Ubuntu – Can I run shell script together with application

background-processbashscripts

I wrote small script to automatically delete Chrome cache after I close it:

#!/bin/bash

while true; do
    if [[ $(pgrep -l chrome) ]]; then
        sleep 20
    else
        rm -rf ~/.cache/google-chrome/Default/Cache/*
        rm -rf ~/.cache/google-chrome/Default/"Media Cache"/*
        notify-send "CCD" "Cache deleted!"
        break    
fi
done

Now I don't want to run this script manually everytime, I want it to run in background automatically when I launch Chrome. I tried to edit Chrome quicklist with Ubuntu Tweak:

But as I expected, it didn't work. So, is there any other way to do it?

Best Answer

Just add full path to your script into Startup Applications to make it start automatically upon login. Open Unity Dash and add it as new command. Of course, make sure you script has executable permissions with chmod +x /path/to/script.sh

To address the issue of chrome deleting cache on startup ( which is undesirable , as mentioned in the comments), use polling with while-loop to wait for chrome to appear.

while true; do

    # Wait for chrome window to appear
    while ! pgrep -l 'chrome' ; do : ; sleep 20; done 

    # Now wait for it to disappear
    while pgrep -l 'chrome' ; do : ; sleep 20; done 

    # Once chrome window disappears, delete cache.
    rm -rf ~/.cache/google-chrome/Default/Cache/*
    rm -rf ~/.cache/google-chrome/Default/"Media Cache"/*
    notify-send "CCD" "Cache deleted!"

    # And at this point we restart the whole process again.

done

Related Solutions

Ubuntu – Shell Script to Run Commands to All Subdirectories

You can iterate subdirs with a for-loop like this

#!/usr/bin/env bash

for dir in /the/path/*/; do
    awk 'tolower($0) ~ /01_node_setup/{line=$0} END{print line}' "$dir/his_file"

    if [[ -f "$dir/sysconfig.out" ]]; then
        awk '/Drive Model/{getline line} END{print line}' "$dir/sysconfig.out"
    else
        awk '/Pair0 DIMM0/{print;exit}' "$dir/node0/trans_file_prev"/*setupsys*
    fi
done >> /home/xtee/sst-logs.out

I changed your greps into awks that should be more portable. I believe they should produce the same output. The important parts are the quotes around the paths.

PHP – How to Call and Run PHP Script from Shell Script

The problem here is that you are quoting the entiore command you are trying to run as a single variable. As a result, you're not running php with foo.php as an argument but instead are attempting to execute a file called php foo.php. Here's a simpler example to show you what I mean:

$ var1="echo "
$ var2="foo"
$ set -x ## debugging info
$ "$var1$var2"
+ 'echo foo'      ### The shell tries to execute both vars as a single command
bash: echo foo: command not found

$ "$var1" "$var2"
+ 'echo ' foo     ### The shell tries to execute 'echo ' (echo with a space)
bash: echo foo: command not found

So, the right way is to remove the space and quote each variable separately:

$ var1="echo"
$ var2="foo"
$ "$var1" "$var2"

If you do that though, you'll hit the next error. The . is the source command. That tries to read a shell script and execute it in the current session. You are giving it a php script instead. That won't work, you need to execute it, not source it.

Finally, always avoid using CAPITAL variable names. The shell's reserved variables are capitalized so it's a good idea to always use lower case variable names for your scripts.

Putting all this together (with a few other minor improvements), what you want is something like:

#!/bin/sh

list="/path/to/my/site/dir"
config="/usr/bin/php"

for i in "$list"
do
    "$config" "$i"/test.php
done

Best Answer

Related Solutions

Ubuntu – Shell Script to Run Commands to All Subdirectories

PHP – How to Call and Run PHP Script from Shell Script

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