bash 在名称以数字开头的目录中查找编号最高的文件名
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Find highest numbered filename in a directory where names start with digits
提问by Jake
I have a directory with files that look like this:
我有一个包含如下文件的目录：
001_something.php 002_something_else.php
004_xyz.php 005_do_good_to_others.php
I ultimately want to create a new, empty PHP file whose name starts with the next number in the series.
我最终想创建一个新的空 PHP 文件，其名称以系列中的下一个数字开头。
LIST=`exec ls $MY_DIR | sed 's/\([0-9]\+\).*//g' | tr '\n' ' '`
The preceding code gives me a string like this:
前面的代码给了我一个这样的字符串：
LIST='001 002 004 005 '
I want to grab that 005, increment by one, and then use that number to generate the new filename. How do I do that in BASH?
我想获取那个 005，加一，然后使用这个数字来生成新的文件名。我如何在 BASH 中做到这一点？
采纳答案by DigitalRoss
$ LIST=($LIST)
$ newfile=`printf %03d-whatever $((10#${LIST[${#LIST}]}+1))`
$ echo $newfile
006-whatever
So, that's bash-specific. Below is a any-posix-shell-including-bashsolution, tho I imagine something simpler is possible.
所以，这是特定于 bash 的。下面是一个any-posix-shell-including-bash解决方案，虽然我想更简单的事情是可能的。
$ cat /tmp/z
f () {
eval echo ${$#} | sed -e 's/^0*//'
}
LIST='001 002 004 005 '
newname=`printf %03d-whatever $(($(f $LIST) + 1))`
echo $newname
$ sh /tmp/z
006-whatever
$
回答by bmb
Do you need the whole LIST?
你需要整个 LIST 吗？
If not
如果不
LAST=`exec ls $MY_DIR | sed 's/\([0-9]\+\).*//g' | sort -n | tail -1`
will give you just the 005 part and
只会给你 005 部分和
printf "%03d" `expr 1 + $LAST`
will print the next number in the sequence.
将打印序列中的下一个数字。
回答by Idelic
Using only standard tools, the following will give you the prefix for the new file (006):
仅使用标准工具，以下内容将为您提供新文件 (006) 的前缀：
ls [0-9]* | sed 's/_/ _/' | sort -rn | awk '{printf "%03d", + 1; exit}'
回答by Vijay
This seems to be more simple.
这似乎更简单。
ls [0-9]* | sort -rn | awk '{FS="_"; printf "%03d_new_file.php\n",+1;exit}'
回答by Paused until further notice.
Fast, subshell-free and loop-free (also handles filenames with spaces)*:
快速、无子外壳和无循环（也处理带空格的文件名）*：
list=([0-9]*)
last=${list[@]: -1}
nextnum=00$((10#${last%%[^0-9]*} + 1))
nextnum=${nextnum: -3}
touch ${nextnum}_a_new_file.php
Given your example files the output would be:
鉴于您的示例文件，输出将是：
006_a_new_file.php
回答by ghostdog74
touch "newfile_$(printf "%03d" $(echo $(ls 00?_*.php|sed 's/_.*//'|sort -rn|head -1)+1|bc))"
2.
num=$(ls 00?*.php|sed 's/.*//'|sort -rn|head -1) touch $(printf "newfile_%03d" $((num+1)))
touch "newfile_$(printf "%03d" $(echo $(ls 00?_*.php|sed 's/_.*//'|sort -rn|head -1)+1|bc))"
2.
num=$(ls 00?*.php|sed 's/.*//'|sort -rn|head -1) touch $(printf "newfile_%03d" $((num+1)))
回答by SebPe
I post this answer to propose a minimal solution to the core question "Find highest numbered filename in a directory where names start with digits":
我发布此答案是为了针对核心问题“在名称以数字开头的目录中查找编号最高的文件名”提出一个最小的解决方案：
ls -1 "$MyDir" | sort -hr | head -n 1
The first statement puts out the files in MyDir as one liners with the -1 option. This is piped to the sort function which sorts in human numeric order with -h and in reverse order (highest number first) with -r. This is piped to the head function which just delivers the -n top most results (1 in the example). The answer considers human numeric sort order i.e. 221 is higher than 67. It is a POSIX shell solution so highly compatible.
第一条语句将 MyDir 中的文件作为带有 -1 选项的单行输出。这是通过管道传送到 sort 函数，该函数使用 -h 以人类数字顺序排序，并使用 -r 以相反顺序（最高数字在前）排序。这通过管道传送到 head 函数，该函数只提供 -n 最高的结果（示例中为 1）。答案考虑了人类的数字排序顺序，即 221 高于 67。它是一种高度兼容的 POSIX shell 解决方案。
回答by ghostdog74
$ for file in *php; do last=${file%%_*} ; done
$ newfilename="test.php"
$ printf "%03d_%s" $((last+1)) $newfilename
006_test.php
i leave it to you to do the creation of new file
我让你来创建新文件
回答by ghostdog74
Here's a solution (you need bc):
这是一个解决方案（您需要 bc）：
#!/bin/bash
LIST='001 002 008 005 004 002 '
a=0
for f in $LIST
do
t=$(echo $f + 1|bc)
if [ $t -ge $a ]
then
a=$t
fi
done
printf "%03d\n" $a
