Shell Programming and Scripting

So the directories won't be there to move ...?
I think it'd much better to read through the file once and perform the work, then for each directory to spawn off awk to find it in a file.. Maybe a style thing. In the end directories that exist will be moved. You can suppress errors or check for it first I guess

Code:

[[ -d $olddir ]] && {
    echo "Moving $olddir"
    mv "$olddir" "$newdir"
}

Correct, I think. Basically between mergers and new policies we are migrating in tons of remote offices into one global standard. These mobile offices have been self managing for years. Someone has pieced together a master list of all the folders on all the servers and shares. They have input all the old names and what the new names should be. So, not every folder exists on every share, there is just one master list. That is why I was building an array of data from the share point, and then looping it through the master list and moving folders only when matches were found.

This isn't a very elegant solution but I am working with what I got here.

Thanks,

---------- Post updated at 12:46 PM ---------- Previous update was at 09:14 AM ----------

So, is the problem that I am passing ${i} in awk, inside single quotes? Which is why it outputs blank because it is interpreting it as a literal character and not a variable?

Code:

newFolder=`awk -F, '/${i}/ { sub(/\r/,"");print $2;exit }' ~/home/folders_list.txt `

Since ${i} is in single quotes inside the awk logic it is not passing it as a variable?

---------- Post updated at 01:01 PM ---------- Previous update was at 12:46 PM ----------

So, after reading through the man page I think this is what I want to do with awk

Code:

newFolder=`/usr/bin/awk -F, -v var=${i} '/var/ { sub(/\r/,"");print $2;exit }' ~/test/folders_list.txt`

That seems to pass the variable properly but doesn't grab the correct value.

https://www.unix.com/302627793-post3.html

also quote your shell expansions, -v var="$i"
if you REALLY want to use use /var/ (which matches anywhere in the line, not just the old directory column like you should) then you cannot use that shorthand, because /var/ treats var as the regex itself... you'd use $0 ~ var which would still treat the variable as a regex (but you've a fixed string, dots should not match ANY character probably). So you'd use index($0,var).. But still, see post #3 I really think you want $1 == var (unless i'm still totally mistaken about whats going on here)

So, yeah I figured it out, and it is working. I guess I wasn't grasping at first you cannot pass a bash variable to an awk program.

here is my new method and the output works as expected. Thanks for all your suggestions man I super appreciate it. This isn't high priority but I cannot let it sit any longer, otherwise I will have managers asking me why it is not done yet.

Code:

newFolder=`/usr/bin/awk -F, '/^'${i}'/ { sub(/\r/,"");print $2;exit }' ~/path/to/master_list.txt`

Yeah I know probably not the most elegant solution, but it does work. Sorry, I am not quite an awk master here and am working with what they gave me.

---------- Post updated at 01:59 PM ---------- Previous update was at 01:52 PM ----------


yeah man, that is exactly what I did. I travel for work right now so I am in different time zones and hotels 6 days a week, and only off one day. I am pretty much half zombie right now.

Haha, I will totally paypal you money for a beer, you deserve it after dealing with me!

Like that, you would still need to double quote the variable expansion:

Code:

newFolder=`/usr/bin/awk -F, '/^'"$i"'/ { ...`

But that would still leave room for partial matches...
Neutronscott made some better suggestions IMO.