Bash script string interpolation leaves curly braces intact

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2
down vote

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The request string in the example below interpolates version variable, but keeps the curly braces and I can't figure out why.

#!/bin/sh

version=2989
request="http://example.com/?version=$version&therest"
echo "$request"

Result:

$ ~/script.sh
http://example.com/?version=2989&therest

Environment:

$ echo $0
-zsh

asked Dec 15 '17 at 20:02

Alexei Danchenkov

207311

You don't run bash in your example, you run sh.
â€“Â user1934428
Dec 18 '17 at 11:12

You have the parenthesis inside a quoted string, so they are output literally. Only $version is replaced, because variable expansion is one the things which does get performed within double-quoted strings.
â€“Â user1934428
Dec 18 '17 at 11:14

add a commentÂ |Â

up vote
2
down vote

favorite

The request string in the example below interpolates version variable, but keeps the curly braces and I can't figure out why.

#!/bin/sh

version=2989
request="http://example.com/?version=$version&therest"
echo "$request"

Result:

$ ~/script.sh
http://example.com/?version=2989&therest

Environment:

$ echo $0
-zsh

asked Dec 15 '17 at 20:02

Alexei Danchenkov

207311

You don't run bash in your example, you run sh.
â€“Â user1934428
Dec 18 '17 at 11:12

You have the parenthesis inside a quoted string, so they are output literally. Only $version is replaced, because variable expansion is one the things which does get performed within double-quoted strings.
â€“Â user1934428
Dec 18 '17 at 11:14

add a commentÂ |Â

up vote
2
down vote

favorite

The request string in the example below interpolates version variable, but keeps the curly braces and I can't figure out why.

#!/bin/sh

version=2989
request="http://example.com/?version=$version&therest"
echo "$request"

Result:

$ ~/script.sh
http://example.com/?version=2989&therest

Environment:

$ echo $0
-zsh

asked Dec 15 '17 at 20:02

Alexei Danchenkov

207311

The request string in the example below interpolates version variable, but keeps the curly braces and I can't figure out why.

#!/bin/sh

version=2989
request="http://example.com/?version=$version&therest"
echo "$request"

Result:

$ ~/script.sh
http://example.com/?version=2989&therest

Environment:

$ echo $0
-zsh

asked Dec 15 '17 at 20:02

Alexei Danchenkov

207311

asked Dec 15 '17 at 20:02

Alexei Danchenkov

207311

asked Dec 15 '17 at 20:02

Alexei Danchenkov

207311

asked Dec 15 '17 at 20:02

Alexei Danchenkov

207311

You don't run bash in your example, you run sh.
â€“Â user1934428
Dec 18 '17 at 11:12

You have the parenthesis inside a quoted string, so they are output literally. Only $version is replaced, because variable expansion is one the things which does get performed within double-quoted strings.
â€“Â user1934428
Dec 18 '17 at 11:14

add a commentÂ |Â

You don't run bash in your example, you run sh.
â€“Â user1934428
Dec 18 '17 at 11:12

You have the parenthesis inside a quoted string, so they are output literally. Only $version is replaced, because variable expansion is one the things which does get performed within double-quoted strings.
â€“Â user1934428
Dec 18 '17 at 11:14

You don't run bash in your example, you run sh.
â€“Â user1934428
Dec 18 '17 at 11:12

You have the parenthesis inside a quoted string, so they are output literally. Only $version is replaced, because variable expansion is one the things which does get performed within double-quoted strings.
â€“Â user1934428
Dec 18 '17 at 11:14

add a commentÂ |Â

2 Answers
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up vote
3
down vote

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The { is before $. It should be $version :)

answered Dec 15 '17 at 20:07

m0dular

63115

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up vote
1
down vote

Compare:

$ version=1.2; echo "http://example.com/?version=$version&therest"
http://example.com/?version=1.2&therest
$ version=1.2; echo "http://example.com/?version=$version&therest"
http://example.com/?version=1.2&therest

Inside quoted, braces are regarded as just normal characters unless they are part of some construct like $variableName.

answered Dec 15 '17 at 20:07

John1024

44.2k4100116

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
3
down vote

accepted

The { is before $. It should be $version :)

answered Dec 15 '17 at 20:07

m0dular

63115

add a commentÂ |Â

up vote
3
down vote

accepted

The { is before $. It should be $version :)

answered Dec 15 '17 at 20:07

m0dular

63115

add a commentÂ |Â

up vote
3
down vote

accepted

The { is before $. It should be $version :)

answered Dec 15 '17 at 20:07

m0dular

63115

The { is before $. It should be $version :)

answered Dec 15 '17 at 20:07

m0dular

63115

answered Dec 15 '17 at 20:07

m0dular

63115

answered Dec 15 '17 at 20:07

m0dular

63115

answered Dec 15 '17 at 20:07

m0dular

63115

add a commentÂ |Â

up vote
1
down vote

Compare:

$ version=1.2; echo "http://example.com/?version=$version&therest"
http://example.com/?version=1.2&therest
$ version=1.2; echo "http://example.com/?version=$version&therest"
http://example.com/?version=1.2&therest

Inside quoted, braces are regarded as just normal characters unless they are part of some construct like $variableName.

answered Dec 15 '17 at 20:07

John1024

44.2k4100116

add a commentÂ |Â

up vote
1
down vote

Compare:

$ version=1.2; echo "http://example.com/?version=$version&therest"
http://example.com/?version=1.2&therest
$ version=1.2; echo "http://example.com/?version=$version&therest"
http://example.com/?version=1.2&therest

Inside quoted, braces are regarded as just normal characters unless they are part of some construct like $variableName.

answered Dec 15 '17 at 20:07

John1024

44.2k4100116

add a commentÂ |Â

up vote
1
down vote

Compare:

$ version=1.2; echo "http://example.com/?version=$version&therest"
http://example.com/?version=1.2&therest
$ version=1.2; echo "http://example.com/?version=$version&therest"
http://example.com/?version=1.2&therest

Inside quoted, braces are regarded as just normal characters unless they are part of some construct like $variableName.

answered Dec 15 '17 at 20:07

John1024

44.2k4100116

Compare:

$ version=1.2; echo "http://example.com/?version=$version&therest"
http://example.com/?version=1.2&therest
$ version=1.2; echo "http://example.com/?version=$version&therest"
http://example.com/?version=1.2&therest

Inside quoted, braces are regarded as just normal characters unless they are part of some construct like $variableName.

answered Dec 15 '17 at 20:07

John1024

44.2k4100116

answered Dec 15 '17 at 20:07

John1024

44.2k4100116

answered Dec 15 '17 at 20:07

John1024

44.2k4100116

answered Dec 15 '17 at 20:07

John1024

44.2k4100116

add a commentÂ |Â

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