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 replace
asked 18 hours ago
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 replace
asked 18 hours ago
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 replace
asked 18 hours ago
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 replace
bash sed replace
asked 18 hours ago
pbies
15110
asked 18 hours ago
pbies
15110
asked 18 hours ago
pbies
15110
asked 18 hours ago
pbies
15110
asked 18 hours ago
pbies
15110
15110
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
9
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 14 hours ago
ilkkachu
54.3k782147
answered 18 hours ago
nohillside
2,077716
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
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 14 hours ago
ilkkachu
54.3k782147
answered 18 hours ago
nohillside
2,077716
add a comment |
up vote
9
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 14 hours ago
ilkkachu
54.3k782147
answered 18 hours ago
nohillside
2,077716
add a comment |
up vote
9
down vote
accepted
up vote
9
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 14 hours ago
ilkkachu
54.3k782147
answered 18 hours ago
nohillside
2,077716
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 14 hours ago
ilkkachu
54.3k782147
answered 18 hours ago
nohillside
2,077716
edited 14 hours ago
ilkkachu
54.3k782147
edited 14 hours ago
ilkkachu
54.3k782147
edited 14 hours ago
ilkkachu
54.3k782147
54.3k782147
answered 18 hours ago
nohillside
2,077716
answered 18 hours ago
nohillside
2,077716
answered 18 hours ago
nohillside
2,077716
2,077716
add a comment |
add a comment |
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