views:

254

answers:

2

Bash script, need help with

+1  A: 

Instead of specifying $1 $2 $3, loop over the inputs like so:

for user ; do

Then, you will need to change all $1 to $user

Bruce Armstrong
You almost never want `$*`. Use `"$@"` instead. Or just remove the sequence to iterate over entirely; the default for `for` is `$@`.
Ignacio Vazquez-Abrams
I hadn't realized the difference between the variables. I'll update to reflect that.
Bruce Armstrong
Do u mean it should be like:
trs
You won't need the shift before done, aside from that, its better. Although, please reformat the code to display as code.
Bruce Armstrong
+3  A: 

code from the initial post seems to have disappeared... rolled up an example.

#!/bin/bash

function mod {
    if [ -d "/Users/$1/share" ]; then
        echo "permissions changes for $1"
        #chmod 750 /Users/$1/share
        #find /Users/$1/share -type f -exec chmod 744 {} \;
        #find /Users/$1/share -type d -exec chmod 750 {} \;
        #find /Users/$1/ -type f -exec chmod 600 {} \;
        #find /Users/$1/ -type d -exec chmod 700 {} \;
    fi   
}

function clean {
    IFS=':' read -ra pw <<< "$1"
    local c=0
    local n=""
    local t=500
    for i in "${pw[@]}"; do 
        if [ $c == 0 ]; then
            n="$i"
        fi
        if [ $c == 2 ]; then
            if [ "$i" -gt "$t" ]; then
                mod $n
            fi
        fi
        c=$[$c+1]
    done
}

function report {
    export user=$(whoami)
    ls -la "/Users/$user" > report.txt
}

if [ -z $1 ]; then"
    while read line; do 
        if [ "${line:0:1}" != "#" ]; then
            clean $line
        fi
    done < /etc/passwd
else
    for arg in "$@"
       do
         mod $arg
       done
fi
report

only requirement that it doesn't meet is printing full path (only prints relative) under #3 in the report. the variable t is the lowest uid the permissions changes will affect. commented out the chmods so no one accidentally does this to their system. oops.

Orbit