Find命令中的Linux LS -T输出

Question

I have prepared a script with ls -t to fetch latest file and compare with duplicates i use below

ls -t *xml |awk 'BEGIN{FS="_"}{if (++dup[$1] >= 2) print}'

However for large size folder ls command not working. so i tried with

find ./ -type f \( -iname "*.xml" \) | sort |awk 'BEGIN{FS="_"}{if (++dup[$1] >= 2) print}'

but newly created files is not extracted first so i am unable to retain newly created file.

i need to change find command in similiar way output of ls -t command.

Answer 1

If your files are guaranteed not to have newlines in their names, try:

find . -type f -printf '%T@ %p\n' | sort -rg | sed -E 's/[^ ]* //' | awk -F_ '{if (++dup[$1] >= 2) print}'

For a more robust solution that accepts all filenames, try (GNU tools required):

find . -type f -printf '%T@ %p\0' | sort -rgz | sed -Ez 's/[^ ]* //' | awk -v RS="\0" -F_ '{if (++dup[$1] >= 2) print}'

How it works

So that we have an example, let's create three files:

$ touch b_1
$ touch b_2
$ touch b_3

We use find to print out the file's timestamp followed by the file's name:

$ find . -type f -printf '%T@ %p\n'
1511234577.7454717760 ./b_3
1511234574.9814419470 ./b_1
1511234576.1054540780 ./b_2

We want the files sorted by timestamp, newest file first, so we use sort -rg to do a numeric reverse sort on the timestamp (expressed as seconds since epoch):

$ find . -type f -printf '%T@ %p\n' | sort -rg
1511234577.7454717760 ./b_3
1511234576.1054540780 ./b_2
1511234574.9814419470 ./b_1

The next step is to get rid of the timestamps. So, we use sed:

$ find . -type f -printf '%T@ %p\n' | sort -rg | sed -E 's/[^ ]* //'
./b_3
./b_2
./b_1

Now, we can use your awk script to identify the older files:

$ find . -type f -printf '%T@ %p\n' | sort -rg | sed -E 's/[^ ]* //' | awk -F_ '{if (++dup[$1] >= 2) print}'
./b_2
./b_1

Compatibility

Very old GNU systems don't support the -E option to sed. On such systems, one may replace -E with -r like:

sed -r 's/[^ ]* //'

Or, for the more robust version:

sed -rz 's/[^ ]* //'