CODEWITHSUNDEEP
bashshellsedls
Jake
Jake

Reputation: 3075

Find highest numbered filename in a directory where names start with digits

I have a directory with files that look like this:

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.

LIST=`exec ls $MY_DIR | sed 's/\([0-9]\+\).*/\1/g' | tr '\n' ' '`

The preceding code gives me a string like this:

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?

Upvotes: 13

Views: 16262

Answers (9)

SebPe
SebPe

Reputation: 11

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":

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.

Upvotes: 1

ghostdog74
ghostdog74

Reputation: 342363

  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)))

Upvotes: 1

Vijay
Vijay

Reputation: 67221

This seems to be more simple.

ls [0-9]* | sort -rn | awk '{FS="_"; printf "%03d_new_file.php\n",$1+1;exit}'

Upvotes: 3

Idelic
Idelic

Reputation: 15582

Using only standard tools, the following will give you the prefix for the new file (006):

ls [0-9]* | sed 's/_/ _/' | sort -rn | awk '{printf "%03d", $1 + 1; exit}'

Upvotes: 8

Dennis Williamson
Dennis Williamson

Reputation: 360055

Fast, subshell-free and loop-free (also handles filenames with spaces)*:

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:

006_a_new_file.php

Upvotes: 2

DigitalRoss
DigitalRoss

Reputation: 146043

$ 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-bash solution, tho I imagine something simpler is possible.

$ 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
$

Upvotes: 10

bmb
bmb

Reputation: 6248

Do you need the whole LIST?

If not

LAST=`exec ls $MY_DIR | sed 's/\([0-9]\+\).*/\1/g' | sort -n | tail -1`

will give you just the 005 part and 

printf "%03d" `expr 1 + $LAST`

will print the next number in the sequence.

Upvotes: 16

ghostdog74
ghostdog74

Reputation: 342363

$ 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

Upvotes: 1

user180100
user180100

Reputation:

Here's a solution (you need 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

Upvotes: 0

Related Questions