[SOLVED] Bash for loop

edwardcode · 08-24-2011, 11:30 AM

I am having some trouble with this script I am writing. I need this script to copy all of the files from one dir to an other. I have to use the "TO" and "FROM" variables for the rest of the script so I thought the "for" loop was the best solution. The problem I am having is I need the for loop to do somthing like this:

Code:

#!/bin/bash
FROM="
/home/test1/hey
/home/test1/hey1
/home/test1/hey2
/home/test1/hey3
/home/test1/hey4
/home/test1/hey5
"
TO="
/home/test2/hey
/home/test2/hey1
/home/test2/hey2
/home/test2/hey3
/home/test2/hey4
/home/test2/hey5
"

for FILE in ${FROM} DEST in ${TO}
do
   cp -v ${FILE} ${DEST}
done

Obviously this will not work but I need something similar to do the same thing.

Thanks

szboardstretcher · 08-24-2011, 11:40 AM

Unless I am misunderstanding the point here, you seem to be needlessly overcomplicating this process. This would work just fine:

Code:

cp /home/test1/* /home/test2/

weibullguy · 08-24-2011, 11:42 AM

Something like this in case you want to pass the source and destination directories as arguments?

Code:

#!/bin/sh

SRCDIR=$1
DESTDIR=$2

cp -v $SRCDIR/* $DESTDIR/

exit 0

Quote:

Originally Posted by szboardstretcher

...you seem to be needlessly overcomplicating this process.

Hehe, I needlessly over-complicate stuff like this all the time before I realize what I've done.

edwardcode · 08-24-2011, 11:47 AM

well this is just a test script for what is going to be used in a different environment. The real environment will have all of the files in different locations and different destinations.

Sorry for the miss understanding.

szboardstretcher · 08-24-2011, 11:49 AM

I believe I see what you are trying to do here. You are trying to increment *both* variables as you go through the loop. Correct?

edwardcode · 08-24-2011, 11:58 AM

David the H. · 08-24-2011, 12:24 PM

When you're working with sequences, such as lists of filenames, you should generally use arrays instead. Remember, a variable is a single string, while an array is a string of strings (a list).

So something like this:

Code:

#!/bin/bash
FROM=(
/home/test1/hey
/home/test1/hey1
/home/test1/hey2
/home/test1/hey3
/home/test1/hey4
/home/test1/hey5
)

TO=(
/home/test2/hey
/home/test2/hey1
/home/test2/hey2
/home/test2/hey3
/home/test2/hey4
/home/test2/hey5
)


for i in "${!FROM[@]}"; do
   cp -v "${FROM[i]}" "${TO[i]}"
done

As long as the index numbers for the files match in both arrays, you should get your desired behavior.

BTW, remember to always quote your variable expansions!

Edit: Just thinking, depending on your purposes, you may be able to shorten the above with parameter substitution. You could use it to modify the directory path directly, so you'd need only one array.

Code:

#!/bin/bash
FILES=(
/home/test1/hey
/home/test1/hey1
/home/test1/hey2
/home/test1/hey3
/home/test1/hey4
/home/test1/hey5
)
NEWPATH="/home/test2"

for i in "${!FILES[@]}"; do
   cp -v "${FILES[i]}" "$NEWPATH/${FILES[i]##*/}"
done

edwardcode · 08-24-2011, 12:48 PM

If it is not to much to ask I would like to ask you a few questions so I understand what is going on in your script.

1. In the statement below what dose the "!" do?
for i in "${!FROM[@]}"; do

2. In the statement below what dose the "[@] do?
for i in "${!FROM[@]}"; do

3.in the statement below how is the "[i]" used?
cp -v "${FROM[i]}" "${TO[i]}"

Thanks

szboardstretcher · 08-24-2011, 12:50 PM

Quote:

2. In the statement below what dose the "[@] do?
for i in "${!FROM[@]}"; do

@ references all the members in the array

David the H. · 08-24-2011, 01:38 PM

The links I gave above should explain most of it, and the rest should be easily locatable in the bash man page or on the web. But anyway:

Quote:

1. In the statement below what dose the "!" do?
for i in "${!FROM[@]}"; do

${!array[@]} outputs a list of all existing array indexes. This is quite useful for iterating over the array in a for loop. Stick an echo in front of it and see what you get.

Quote:

2. In the statement below what dose the "[@] do?
for i in "${!FROM[@]}"; do

As mentioned above, it references the entire array. @ considers each entry as a separate element (where applicable), particularly when quoted. There's also *, which references the entire array as a single unit. Again, try echoing them and see what you get.

Quote:

3.in the statement below how is the "[i]" used?
cp -v "${FROM[i]}" "${TO[i]}"

i is simply the for loop's variable. It contains each index number from "${!FROM[@]}" in turn, so it's "${FROM[0]}" first, then "${FROM[1]}", then "${FROM[2]}", etc.

Note that in regular arrays the [] index field has an arithmetic context, so the only things allowed in it are numbers, mathematical operators, and variable names, which are automatically expanded, even without $ in front of them.

edwardcode · 08-25-2011, 08:00 AM

thanks

MTK358 · 08-26-2011, 08:51 AM

For more information, see "man bash" and search for "Parameter Expansion", by typing this:

Code:

/Parameter Expansion

and pressing Enter.