删除名称相似的文件，除了一个？

Question

I want to remove all previous versions of Tableau Reader.app from applications apart from the latest one. I want to do this through command line. The problem that I am having is I want to remove all files containing tableau reader except for the file tableau reader 2019.2 .

I have tried multiple ways but they aren't working. I am not very experienced with this. Any help is appreciated. The script below: Checks if the Application is running and if not goes about deleting the different versions(bit I need help on)

#!/bin/bash
shopt -s extglob

process="$4"
processrunning=$( ps axc | grep "${process}" )

if [ "$processrunning" != "" ] ; then

echo "$process IS running, do nothing" 
echo "error: script failed"

exit 1
else 

echo "$process is not running and will remove the old versions"

find $HOME/Applications/ -type f -not -name 'Tableau Reader 2019.2.app' | grep "tableau reader" | xargs rm

#rm -r $HOME/Applications/Tableau Reader?*.app![$HOME/Applications/"Tableau Reader 2019.2.app"]

#find. -type f !( -name 'Tableau Reader 2019.2.app') -exec rm -f "Tableau Reader"*.app {} +

fi

Answer 1

尝试

ls "tableau reader"* | sort | head -n -1 | xargs -d '\n' rm

Answer 2

You can try this:

find $HOME/Applications -type f -not -name 'Tableau Reader 2019.2.app' | grep "Tableau Reader" | xargs -I {} rm {}

The scripts finds all the files except with the name Tableau Reader 2019.2.app and from that using grep all files containing Tableau Reader are separated and then deletes it.

Answer 3

This is achievable with the find command only, but lets test it before it deletes something:

find "$HOME/Applications/" -maxdepth 1 -type f -name 'Tableau Reader*.app' -not -name 'Tableau Reader 2019.2.app'

See if it returns the list of the files you want to delete only:

The list of files should look like this, with your home directory instead of /tmp :

/tmp/Applications/Tableau Reader 10.5.app
/tmp/Applications/Tableau Reader 2012.app
/tmp/Applications/Tableau Reader 10.3.app
/tmp/Applications/Tableau Reader 1.app

Lets see the options used to drive find :

• find : the command to find files
• "$HOME/Applications/" : the directory to start the search
• -maxdepth 1 : stay in the directory, do not go into sub-directories
• -type f : find regular files
• -name 'Tableau Reader*.app' : find files with names matching this pattern
• -not -name 'Tableau Reader 2019.2.app' and not matching this other pattern

Now if you are satisfied with the list and are sure that you want to delete these files; find can do it for you if you add it the -delete option like this:

find "$HOME/Applications/" -maxdepth 1 -type f -name 'Tableau Reader*.app' -not -name 'Tableau Reader 2019.2.app' -delete

Answer 4

How about something like

rm -f $(find . -name "tableau reader*" -type f -exec ls -t {} \\; | head -n +2)

Essentialy, run a subshell (as can be seen between $( and ) to obtain what to remove. Within that subshell, use find in the current directory only ( . ) with result type of file ( f ) exec an ls in time modified order using the -t flag. Then pipe to head starting at the 2nd result ( -n +2 ) (which ensures your latest file is excluded from the rm command.

Note - you have to be careful with ls though as you'll get strange results if filenames contain some special characters, but for your use-case it looks like it should be fine.

What I do with these tasks is copy some test files to a new directory to get my expression perfect before scripting it. Make sure you use cp -a to preserve the file dates though.