以冒号结尾的行，去掉冒号并打印该行|1930996|通过遍历目录列表，我们也应该能够找到目录。还在努力呢。通过xargs获取通配符是一个挑战。|1930997|entityMap^0|0|0|0|0|8|0|0|F|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]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@$K|Y|L|Z|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|10|8|@$K|11|L|12|M|N]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|T|$]]

Here is a partial answer that shows the directory names. 

<pre><code>ls -mR * | sed -n 's/://p'
</code></pre>

Explanation:

<code>ls -mR *</code> lists the full directory names ending in a ':', then lists the files in that directory separately

<code>sed -n 's/://p'</code> finds lines that end in a colon, strip off the colon and print the line

By iterating over the list of directories, we should be able to find the directories as well. Still workin on it. It is a challenge to get the wildcards through xargs.

blocks|key|528473|text|将通配符添加到ls目录的末尾将强制使用完整路径。现在你可以这样做了：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|528474|$+ls+/home/dreftymac/
foo.txt
bar.txt
stackoverflow
stackoverflow/alpha.txt
stackoverflow/bravo.txt
stackoverflow/charlie.txt|code-block|syntax|javascript|528475|你可以这样做：|528476|$+ls+/home/dreftymac/*
/home/dreftymac/.
/home/dreftymac/foo.txt
/home/dreftymac/bar.txt
/home/dreftymac/stackoverflow:
alpha.txt
bravo.txt
charlie.txt|528477|不幸的是，这并没有打印出递归到的目录的完整路径，因此它可能不是您正在寻找的完整解决方案。|528478|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|$]]

Adding a wildcard to the end of an ls directory forces full paths. Right now you have this:

<pre><code>$ ls /home/dreftymac/
foo.txt
bar.txt
stackoverflow
stackoverflow/alpha.txt
stackoverflow/bravo.txt
stackoverflow/charlie.txt
</code></pre>

You could do this instead:

<pre><code>$ ls /home/dreftymac/*
/home/dreftymac/.
/home/dreftymac/foo.txt
/home/dreftymac/bar.txt
/home/dreftymac/stackoverflow:
alpha.txt
bravo.txt
charlie.txt
</code></pre>

Unfortunately this does not print the full path for directories recursed into, so it may not be the full solution you're looking for.

blocks|key|523180|text|我看到了很多答案。这是我的，如果你在Mac上工作，我认为非常有用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|523181|我相信您知道有一些“包”文件(.app、.rtfd、.workflow等)。在Finder的窗口中，它们看起来像是单个文件。但事实并非如此。而$+ls或$+find将它们视为目录...因此，除非您还需要列出它们的内容，否则这对我来说是可行的：|offset|length|style|CODE|523182|find+.+-not+-name+".*"+-not+-name+"."+%7C+egrep+-v+"\.rtfd/%7C\.app/%7C\.lpdf/%7C\.workflow/"|code-block|syntax|javascript|523183|当然，这是针对工作目录的，您可以添加其他包的扩展名(但后面总是有一个/+)。或任何其他扩展，如果不是捆绑的没有/的。|523184|相当有趣的".lpdf/“(多语言pdf)。它有正常的".pdf“扩展名(!!)或者在Finder中没有。这样你就得到了(或者说只有1个文件)这个pdf，而不是一堆…|523185|entityMap^0|0|1Z|4|24|6|0|0|Y|1|1J|1|0|6|6|S|4|21|3|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]|$D|W|E|X|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|Z|8|@$D|10|E|11|F|G]|$D|12|E|13|F|G]]|9|@]|A|$]]|$1|O|3|P|5|6|7|14|8|@$D|15|E|16|F|G]|$D|17|E|18|F|G]|$D|19|E|1A|F|G]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|R|$]]

A lot of answers I see. This is mine, and I think quite useful if you are working on Mac.

I'm sure you know there are some "bundle" files (.app, .rtfd, .workflow, and so on). And looking at Finder's window they seem single files. But they are not. And <code>$ ls</code> or <code>$ find</code> see them as directories... So, unless you need list their contents as well, this works for me:

<pre><code>find . -not -name ".*" -not -name "." | egrep -v "\.rtfd/|\.app/|\.lpdf/|\.workflow/"
</code></pre>

Of course this is for the working dir, and you could add other bundles' extensions (but always with a <code>/</code> after them). Or any other extensions if not bundle's without the <code>/</code>.

Rather interesting the "<code>.lpdf/</code>" (multilingual pdf). It has normal "<code>.pdf</code>" extension (!!) or none in Finder. This way you get (or it just counts 1 file) for this <code>pdf</code> and not a bunch of stuff…

blocks|key|1830230|text|ls+-lR就是你要找的，至少我是。干杯|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1830231|entityMap^0|0|6|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|$]]

<code>ls -lR</code> is what you were looking for, or atleast I was. cheers

blocks|key|1829963|text|如果目录是作为相对路径传递的，则需要在调用find之前将其转换为绝对路径。在下面的示例中，目录作为第一个参数传递给脚本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1829964|#!/bin/bash

#+get+absolute+path
directory=`cd+$1;+pwd`
#+print+out+list+of+files+and+directories
find+"$directory"|code-block|syntax|javascript|1829965|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|$]]

If the directory is passed as a relative path and you will need to convert it to an absolute path before calling find. In the following example, the directory is passed as the first parameter to the script:
<pre><code>#!/bin/bash

# get absolute path
directory=`cd $1; pwd`
# print out list of files and directories
find &quot;$directory&quot;
</code></pre>

blocks|key|523098|text|tar+cf+-+$PWD%7Ctar+tvf+-+++++++++++++|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|523099|这是缓慢的，但递归地工作，并打印目录和文件。如果您只想要文件名而不需要所有其他信息/目录，则可以使用awk/grep作为管道：|unstyled|523100|tar+cf+-+$PWD%7Ctar+tvf+-%7Cawk+'{print+$6}'%7Cgrep+-v+"/$"++++++++++|523101|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>tar cf - $PWD|tar tvf - 
</code></pre>

This is slow but works recursively and prints both directories and files. You can pipe it with awk/grep if you just want the file names without all the other info/directories:

<pre><code>tar cf - $PWD|tar tvf -|awk '{print $6}'|grep -v "/$" 
</code></pre>

blocks|key|523515|text|当前位置的所有文件的递归列表：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|523516|ls+-l+$(find+.+-type+f)|offset|length|style|CODE|523517|entityMap^0|0|0|N|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|$]]

Recursive list of all files from current location:
<code>ls -l $(find . -type f)</code>

blocks|key|1931251|text|@ghostdog74+74:稍微调整一下你的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1931252|下面的代码可以用来搜索文件的完整绝对路径。|1931253|sudo+ls+-R+/+%7C+awk+'
/:$/&&f{s=$0;f=0}
/:$/&&!f{sub(/:$/,"");s=$0;f=1;next}
NF&&f{+print+s"/"$0+}'+%7C+grep+[file_to_search]|code-block|syntax|javascript|1931254|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

@ghostdog74:
Little tweak with your solution. 
Following code can be used to search file with its full absolute path.
<pre class="lang-bash prettyprint-override"><code>sudo ls -R / | awk '
/:$/&amp;&amp;f{s=$0;f=0}
/:$/&amp;&amp;!f{sub(/:$/,&quot;&quot;);s=$0;f=1;next}
NF&amp;&amp;f{ print s&quot;/&quot;$0 }' | grep [file_to_search]
</code></pre>

blocks|key|1931283|text|我知道文件名，但也想要目录。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1931284|find+$PWD+%7C+fgrep+filename|code-block|syntax|javascript|1931285|在Mac+OS+10.12.1中完美工作|1931286|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|$]]

I knew the file name but wanted the directory as well.
<pre><code>find $PWD | fgrep filename
</code></pre>
worked perfectly in Mac OS 10.12.1

blocks|key|528578|text|realpath命令打印解析后的路径：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|528579|realpath+*|code-block|syntax|javascript|528580|要包含点文件，请将ls+-a的输出通过管道传输到realpath：|528581|ls+-a+%7C+xargs+realpath|528582|要递归列出子目录，请执行以下操作：|528583|ls+-aR+%7C+xargs+realpath|528584|如果文件名中有空格，man+xargs建议使用-o选项来防止文件名被错误处理，这对find+-print0的输出效果最好，而且它看起来比其他答案复杂得多：|528585|find+-print0+%7Cxargs+-0+realpath|528586|另请参阅Unix和Linux+stackexchange+question+on+how+to+list+all+files+in+a+directory+with+absolute+path。|528587|entityMap|0|LINK|mutability|MUTABLE|url|https://unix.stackexchange.com/questions/268474/how-to-list-all-files-in-a-directory-with-absolute-paths^0|0|8|0|0|9|5|0|0|0|0|A|9|N|2|15|C|0|0|T|1V|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@$9|17|A|18|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1A|8|@$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1E|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]]|D|@]|E|$]]|$1|U|3|V|5|H|7|1N|8|@]|D|@]|E|$I|J]]|$1|W|3|X|5|6|7|1O|8|@]|D|@$9|1P|A|1Q|1|1R]]|E|$]]|$1|Y|3|-4|5|6|7|1S|8|@]|D|@]|E|$]]]|Z|$10|$5|11|12|13|E|$14|15]]]]

The <code>realpath</code> command prints the resolved path:
<pre><code>realpath *
</code></pre>
To include dot files, pipe the output of <code>ls -a</code> to realpath:
<pre><code>ls -a | xargs realpath
</code></pre>
To list subdirectories recursively:
<pre><code>ls -aR | xargs realpath
</code></pre>
In case you have spaces in file names, <code>man xargs</code> recommends using the <code>-o</code> option to prevent file names from being processed incorrectly, this works best with the output of <code>find -print0</code> and it starts to look a lot more complex than other answers:
<pre><code>find -print0 |xargs -0 realpath
</code></pre>
See also Unix and Linux stackexchange <a href="https://unix.stackexchange.com/questions/268474/how-to-list-all-files-in-a-directory-with-absolute-paths">question on how to list all files in a directory with absolute path</a>.

How can I get ls to spit out a flat list of recursive one-per-line paths?

For example, I just want a flat listing of files with their full paths:

<pre><code>/home/dreftymac/.
/home/dreftymac/foo.txt
/home/dreftymac/bar.txt
/home/dreftymac/stackoverflow
/home/dreftymac/stackoverflow/alpha.txt
/home/dreftymac/stackoverflow/bravo.txt
/home/dreftymac/stackoverflow/charlie.txt
</code></pre>

<code>ls -a1</code> almost does what I need, but I do not want path fragments, I want full paths.

ls command: how can I get a recursive full-path listing, one line per file?

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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如何让ls输出一个递归的单行路径的平面列表？例如，我只想要一个包含完整路径的文件的平面列表：/home/dreftymac/./home/dreftymac/foo.txt/home/dreftymac/bar.txt/home/dreftymac/stackoverflow/home/dreftymac/stackoverflow/alpha.txt/home/dreftymac/stackov

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