将多个文件分类为一个文件，文件名位于数据之前

Question

I have several log files with data in them. What i want to do is cat all these files into one file. But before the data goes in i want the file name to be there without the extension. For Example: Files I have:

file1.log file2.log file3.log

The file that i want: all.log

all.log to have in it:

file1
    file1's data
file2
    file2's data
file3
    file3's data

Answer 1

Using awk

awk 'FNR==1{sub(/[.][^.]*$/, "", FILENAME); print FILENAME} 1' file*.log >all.log

FNR is the file record number. It is one at the beginning of each file. Thus, the test FNR==1 tells us if we are at the beginning of a file. If we are, then we remove the extension from the filename using sub(/[.][^.]*$/, "", FILENAME) and then we print it.

The final 1 in the program is awk's cryptic way of saying print-this-line.

The redirection >all.log saves all the output in file all.log .

Using shell

for f in file*.log; do echo "${f%.*}"; cat "$f"; done >all.log

Or:

for f in file*.log
do 
    echo "${f%.*}"
    cat "$f"
done >all.log

In shell, for f in file*.log; do for f in file*.log; do starts a loop over all files matching the glob file*.log . The statement echo "${f%.*}" prints the file name minus the extension. ${f%.*} is an example of suffix removal . cat "$f" prints the contents of the file. done >all.log terminates the loop and saves all the output in all.log .

This loop will work correctly even if file names contain spaces, tabs, newlines, or other difficult characters.

Answer 2

Suppose you have two files:

foo:

a
b
c

bar:

d
e
f

Using Perl:

perl -lpe 'print $ARGV if $. == 1; close(ARGV) if eof' foo bar > all.log

foo
a
b
c
bar
d
e
f

$. is the line number
$ARGV is the name of the current file
close(ARGV) if eof resets the line number at the end of each file

Using grep:

grep '' foo bar > all.log

foo:a
foo:b
foo:c
bar:d
bar:e
bar:f

Answer 3

for i in `ls file*`; do `echo $i | awk -F"." '{print $1}' >> all.log; cat $i >> all.log`; done