Bash string range and replace
up vote
2
down vote
favorite
To print part of the string, replace , with . I use command:
echo "${q:16:6}" | sed 's/,/./'
Is it possible to use something like:
echo "${q:16:6/,/.}"
because it does not work?
bash sed variable replace
edited 2 days ago
Jeff Schaller
37.6k1052121
asked Dec 8 at 10:21
pbies
15110
add a comment |
up vote
2
down vote
favorite
To print part of the string, replace , with . I use command:
echo "${q:16:6}" | sed 's/,/./'
Is it possible to use something like:
echo "${q:16:6/,/.}"
because it does not work?
bash sed variable replace
edited 2 days ago
Jeff Schaller
37.6k1052121
asked Dec 8 at 10:21
pbies
15110
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
To print part of the string, replace , with . I use command:
echo "${q:16:6}" | sed 's/,/./'
Is it possible to use something like:
echo "${q:16:6/,/.}"
because it does not work?
bash sed variable replace
edited 2 days ago
Jeff Schaller
37.6k1052121
asked Dec 8 at 10:21
pbies
15110
To print part of the string, replace , with . I use command:
echo "${q:16:6}" | sed 's/,/./'
Is it possible to use something like:
echo "${q:16:6/,/.}"
because it does not work?
bash sed variable replace
bash sed variable replace
edited 2 days ago
Jeff Schaller
37.6k1052121
asked Dec 8 at 10:21
pbies
15110
edited 2 days ago
Jeff Schaller
37.6k1052121
asked Dec 8 at 10:21
pbies
15110
edited 2 days ago
Jeff Schaller
37.6k1052121
edited 2 days ago
Jeff Schaller
37.6k1052121
edited 2 days ago
Jeff Schaller
37.6k1052121
37.6k1052121
asked Dec 8 at 10:21
pbies
15110
asked Dec 8 at 10:21
pbies
15110
asked Dec 8 at 10:21
pbies
15110
15110
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
10
down vote
accepted
You can't stack/nest parameter expansion in Bash, so not even
echo "${${q:16:6}/,/.}"
will work. (Nested expansions like that do work in Zsh, though.)
If you want to stay within Bash, you'll need to use a temporary variable:
foo="${q:16:6}"
echo "${foo/,/.}"
edited Dec 8 at 14:35
ilkkachu
54.4k782148
answered Dec 8 at 10:30
nohillside
2,087817
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
You can't stack/nest parameter expansion in Bash, so not even
echo "${${q:16:6}/,/.}"
will work. (Nested expansions like that do work in Zsh, though.)
If you want to stay within Bash, you'll need to use a temporary variable:
foo="${q:16:6}"
echo "${foo/,/.}"
edited Dec 8 at 14:35
ilkkachu
54.4k782148
answered Dec 8 at 10:30
nohillside
2,087817
add a comment |
up vote
10
down vote
accepted
You can't stack/nest parameter expansion in Bash, so not even
echo "${${q:16:6}/,/.}"
will work. (Nested expansions like that do work in Zsh, though.)
If you want to stay within Bash, you'll need to use a temporary variable:
foo="${q:16:6}"
echo "${foo/,/.}"
edited Dec 8 at 14:35
ilkkachu
54.4k782148
answered Dec 8 at 10:30
nohillside
2,087817
add a comment |
up vote
10
down vote
accepted
up vote
10
down vote
accepted
You can't stack/nest parameter expansion in Bash, so not even
echo "${${q:16:6}/,/.}"
will work. (Nested expansions like that do work in Zsh, though.)
If you want to stay within Bash, you'll need to use a temporary variable:
foo="${q:16:6}"
echo "${foo/,/.}"
edited Dec 8 at 14:35
ilkkachu
54.4k782148
answered Dec 8 at 10:30
nohillside
2,087817
You can't stack/nest parameter expansion in Bash, so not even
echo "${${q:16:6}/,/.}"
will work. (Nested expansions like that do work in Zsh, though.)
If you want to stay within Bash, you'll need to use a temporary variable:
foo="${q:16:6}"
echo "${foo/,/.}"
edited Dec 8 at 14:35
ilkkachu
54.4k782148
answered Dec 8 at 10:30
nohillside
2,087817
edited Dec 8 at 14:35
ilkkachu
54.4k782148
edited Dec 8 at 14:35
ilkkachu
54.4k782148
edited Dec 8 at 14:35
ilkkachu
54.4k782148
54.4k782148
answered Dec 8 at 10:30
nohillside
2,087817
answered Dec 8 at 10:30
nohillside
2,087817
answered Dec 8 at 10:30
nohillside
2,087817
2,087817
add a comment |
add a comment |
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