How to send output of convert to a writable directory in bash?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I want to convert .tbl
files to .csv
in /data/
directory which I've got read-only access.
I'm using this command:
for i in `ls *.tbl`; do
sed 's/|$//' $i > $i/tbl/csv
echo $i
done
But I'm getting:
-bash: supplier.csv: Permission denied
And I don't know how to change the script such that it could save the output in a writable directory.
Any idea how to solve it?
command-line bash
add a comment |
up vote
2
down vote
favorite
I want to convert .tbl
files to .csv
in /data/
directory which I've got read-only access.
I'm using this command:
for i in `ls *.tbl`; do
sed 's/|$//' $i > $i/tbl/csv
echo $i
done
But I'm getting:
-bash: supplier.csv: Permission denied
And I don't know how to change the script such that it could save the output in a writable directory.
Any idea how to solve it?
command-line bash
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I want to convert .tbl
files to .csv
in /data/
directory which I've got read-only access.
I'm using this command:
for i in `ls *.tbl`; do
sed 's/|$//' $i > $i/tbl/csv
echo $i
done
But I'm getting:
-bash: supplier.csv: Permission denied
And I don't know how to change the script such that it could save the output in a writable directory.
Any idea how to solve it?
command-line bash
I want to convert .tbl
files to .csv
in /data/
directory which I've got read-only access.
I'm using this command:
for i in `ls *.tbl`; do
sed 's/|$//' $i > $i/tbl/csv
echo $i
done
But I'm getting:
-bash: supplier.csv: Permission denied
And I don't know how to change the script such that it could save the output in a writable directory.
Any idea how to solve it?
command-line bash
command-line bash
edited Nov 23 at 16:11
wjandrea
7,96042258
7,96042258
asked Nov 23 at 15:51
Marzieh Heidari
1136
1136
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Just specify a different path:
for tbl in *.tbl ; do
sed 's/|$//' "$tbl" > /another/path/"$tbl%.tbl".csv
echo "$tbl"
done
Few minor fixes:
Don't parse the output of
ls
, just iterate over an expanded wildcard pattern.Double quote variables (filenames can contain whitespace).
I used Remove Matching Suffix instead of Substitution.
add a comment |
up vote
4
down vote
Just specify the path (here /path/to/writable/dir/
) before the filename:
for i in *.tbl; do
sed 's/|$//' "$i" >"/path/to/writable/dir/$i/%tbl/csv"
echo "$i"
done
You should never parse the output of ls
(Why?) and always quote variables like $i
here. Adding %
at the pattern’s beginning makes it match against the end of the string so that a filename like tbl.tbl
becomes tbl.csv
rather than csv.tbl
. Read more about Parameter expansion here on bash-hackers.org.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Just specify a different path:
for tbl in *.tbl ; do
sed 's/|$//' "$tbl" > /another/path/"$tbl%.tbl".csv
echo "$tbl"
done
Few minor fixes:
Don't parse the output of
ls
, just iterate over an expanded wildcard pattern.Double quote variables (filenames can contain whitespace).
I used Remove Matching Suffix instead of Substitution.
add a comment |
up vote
4
down vote
accepted
Just specify a different path:
for tbl in *.tbl ; do
sed 's/|$//' "$tbl" > /another/path/"$tbl%.tbl".csv
echo "$tbl"
done
Few minor fixes:
Don't parse the output of
ls
, just iterate over an expanded wildcard pattern.Double quote variables (filenames can contain whitespace).
I used Remove Matching Suffix instead of Substitution.
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Just specify a different path:
for tbl in *.tbl ; do
sed 's/|$//' "$tbl" > /another/path/"$tbl%.tbl".csv
echo "$tbl"
done
Few minor fixes:
Don't parse the output of
ls
, just iterate over an expanded wildcard pattern.Double quote variables (filenames can contain whitespace).
I used Remove Matching Suffix instead of Substitution.
Just specify a different path:
for tbl in *.tbl ; do
sed 's/|$//' "$tbl" > /another/path/"$tbl%.tbl".csv
echo "$tbl"
done
Few minor fixes:
Don't parse the output of
ls
, just iterate over an expanded wildcard pattern.Double quote variables (filenames can contain whitespace).
I used Remove Matching Suffix instead of Substitution.
edited Nov 23 at 16:09
answered Nov 23 at 16:04
choroba
6,56411730
6,56411730
add a comment |
add a comment |
up vote
4
down vote
Just specify the path (here /path/to/writable/dir/
) before the filename:
for i in *.tbl; do
sed 's/|$//' "$i" >"/path/to/writable/dir/$i/%tbl/csv"
echo "$i"
done
You should never parse the output of ls
(Why?) and always quote variables like $i
here. Adding %
at the pattern’s beginning makes it match against the end of the string so that a filename like tbl.tbl
becomes tbl.csv
rather than csv.tbl
. Read more about Parameter expansion here on bash-hackers.org.
add a comment |
up vote
4
down vote
Just specify the path (here /path/to/writable/dir/
) before the filename:
for i in *.tbl; do
sed 's/|$//' "$i" >"/path/to/writable/dir/$i/%tbl/csv"
echo "$i"
done
You should never parse the output of ls
(Why?) and always quote variables like $i
here. Adding %
at the pattern’s beginning makes it match against the end of the string so that a filename like tbl.tbl
becomes tbl.csv
rather than csv.tbl
. Read more about Parameter expansion here on bash-hackers.org.
add a comment |
up vote
4
down vote
up vote
4
down vote
Just specify the path (here /path/to/writable/dir/
) before the filename:
for i in *.tbl; do
sed 's/|$//' "$i" >"/path/to/writable/dir/$i/%tbl/csv"
echo "$i"
done
You should never parse the output of ls
(Why?) and always quote variables like $i
here. Adding %
at the pattern’s beginning makes it match against the end of the string so that a filename like tbl.tbl
becomes tbl.csv
rather than csv.tbl
. Read more about Parameter expansion here on bash-hackers.org.
Just specify the path (here /path/to/writable/dir/
) before the filename:
for i in *.tbl; do
sed 's/|$//' "$i" >"/path/to/writable/dir/$i/%tbl/csv"
echo "$i"
done
You should never parse the output of ls
(Why?) and always quote variables like $i
here. Adding %
at the pattern’s beginning makes it match against the end of the string so that a filename like tbl.tbl
becomes tbl.csv
rather than csv.tbl
. Read more about Parameter expansion here on bash-hackers.org.
edited Nov 23 at 16:38
answered Nov 23 at 16:04
dessert
21.3k55896
21.3k55896
add a comment |
add a comment |
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