blocks|key|540125|text|去掉引号。shell将它们视为一个文件，因此它查找"ls+-l“。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|540126|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|$]]

Get rid of the quotes. The shell is treating them like one file, so it's looking for "ls -l".

blocks|key|323232|text|REmove这些qoutes，你就会好起来的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|323233|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|$]]

REmove the qoutes and you will be fine

blocks|key|323292|text|引号导致了错误消息。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|323293|要获取目录中的文件数，请执行以下操作：|323294|shopt+-s+nullglob
numfiles=(*)
numfiles=${#numfiles[@]}|code-block|syntax|javascript|323295|它创建一个数组，然后将其替换为元素的计数。这将包括文件和目录，但不包括点文件、.、..或其他点分目录。|offset|length|style|CODE|323296|使用nullglob，这样空目录的计数就是0而不是1。|323297|您可以改用find+-type+f，也可以计算目录数并减去：|323298|#+continuing+from+above
numdirs=(*/)
numdirs=${#numdirs[@]}
((+numfiles+-=+numdirs+))|323299|另请参阅"How+can+I+find+the+latest+(newest,+earliest,+oldest)+file+in+a+directory?“|323300|你可以在一个执行块中有任意多的空格。它们通常有助于提高可读性。唯一的缺点是它们会使文件变得更大，并且可能会稍微减慢初始解析速度。有一些地方必须有空格(例如在比较中的[，[[，]，]]和=周围)，还有一些地方不能有空格(例如在赋值中的=周围)。|323301|entityMap|0|LINK|mutability|MUTABLE|url|http://mywiki.wooledge.org/BashFAQ/003^0|0|0|0|13|1|15|2|0|2|8|0|5|C|0|0|5|21|0|0|2A|1|2C|2|2F|1|2H|2|2K|1|38|1|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|17|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|18|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|19|8|@$K|1A|L|1B|M|N]|$K|1C|L|1D|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1E|8|@$K|1F|L|1G|M|N]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1H|8|@$K|1I|L|1J|M|N]]|9|@]|A|$]]|$1|S|3|T|5|F|7|1K|8|@]|9|@]|A|$G|H]]|$1|U|3|V|5|6|7|1L|8|@]|9|@$K|1M|L|1N|1|1O]]|A|$]]|$1|W|3|X|5|6|7|1P|8|@$K|1Q|L|1R|M|N]|$K|1S|L|1T|M|N]|$K|1U|L|1V|M|N]|$K|1W|L|1X|M|N]|$K|1Y|L|1Z|M|N]|$K|20|L|21|M|N]]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|22|8|@]|9|@]|A|$]]]|Z|$10|$5|11|12|13|A|$14|15]]]]

The quotes are causing the error messages.

To get a count of files in the directory:

<pre><code>shopt -s nullglob
numfiles=(*)
numfiles=${#numfiles[@]}
</code></pre>

which creates an array and then replaces it with the count of its elements. This will include files and directories, but not dotfiles or <code>.</code> or <code>..</code> or other dotted directories.

Use <code>nullglob</code> so an empty directory gives a count of 0 instead of 1.

You can instead use <code>find -type f</code> or you can count the directories and subtract:

<pre><code># continuing from above
numdirs=(*/)
numdirs=${#numdirs[@]}
(( numfiles -= numdirs ))
</code></pre>

Also see "<a href="http://mywiki.wooledge.org/BashFAQ/003" rel="noreferrer">How can I find the latest (newest, earliest, oldest) file in a directory?</a>"

You can have as many spaces as you want inside an execution block. They often aid in readability. The only downside is that they make the file a little larger and may slow initial parsing (only) slightly. There are a few places that must have spaces (e.g. around <code>[</code>, <code>[[</code>, <code>]</code>, <code>]]</code> and <code>=</code> in comparisons) and a few that must not (e.g. around <code>=</code> in an assignment.

blocks|key|540241|text|这样如何：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|540242|count=$(find+..+-maxdepth+1+-type+f%7Cwc+-l)
echo+$count
let+count=count%2B1+#+Increase+by+one,+for+the+next+file+number
echo+$count|code-block|syntax|javascript|540243|请注意，此解决方案效率不高:它为find和wc命令生成子shells，但它应该可以工作。|offset|length|style|CODE|540244|entityMap^0|0|0|G|4|L|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]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

How about:

<pre><code>count=$(find .. -maxdepth 1 -type f|wc -l)
echo $count
let count=count+1 # Increase by one, for the next file number
echo $count
</code></pre>

Note that this solution is not efficient: it spawns sub shells for the <code>find</code> and <code>wc</code> commands, but it should work.

blocks|key|323398|text|file_num=$(ls+-1+--file-type+%7C+grep+-v+'/$'+%7C+wc+-l)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|323399|这比find命令要轻量级一点，可以计算当前目录下的所有文件。|323400|entityMap^0|0|1G|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|N|8|@]|D|@]|E|$]]]|I|$]]

<code>file_num=$(ls -1 --file-type | grep -v '/$' | wc -l)</code>

this is a bit lightweight than a find command, and count all files of the current directory.

blocks|key|540316|text|简单有效的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|540317|#!/bin/bash
RES=$(find+${SOURCE}+-type+f+%7C+wc+-l)|code-block|syntax|javascript|540318|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|$]]

Simple efficient method:

<pre><code>#!/bin/bash
RES=$(find ${SOURCE} -type f | wc -l)
</code></pre>

blocks|key|323483|text|ls+-l+%7C+grep+-v+%5Ed+%7C+wc+-l|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|323484|一句话。|unstyled|323485|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>ls -l | grep -v ^d | wc -l
</code></pre>

One line.

blocks|key|540430|text|扩展被接受的答案(由Dennis+W)：当我尝试这种方法时，我得到了Bash+4.4.5中不带子目录的dirs的错误计数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|540431|问题是，在默认情况下，Bash中没有设置nullglob，而numdirs=(*/)设置了一个具有全局模式*/的1元素数组。同样，我怀疑numfiles=(*)会有一个空文件夹的1个元素。|offset|length|style|CODE|540432|将shopt+-s+nullglob设置为禁用nullglobbing可以为我解决这个问题。有关为什么在Bash上没有默认设置nullglob的精彩讨论，请参阅此处的答案：Why+is+nullglob+not+default?|540433|注:我会直接对答案进行评论，但缺乏口碑。|540434|entityMap|0|LINK|mutability|MUTABLE|url|https://unix.stackexchange.com/questions/204803/why-is-nullglob-not-default/204944#204944^0|0|U|C|1H|2|1W|C|0|1|H|2E|S|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@$D|V|E|W|F|G]|$D|X|E|Y|F|G]|$D|Z|E|10|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|11|8|@$D|12|E|13|F|G]]|9|@$D|14|E|15|1|16]]|A|$]]|$1|J|3|K|5|6|7|17|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|M|$N|$5|O|P|Q|A|$R|S]]]]

Expanding on the accepted answer (by Dennis W): when I tried this approach I got incorrect counts for dirs without subdirs in Bash 4.4.5. 

The issue is that by default nullglob is not set in Bash and <code>numdirs=(*/)</code> sets an 1 element array with the glob pattern <code>*/</code>. Likewise I suspect <code>numfiles=(*)</code> would have 1 element for an empty folder. 

Setting <code>shopt -s nullglob</code> to disable nullglobbing resolves the issue for me. For an excellent discussion on why nullglob is not set by default on Bash see the answer here: <a href="https://unix.stackexchange.com/questions/204803/why-is-nullglob-not-default/204944#204944">Why is nullglob not default?</a>

Note: I would have commented on the answer directly but lack the reputation points.

blocks|key|540516|text|这里有一种方法，你可以把它作为一个函数来做。注意:您可以传递此示例，dirs表示(目录计数)，files表示文件计数，"all“表示目录中所有内容的计数。不遍历树，因为我们并不打算这样做。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|540517|function+get_counts_dir()+{

++++#+--+handle+inputs+(e.g.+get_counts_dir+"files"+/path/to/folder)
++++[[+-z+"${1,,}"+]]+&&+type="files"+%7C%7C+type="${1,,}"
++++[[+-z+"${2,,}"+]]+&&+dir="$(pwd)"+%7C%7C+dir="${2,,}"

++++shopt+-s+nullglob
++++PWD=$(pwd)
++++cd+${dir}

++++numfiles=(*)
++++numfiles=${#numfiles[@]}
++++numdirs=(*/)
++++numdirs=${#numdirs[@]}

++++#+--+handle+input+types+files/dirs/or+both
++++result=0
++++case+"${type,,}"+in
++++++++"files")
++++++++++++result=$(((+numfiles+-=+numdirs+)))
++++++++;;
++++++++"dirs")
++++++++++++result=${numdirs}
++++++++;;
++++++++*)++#+--+returns+all+files/dirs
++++++++++++result=${numfiles}
++++++++;;

++++esac

++++cd+${PWD}
++++shopt+-u+nullglob

++++#+--+return+result+--
++++[[+-z+${result}+]]+&&+echo+0+%7C%7C+echo+${result}
}|code-block|syntax|javascript|540518|使用函数的示例：|540519|folder="/home"
get_counts_dir+"files"+"${folder}"
get_counts_dir+"dirs"+"${folder}"
get_counts_dir+"both"+"${folder}"|540520|将打印如下内容：|540521|2
4
6|540522|entityMap^0|0|0|0|0|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]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Here's one way you could do it as a function. Note: you can pass this example, dirs for (directory count), files for files count or "all" for count of everything in a directory. Does not traverse tree as we aren't looking to do that.

<pre><code>function get_counts_dir() {

 # -- handle inputs (e.g. get_counts_dir "files" /path/to/folder)
 [[ -z "${1,,}" ]] &amp;&amp; type="files" || type="${1,,}"
 [[ -z "${2,,}" ]] &amp;&amp; dir="$(pwd)" || dir="${2,,}"

 shopt -s nullglob
 PWD=$(pwd)
 cd ${dir}

 numfiles=(*)
 numfiles=${#numfiles[@]}
 numdirs=(*/)
 numdirs=${#numdirs[@]}

 # -- handle input types files/dirs/or both
 result=0
 case "${type,,}" in
 "files")
 result=$((( numfiles -= numdirs )))
 ;;
 "dirs")
 result=${numdirs}
 ;;
 *) # -- returns all files/dirs
 result=${numfiles}
 ;;

 esac

 cd ${PWD}
 shopt -u nullglob

 # -- return result --
 [[ -z ${result} ]] &amp;&amp; echo 0 || echo ${result}
}
</code></pre>

Examples of using the function :

<pre><code>folder="/home"
get_counts_dir "files" "${folder}"
get_counts_dir "dirs" "${folder}"
get_counts_dir "both" "${folder}"
</code></pre>

Will print something like :

<pre><code>2
4
6
</code></pre>

blocks|key|540574|text|我能想到的最直接、最可靠的方法是使用find命令创建可靠的可计数输出。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|540575|用wc统计find的字符输出|540576|find+.+-maxdepth+1+-type+f+-printf+'.'+%7C+wc+--char|code-block|syntax|javascript|540577|或find输出的字符串长度：|540578|a=$(find+.+-maxdepth+1+-type+f+-printf+'.')
echo+${#a}|540579|或者使用find输出填充算术表达式：|540580|echo+$(($(find+.+-maxdepth+1+-type+f+-printf+'%2B1')))|540581|entityMap^0|I|4|0|1|2|5|4|0|0|1|4|0|0|4|4|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|18|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|19|8|@$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|S|3|T|5|J|7|1C|8|@]|D|@]|E|$K|L]]|$1|U|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|V|$]]

The most straightforward, reliable way I can think of is using the <code>find</code> command to create a reliably countable output.

Counting characters output of <code>find</code> with <code>wc</code>:

<pre class="lang-sh prettyprint-override"><code>find . -maxdepth 1 -type f -printf '.' | wc --char
</code></pre>

or string length of the <code>find</code> output:

<pre class="lang-sh prettyprint-override"><code>a=$(find . -maxdepth 1 -type f -printf '.')
echo ${#a}
</code></pre>

or using <code>find</code> output to populate an arithmetic expression:

<pre class="lang-sh prettyprint-override"><code>echo $(($(find . -maxdepth 1 -type f -printf '+1')))
</code></pre>

Using bash, how can one get the number of files in a folder, excluding directories from a shell script without the interpreter complaining?

With the help of a friend, I've tried

<pre><code>$files=$(find ../ -maxdepth 1 -type f | sort -n)
$num=$("ls -l" | "grep ^-" | "wc -l")
</code></pre>

which returns from the command line:

<pre><code>../1-prefix_blended_fused.jpg: No such file or directory
ls -l : command not found
grep ^-: command not found
wc -l: command not found
</code></pre>

respectively. These commands work on the command line, but NOT with a bash script.

Given a file filled with image files formatted like <code>1-pano.jpg</code>, I want to grab all the images in the directory to get the largest numbered file to tack onto the next image being processed.

Why the discrepancy?

How to get the number of files in a folder as a variable?

解释器

使用bash，如何才能在没有解释器抱怨的情况下获得文件夹中的文件数量，而不是shell脚本中的目录？在朋友的帮助下，我试着$files=$(find ../ -maxdepth 1 -type f | sort -n)$num=$("ls -l" | "grep ^-" | "wc -l")它从命令行返回：../1-p...

问如何将文件夹中的文件数作为变量获取？
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