Bash string range and replace
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 Dec 9 '18 at 14:37
Jeff Schaller
39.9k1054126
asked Dec 8 '18 at 10:21
pbiespbies
15610
add a comment |
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 Dec 9 '18 at 14:37
Jeff Schaller
39.9k1054126
asked Dec 8 '18 at 10:21
pbiespbies
15610
add a comment |
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 Dec 9 '18 at 14:37
Jeff Schaller
39.9k1054126
asked Dec 8 '18 at 10:21
pbiespbies
15610
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 Dec 9 '18 at 14:37
Jeff Schaller
39.9k1054126
asked Dec 8 '18 at 10:21
pbiespbies
15610
edited Dec 9 '18 at 14:37
Jeff Schaller
39.9k1054126
asked Dec 8 '18 at 10:21
pbiespbies
15610
edited Dec 9 '18 at 14:37
Jeff Schaller
39.9k1054126
edited Dec 9 '18 at 14:37
Jeff Schaller
39.9k1054126
edited Dec 9 '18 at 14:37
Jeff Schaller
39.9k1054126
39.9k1054126
asked Dec 8 '18 at 10:21
pbiespbies
15610
asked Dec 8 '18 at 10:21
pbiespbies
15610
asked Dec 8 '18 at 10:21
pbiespbies
15610
15610
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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 '18 at 14:35
ilkkachu
57k785158
answered Dec 8 '18 at 10:30
nohillsidenohillside
2,382919
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 '18 at 14:35
ilkkachu
57k785158
answered Dec 8 '18 at 10:30
nohillsidenohillside
2,382919
add a comment |
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 '18 at 14:35
ilkkachu
57k785158
answered Dec 8 '18 at 10:30
nohillsidenohillside
2,382919
add a comment |
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 '18 at 14:35
ilkkachu
57k785158
answered Dec 8 '18 at 10:30
nohillsidenohillside
2,382919
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 '18 at 14:35
ilkkachu
57k785158
answered Dec 8 '18 at 10:30
nohillsidenohillside
2,382919
edited Dec 8 '18 at 14:35
ilkkachu
57k785158
edited Dec 8 '18 at 14:35
ilkkachu
57k785158
edited Dec 8 '18 at 14:35
ilkkachu
57k785158
57k785158
answered Dec 8 '18 at 10:30
nohillsidenohillside
2,382919
answered Dec 8 '18 at 10:30
nohillsidenohillside
2,382919
answered Dec 8 '18 at 10:30
nohillsidenohillside
2,382919
2,382919
add a comment |
add a comment |
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