how to keep line feed in result returned by bash command
let's say, I can rsh to machine XXX as user foo, then after executing:
$result=$(rsh -l foo XXX "ls");
echo $result;
I found that the line feed is removed in the result, and I can't see the result in a line by line way. so what I should do if I'd like to have the line feed int the returned result?
bash
asked Mar 12 '12 at 7:27
HaiYuan ZhangHaiYuan Zhang
1,49452330
add a comment |
let's say, I can rsh to machine XXX as user foo, then after executing:
$result=$(rsh -l foo XXX "ls");
echo $result;
I found that the line feed is removed in the result, and I can't see the result in a line by line way. so what I should do if I'd like to have the line feed int the returned result?
bash
asked Mar 12 '12 at 7:27
HaiYuan ZhangHaiYuan Zhang
1,49452330
You should not have a dollar sign in front of the varible name when you assign to it.rsh=$(rsh -l foo XXX "ls"). The terminating semicolons are redundant but harmless.
– tripleee
Mar 12 '12 at 8:05
add a comment |
let's say, I can rsh to machine XXX as user foo, then after executing:
$result=$(rsh -l foo XXX "ls");
echo $result;
I found that the line feed is removed in the result, and I can't see the result in a line by line way. so what I should do if I'd like to have the line feed int the returned result?
bash
asked Mar 12 '12 at 7:27
HaiYuan ZhangHaiYuan Zhang
1,49452330
let's say, I can rsh to machine XXX as user foo, then after executing:
$result=$(rsh -l foo XXX "ls");
echo $result;
I found that the line feed is removed in the result, and I can't see the result in a line by line way. so what I should do if I'd like to have the line feed int the returned result?
bash
bash
asked Mar 12 '12 at 7:27
HaiYuan ZhangHaiYuan Zhang
1,49452330
asked Mar 12 '12 at 7:27
HaiYuan ZhangHaiYuan Zhang
1,49452330
asked Mar 12 '12 at 7:27
HaiYuan ZhangHaiYuan Zhang
1,49452330
asked Mar 12 '12 at 7:27
HaiYuan ZhangHaiYuan Zhang
1,49452330
asked Mar 12 '12 at 7:27
HaiYuan ZhangHaiYuan Zhang
1,49452330
1,49452330
You should not have a dollar sign in front of the varible name when you assign to it.rsh=$(rsh -l foo XXX "ls"). The terminating semicolons are redundant but harmless.
– tripleee
Mar 12 '12 at 8:05
add a comment |
You should not have a dollar sign in front of the varible name when you assign to it.rsh=$(rsh -l foo XXX "ls"). The terminating semicolons are redundant but harmless.
– tripleee
Mar 12 '12 at 8:05
You should not have a dollar sign in front of the varible name when you assign to it.
rsh=$(rsh -l foo XXX "ls"). The terminating semicolons are redundant but harmless.– tripleee
Mar 12 '12 at 8:05
You should not have a dollar sign in front of the varible name when you assign to it.
rsh=$(rsh -l foo XXX "ls"). The terminating semicolons are redundant but harmless.– tripleee
Mar 12 '12 at 8:05
add a comment |
2 Answers
2
active
oldest
votes
The newline is there; the error is in using echo to examine it (without double quotes around the variable, too!)
result=$(rsh -l foo XXX "ls")
echo "$result"
In trivial cases, you can get away without quotes, but the interesting cases might even contain security issues.
If you are only capturing the standard output of rsh so that you can print it to standard output yourself, this is a useless use of echo
answered Mar 12 '12 at 8:04
tripleeetripleee
1,82632130
Probably also prefersshoverrshand don't uselsin scripts.
– tripleee
Jan 30 at 11:33
add a comment |
Edit: If you mean the newlines between each file, just make sure to quote the result:
result="$(rsh -l foo XXX "ls");"
This will keep any characters except NUL ($''), which can't be stored in a Bash variable.
If you mean that the trailing newline is lost, then this is the simplest solution I know:
resultx="$(commands...; echo x)"
result="${resultx%x}"
answered Mar 12 '12 at 9:50
l0b0l0b0
5,51622341
add a comment |
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2 Answers
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2 Answers
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The newline is there; the error is in using echo to examine it (without double quotes around the variable, too!)
result=$(rsh -l foo XXX "ls")
echo "$result"
In trivial cases, you can get away without quotes, but the interesting cases might even contain security issues.
If you are only capturing the standard output of rsh so that you can print it to standard output yourself, this is a useless use of echo
answered Mar 12 '12 at 8:04
tripleeetripleee
1,82632130
Probably also prefersshoverrshand don't uselsin scripts.
– tripleee
Jan 30 at 11:33
add a comment |
The newline is there; the error is in using echo to examine it (without double quotes around the variable, too!)
result=$(rsh -l foo XXX "ls")
echo "$result"
In trivial cases, you can get away without quotes, but the interesting cases might even contain security issues.
If you are only capturing the standard output of rsh so that you can print it to standard output yourself, this is a useless use of echo
answered Mar 12 '12 at 8:04
tripleeetripleee
1,82632130
Probably also prefersshoverrshand don't uselsin scripts.
– tripleee
Jan 30 at 11:33
add a comment |
The newline is there; the error is in using echo to examine it (without double quotes around the variable, too!)
result=$(rsh -l foo XXX "ls")
echo "$result"
In trivial cases, you can get away without quotes, but the interesting cases might even contain security issues.
If you are only capturing the standard output of rsh so that you can print it to standard output yourself, this is a useless use of echo
answered Mar 12 '12 at 8:04
tripleeetripleee
1,82632130
The newline is there; the error is in using echo to examine it (without double quotes around the variable, too!)
result=$(rsh -l foo XXX "ls")
echo "$result"
In trivial cases, you can get away without quotes, but the interesting cases might even contain security issues.
If you are only capturing the standard output of rsh so that you can print it to standard output yourself, this is a useless use of echo
answered Mar 12 '12 at 8:04
tripleeetripleee
1,82632130
edited Jan 30 at 6:23
answered Mar 12 '12 at 8:04
tripleeetripleee
1,82632130
answered Mar 12 '12 at 8:04
tripleeetripleee
1,82632130
answered Mar 12 '12 at 8:04
tripleeetripleee
1,82632130
1,82632130
Probably also prefersshoverrshand don't uselsin scripts.
– tripleee
Jan 30 at 11:33
add a comment |
Probably also prefersshoverrshand don't uselsin scripts.
– tripleee
Jan 30 at 11:33
Probably also prefer
ssh over rsh and don't use ls in scripts.– tripleee
Jan 30 at 11:33
Probably also prefer
ssh over rsh and don't use ls in scripts.– tripleee
Jan 30 at 11:33
add a comment |
Edit: If you mean the newlines between each file, just make sure to quote the result:
result="$(rsh -l foo XXX "ls");"
This will keep any characters except NUL ($''), which can't be stored in a Bash variable.
If you mean that the trailing newline is lost, then this is the simplest solution I know:
resultx="$(commands...; echo x)"
result="${resultx%x}"
answered Mar 12 '12 at 9:50
l0b0l0b0
5,51622341
add a comment |
Edit: If you mean the newlines between each file, just make sure to quote the result:
result="$(rsh -l foo XXX "ls");"
This will keep any characters except NUL ($''), which can't be stored in a Bash variable.
If you mean that the trailing newline is lost, then this is the simplest solution I know:
resultx="$(commands...; echo x)"
result="${resultx%x}"
answered Mar 12 '12 at 9:50
l0b0l0b0
5,51622341
add a comment |
Edit: If you mean the newlines between each file, just make sure to quote the result:
result="$(rsh -l foo XXX "ls");"
This will keep any characters except NUL ($''), which can't be stored in a Bash variable.
If you mean that the trailing newline is lost, then this is the simplest solution I know:
resultx="$(commands...; echo x)"
result="${resultx%x}"
answered Mar 12 '12 at 9:50
l0b0l0b0
5,51622341
Edit: If you mean the newlines between each file, just make sure to quote the result:
result="$(rsh -l foo XXX "ls");"
This will keep any characters except NUL ($''), which can't be stored in a Bash variable.
If you mean that the trailing newline is lost, then this is the simplest solution I know:
resultx="$(commands...; echo x)"
result="${resultx%x}"
answered Mar 12 '12 at 9:50
l0b0l0b0
5,51622341
edited Mar 12 '12 at 10:04
answered Mar 12 '12 at 9:50
l0b0l0b0
5,51622341
answered Mar 12 '12 at 9:50
l0b0l0b0
5,51622341
answered Mar 12 '12 at 9:50
l0b0l0b0
5,51622341
5,51622341
add a comment |
add a comment |
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You should not have a dollar sign in front of the varible name when you assign to it.
rsh=$(rsh -l foo XXX "ls"). The terminating semicolons are redundant but harmless.– tripleee
Mar 12 '12 at 8:05