perlwannabe wrote:
I have a file I download everyday, let's call it "output.txt." The file
"output.txt" is saved in a directory by date, for example 10012005 (for Oct.
1, 2005). I have a years worth of output files in my c:\ drive. Now I have
to rename each output file so that I can copy all of the output to a single
directory. It's a nightmare to do manually. I really don't care what the
files are named as long as each file gets a unique name. Imaging this:
C:\01012005
C:\01022005
C:\01032005
C:\01042005
.. . .
Now, each of those directories has a file in it called "output.txt." I want
to get everyone of those "output.txt" and copy it to a single directory
(call it c:\renoutput) and each output will have a unique name. So when I
do a "dir" of C:\RENOUTPUT is looks like:
output1.txt
output2.txt
output3.txt
output4.txt
.. . .
So I suppose I want to do a locate, rename, move. 1) locate all
"output.txt" files on the hard drive; 2) rename each "output.txt" to
something unique, and 3) move each renamed file from its original location
to a single directory.
I have tried a few ways with no success. Thanks for the help.
It's hard to imagine that you spend multiple days looking at such an
easy problem and can't come up with a single line of code to show the
list. To get you started here's a solution of your problem in bash (you
do have bash installed on your windows box ?)
Create a test environment:
[EMAIL PROTECTED]:~/vuilbak$ for i in 1 2 3 4 5 6 7; do mkdir $i; touch
$i/output.txt; done
Create directory for output
[EMAIL PROTECTED]:~/vuilbak$ mkdir result;
Use the following bash oneliner:
[EMAIL PROTECTED]:~/vuilbak$ (( i=0 )); for j in `find . -name
"output.txt"`; do cp $j result/output$i.txt ; (( i++ )); done
Verify result:
[EMAIL PROTECTED]:~/vuilbak$ ls result/
output0.txt output2.txt output4.txt output6.txt
output1.txt output3.txt output5.txt
gr.
E.
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