How to send output of convert to a writable directory in 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 $i
done
But I'm getting:
-bash: supplier.csv: Permission denied
And I don't know how to change the script such that it could save the output in a writable directory.
Any idea how to solve it?
command-line bash
edited Nov 23 '18 at 16:11
wjandrea
8,89042260
asked Nov 23 '18 at 15:51
Marzieh HeidariMarzieh Heidari
1136
add a comment |
I want to convert .tbl files to .csv in /data/ directory which I've got read-only access.
I'm using this command:
for i in `ls *.tbl`; do
sed 's/|$//' $i > ${i/tbl/csv}
echo $i
done
But I'm getting:
-bash: supplier.csv: Permission denied
And I don't know how to change the script such that it could save the output in a writable directory.
Any idea how to solve it?
command-line bash
edited Nov 23 '18 at 16:11
wjandrea
8,89042260
asked Nov 23 '18 at 15:51
Marzieh HeidariMarzieh Heidari
1136
add a comment |
I want to convert .tbl files to .csv in /data/ directory which I've got read-only access.
I'm using this command:
for i in `ls *.tbl`; do
sed 's/|$//' $i > ${i/tbl/csv}
echo $i
done
But I'm getting:
-bash: supplier.csv: Permission denied
And I don't know how to change the script such that it could save the output in a writable directory.
Any idea how to solve it?
command-line bash
edited Nov 23 '18 at 16:11
wjandrea
8,89042260
asked Nov 23 '18 at 15:51
Marzieh HeidariMarzieh Heidari
1136
I want to convert .tbl files to .csv in /data/ directory which I've got read-only access.
I'm using this command:
for i in `ls *.tbl`; do
sed 's/|$//' $i > ${i/tbl/csv}
echo $i
done
But I'm getting:
-bash: supplier.csv: Permission denied
And I don't know how to change the script such that it could save the output in a writable directory.
Any idea how to solve it?
command-line bash
command-line bash
edited Nov 23 '18 at 16:11
wjandrea
8,89042260
asked Nov 23 '18 at 15:51
Marzieh HeidariMarzieh Heidari
1136
edited Nov 23 '18 at 16:11
wjandrea
8,89042260
asked Nov 23 '18 at 15:51
Marzieh HeidariMarzieh Heidari
1136
edited Nov 23 '18 at 16:11
wjandrea
8,89042260
edited Nov 23 '18 at 16:11
wjandrea
8,89042260
edited Nov 23 '18 at 16:11
wjandrea
8,89042260
8,89042260
asked Nov 23 '18 at 15:51
Marzieh HeidariMarzieh Heidari
1136
asked Nov 23 '18 at 15:51
Marzieh HeidariMarzieh Heidari
1136
asked Nov 23 '18 at 15:51
Marzieh HeidariMarzieh Heidari
1136
1136
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Just specify a different path:
for tbl in *.tbl ; do
sed 's/|$//' "$tbl" > /another/path/"${tbl%.tbl}".csv
echo "$tbl"
done
Few minor fixes:
Don't parse the output of
ls, just iterate over an expanded wildcard pattern.Double quote variables (filenames can contain whitespace).
I used Remove Matching Suffix instead of Substitution.
answered Nov 23 '18 at 16:04
chorobachoroba
6,63411730
add a comment |
Just specify the path (here /path/to/writable/dir/) before the filename:
for i in *.tbl; do
sed 's/|$//' "$i" >"/path/to/writable/dir/${i/%tbl/csv}"
echo "$i"
done
You should never parse the output of ls (Why?) and always quote variables like $i here. Adding % at the pattern’s beginning makes it match against the end of the string so that a filename like tbl.tbl becomes tbl.csv rather than csv.tbl. Read more about Parameter expansion here on bash-hackers.org.
answered Nov 23 '18 at 16:04
dessertdessert
22.8k56399
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Just specify a different path:
for tbl in *.tbl ; do
sed 's/|$//' "$tbl" > /another/path/"${tbl%.tbl}".csv
echo "$tbl"
done
Few minor fixes:
Don't parse the output of
ls, just iterate over an expanded wildcard pattern.Double quote variables (filenames can contain whitespace).
I used Remove Matching Suffix instead of Substitution.
answered Nov 23 '18 at 16:04
chorobachoroba
6,63411730
add a comment |
Just specify a different path:
for tbl in *.tbl ; do
sed 's/|$//' "$tbl" > /another/path/"${tbl%.tbl}".csv
echo "$tbl"
done
Few minor fixes:
Don't parse the output of
ls, just iterate over an expanded wildcard pattern.Double quote variables (filenames can contain whitespace).
I used Remove Matching Suffix instead of Substitution.
answered Nov 23 '18 at 16:04
chorobachoroba
6,63411730
add a comment |
Just specify a different path:
for tbl in *.tbl ; do
sed 's/|$//' "$tbl" > /another/path/"${tbl%.tbl}".csv
echo "$tbl"
done
Few minor fixes:
Don't parse the output of
ls, just iterate over an expanded wildcard pattern.Double quote variables (filenames can contain whitespace).
I used Remove Matching Suffix instead of Substitution.
answered Nov 23 '18 at 16:04
chorobachoroba
6,63411730
Just specify a different path:
for tbl in *.tbl ; do
sed 's/|$//' "$tbl" > /another/path/"${tbl%.tbl}".csv
echo "$tbl"
done
Few minor fixes:
Don't parse the output of
ls, just iterate over an expanded wildcard pattern.Double quote variables (filenames can contain whitespace).
I used Remove Matching Suffix instead of Substitution.
answered Nov 23 '18 at 16:04
chorobachoroba
6,63411730
edited Nov 23 '18 at 16:09
answered Nov 23 '18 at 16:04
chorobachoroba
6,63411730
answered Nov 23 '18 at 16:04
chorobachoroba
6,63411730
answered Nov 23 '18 at 16:04
chorobachoroba
6,63411730
6,63411730
add a comment |
add a comment |
Just specify the path (here /path/to/writable/dir/) before the filename:
for i in *.tbl; do
sed 's/|$//' "$i" >"/path/to/writable/dir/${i/%tbl/csv}"
echo "$i"
done
You should never parse the output of ls (Why?) and always quote variables like $i here. Adding % at the pattern’s beginning makes it match against the end of the string so that a filename like tbl.tbl becomes tbl.csv rather than csv.tbl. Read more about Parameter expansion here on bash-hackers.org.
answered Nov 23 '18 at 16:04
dessertdessert
22.8k56399
add a comment |
Just specify the path (here /path/to/writable/dir/) before the filename:
for i in *.tbl; do
sed 's/|$//' "$i" >"/path/to/writable/dir/${i/%tbl/csv}"
echo "$i"
done
You should never parse the output of ls (Why?) and always quote variables like $i here. Adding % at the pattern’s beginning makes it match against the end of the string so that a filename like tbl.tbl becomes tbl.csv rather than csv.tbl. Read more about Parameter expansion here on bash-hackers.org.
answered Nov 23 '18 at 16:04
dessertdessert
22.8k56399
add a comment |
Just specify the path (here /path/to/writable/dir/) before the filename:
for i in *.tbl; do
sed 's/|$//' "$i" >"/path/to/writable/dir/${i/%tbl/csv}"
echo "$i"
done
You should never parse the output of ls (Why?) and always quote variables like $i here. Adding % at the pattern’s beginning makes it match against the end of the string so that a filename like tbl.tbl becomes tbl.csv rather than csv.tbl. Read more about Parameter expansion here on bash-hackers.org.
answered Nov 23 '18 at 16:04
dessertdessert
22.8k56399
Just specify the path (here /path/to/writable/dir/) before the filename:
for i in *.tbl; do
sed 's/|$//' "$i" >"/path/to/writable/dir/${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.
answered Nov 23 '18 at 16:04
dessertdessert
22.8k56399
edited Nov 23 '18 at 16:38
answered Nov 23 '18 at 16:04
dessertdessert
22.8k56399
answered Nov 23 '18 at 16:04
dessertdessert
22.8k56399
answered Nov 23 '18 at 16:04
dessertdessert
22.8k56399
22.8k56399
add a comment |
add a comment |
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