Given files named 1.csv, 2.csv, 3.csv, ...89.csv... n.csv how do I append them together in numeric order)(1 to n) in bash shell script? is there a one-liner for the solution?
Add a comment
|
2 Answers
If your files where named with leading zeros, it would be easier, i.e.
cat [0-9].csv [0-9][0-9].csv .... > new.csv
But its not too hard to get a true numeric order, given
ls -1
1
10
11
12
13
2
20
21
3
7
8
9
(in both samples, note that the option to ls is the number one, (1), not the letter L (l))
AND
ls -1 [0-9]* | sort -n
1
2
3
7
8
9
10
11
12
13
20
21
THEN
cat $( ls -1 *.csv | sort -n ) > new.csv
Assuming all your csv files are numbered.
If you have more than 1000 files, file arg processing in the shell may break down, and you should post a new question for correct use of xargs.
To see what is happening add shell debugging/trace use
set -vx # to turn on
set +vx # to turn it off
.
IHTH.
2 Comments
user121196
ls -l *.csv|sort -n seems to be out of order: ls -l *.csv |sort -n -rw-r--r-- 1 root root 110427099 Sep 4 20:37 94.csv -rw-r--r-- 1 root root 110439250 Sep 4 20:15 93.csv -rw-r--r-- 1 root root 111962812 Sep 4 19:53 92.csv -rw-r--r-- 1 root root 114716696 Sep 4 14:35 65.csv -rw-r--r-- 1 root root 115487626 Sep 4 08:05 35.csv -rw-r--r-- 1 root root 115777573 Sep 3 20:25 1.csv
shellter
Note that is -1 (one), not -l (the letter L). Good luck.
One liner with inline params (you should replace N with maximal file number). Script will skip unexisted files, so if you miss few files in sequence you wouldn't get error :)
for i in `seq 0 N`; do test -f $i.csv && cat $i.csv; done >> result.csv
Or script with params
#!/bin/bash
for i in `seq 0 $1`
do
if [ -f "$i.csv" ]
then
cat $i.csv >> $2
fi
done
Usage: ./script.sh MAX_CSV_ID RESULT_FILE
Example: ./script.sh 89 new.csv
