UNIX for Dummies Questions & Answers

Hi,
I am trying to organize my data files and I would like to know how to cat files with path. Note that the correct command may not be "cat" though.
For example, I have:
abc/data/1/title
dca/data/2/title
gpd/data/1/title
....

I'd like to have a file listing the title info from different directories, as well as directory paths.

Thanks

Simon

How should the o/p look like

The output should be like:
48/pdata/1/title:ub2k63r, 1mM in PBS buffer, crp, for nemo titration
49/pdata/1/title:ub2k63r nemo titration, 1/5 nemo
50/pdata/1/title:ub2k63r nemo titration, 1/5 nemo
83/pdata/1/title:quick T2_2 of n15-NEMO+k63-Ub2

The text after colon( is the text written in the title file.
Actually I succeeded once before, but I forgot what command(s) I used. I think I used "find", but can't make it work this time.

Thanks,

Code:

for file in `find /base/path -type f -name "title"` ; do echo "${file}:`head -1 $file`"; done

Hi, Thanks for the suggestion.
However, I am still confused. Could you be more detailed?
Thanks

Not sure what you mean... Does the code not work? What error do you get?

I think you want to cat the files, but you want the filename printed for each cat (the files contain but a single line each?). AFAIK cat will not do this, but other commands will:

Code:

# Prints filename in front of each line of all files in current dir
 grep . *

# Does the same as above, but finding all files named "title"
 find . -name "title" -type f -exec grep . {} /dev/null \;

Why the "/dev/null" you ask? Because grep will only print the filename if there is more than one file listed, so I use /dev/null as a second file.

The dot (".") RegEx matches all nonblank lines.