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Assign the same string to multiple variables

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5

2

I want to assign the string contained in $value to multiple variables.

Actually the example I gave before (var1=var2=...=$value) was not reflecting exactly what I wanted.
So far, I found this but it only works if $value is an integer:

$ let varT=varz=var3=$value

How can I do that?

bash assignment

share|improve this question

edited May 15 at 9:39

muru

38.4k591168

asked May 15 at 7:06

SebMaSebMa

3351413

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Shouldn't the question be: Assign the same string value to multiple variables? (subject and object switched)?
  
  – rexkogitans
  May 15 at 9:30
  

  
  
  
  

add a comment | 

5

2

I want to assign the string contained in $value to multiple variables.

Actually the example I gave before (var1=var2=...=$value) was not reflecting exactly what I wanted.
So far, I found this but it only works if $value is an integer:

$ let varT=varz=var3=$value

How can I do that?

bash assignment

share|improve this question

edited May 15 at 9:39

muru

38.4k591168

asked May 15 at 7:06

SebMaSebMa

3351413

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Shouldn't the question be: Assign the same string value to multiple variables? (subject and object switched)?
  
  – rexkogitans
  May 15 at 9:30
  

  
  
  
  

add a comment | 

I want to assign the string contained in $value to multiple variables.

Actually the example I gave before (var1=var2=...=$value) was not reflecting exactly what I wanted.
So far, I found this but it only works if $value is an integer:

$ let varT=varz=var3=$value

How can I do that?

bash assignment

share|improve this question

edited May 15 at 9:39

muru

38.4k591168

asked May 15 at 7:06

SebMaSebMa

3351413

$ let varT=varz=var3=$value

share|improve this question

edited May 15 at 9:39

muru

38.4k591168

asked May 15 at 7:06

SebMaSebMa

3351413

share|improve this question

edited May 15 at 9:39

muru

38.4k591168

asked May 15 at 7:06

SebMaSebMa

3351413

edited May 15 at 9:39

muru

38.4k591168

edited May 15 at 9:39

muru

38.4k591168

asked May 15 at 7:06

SebMaSebMa

3351413

asked May 15 at 7:06

SebMaSebMa

3351413

4 Answers
4

active

oldest

votes

6

In my opinion you're better of just doing the more readable:

var1="$value" var2="$value" var3="$value" var4="$value" var5="$value" var6="$value" var7="$value" var8="$value" var9="$value" var10="$value"

But if you want a very short way of accomplishing this then try:

declare var1..10="$value"

Edited: using brace expansions instead of printf and declare instead of eval, which could be dangerous depending on what's in $value.

Cf. EDIT1: You could still use brace expansions in the new case:

declare varT,z,3="$value"

It's safer than the printf approach in the comments because it can handle spaces in $value.

share|improve this answer

edited May 15 at 8:08

answered May 15 at 7:30

laenkeiolaenkeio

4617

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Please see my EDIT1. The printf method you gave was just fine if you put the declare in front : declare $(printf .....)
  
  – SebMa
  May 15 at 7:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  With EDIT1 using printf that would be: declare $(printf "var%s=$value " T z 3). But be careful with spaces in $value.
  
  – laenkeio
  May 15 at 8:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Note that the declare approach is still dangerous if any of the variables were previously declared as arrays. Try after var3=() value='([0$(echo Gotcha>&2)]=)'
  
  – Stéphane Chazelas
  May 15 at 12:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  eval is fine if used as eval var1..10=$value, that is as long as the (arbitrary) content of $value is not evaluated as shell code.
  
  – Stéphane Chazelas
  May 15 at 12:19
  

  
  
  
  

add a comment | 

6

let is doing an arithmetic evaluation. In bash, this is equivalent to (( ... )). This is why your code only works for integers.

Using an array instead of specially named variables:

var=( "$value" "$value" "$value" "$value" "$value" "$value" "$value" "$value" "$value" "$value" )

printf 'var[5]=%sn' "$var[5]"

or an associative array,

declare -A var

var=( ["T"]=$value ["z"]=$value [3]=$value )

printf 'var[T]=%sn' "$var["T"]"

You could also do a loop:

for varname in var0 var1 varT varfoo; do
 declare -n var="$varname"
 var=$value
done

This loop loops over names of variables. In the loop body, a name reference variable is created that references the current loop variable. When the value of the name reference variable is set, the named variable is set.

share|improve this answer

edited May 15 at 8:07

answered May 15 at 7:50

Kusalananda♦Kusalananda

147k18278462

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What does the -n option do? I get an error with it. Ah... namerefs (>= 4.3, IIRC?)
  
  – muru
  May 15 at 8:08
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @muru Yes, declare -n creates a name reference variable. You would have to run bash 4.3+ to use name references.
  
  – Kusalananda♦
  May 15 at 8:09
  
  

  
  
  
  

add a comment | 

3

value=balabala
eval var1..10=$value
echo $var1..10

share|improve this answer

edited May 15 at 12:16

Stéphane Chazelas

319k57603969

answered May 15 at 7:39

dedowsdidedowsdi

2993

add a comment | 

2

Value='-%% this is a test %%-'
for VarName in myVar1 myVar2 myOtherVar; do printf -v "$VarName" %s "$Value"; done

share|improve this answer

edited May 15 at 12:20

Stéphane Chazelas

319k57603969

answered May 15 at 11:08

BuygrushBuygrush

211

New contributor

Buygrush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

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6

In my opinion you're better of just doing the more readable:

var1="$value" var2="$value" var3="$value" var4="$value" var5="$value" var6="$value" var7="$value" var8="$value" var9="$value" var10="$value"

But if you want a very short way of accomplishing this then try:

declare var1..10="$value"

Edited: using brace expansions instead of printf and declare instead of eval, which could be dangerous depending on what's in $value.

Cf. EDIT1: You could still use brace expansions in the new case:

declare varT,z,3="$value"

It's safer than the printf approach in the comments because it can handle spaces in $value.

share|improve this answer

edited May 15 at 8:08

answered May 15 at 7:30

laenkeiolaenkeio

4617

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Please see my EDIT1. The printf method you gave was just fine if you put the declare in front : declare $(printf .....)
  
  – SebMa
  May 15 at 7:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  With EDIT1 using printf that would be: declare $(printf "var%s=$value " T z 3). But be careful with spaces in $value.
  
  – laenkeio
  May 15 at 8:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Note that the declare approach is still dangerous if any of the variables were previously declared as arrays. Try after var3=() value='([0$(echo Gotcha>&2)]=)'
  
  – Stéphane Chazelas
  May 15 at 12:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  eval is fine if used as eval var1..10=$value, that is as long as the (arbitrary) content of $value is not evaluated as shell code.
  
  – Stéphane Chazelas
  May 15 at 12:19
  

  
  
  
  

add a comment | 

6

In my opinion you're better of just doing the more readable:

var1="$value" var2="$value" var3="$value" var4="$value" var5="$value" var6="$value" var7="$value" var8="$value" var9="$value" var10="$value"

But if you want a very short way of accomplishing this then try:

declare var1..10="$value"

Edited: using brace expansions instead of printf and declare instead of eval, which could be dangerous depending on what's in $value.

Cf. EDIT1: You could still use brace expansions in the new case:

declare varT,z,3="$value"

It's safer than the printf approach in the comments because it can handle spaces in $value.

share|improve this answer

edited May 15 at 8:08

answered May 15 at 7:30

laenkeiolaenkeio

4617

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Please see my EDIT1. The printf method you gave was just fine if you put the declare in front : declare $(printf .....)
  
  – SebMa
  May 15 at 7:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  With EDIT1 using printf that would be: declare $(printf "var%s=$value " T z 3). But be careful with spaces in $value.
  
  – laenkeio
  May 15 at 8:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Note that the declare approach is still dangerous if any of the variables were previously declared as arrays. Try after var3=() value='([0$(echo Gotcha>&2)]=)'
  
  – Stéphane Chazelas
  May 15 at 12:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  eval is fine if used as eval var1..10=$value, that is as long as the (arbitrary) content of $value is not evaluated as shell code.
  
  – Stéphane Chazelas
  May 15 at 12:19
  

  
  
  
  

add a comment | 

In my opinion you're better of just doing the more readable:

var1="$value" var2="$value" var3="$value" var4="$value" var5="$value" var6="$value" var7="$value" var8="$value" var9="$value" var10="$value"

But if you want a very short way of accomplishing this then try:

declare var1..10="$value"

Edited: using brace expansions instead of printf and declare instead of eval, which could be dangerous depending on what's in $value.

Cf. EDIT1: You could still use brace expansions in the new case:

declare varT,z,3="$value"

It's safer than the printf approach in the comments because it can handle spaces in $value.

share|improve this answer

edited May 15 at 8:08

answered May 15 at 7:30

laenkeiolaenkeio

4617

var1="$value" var2="$value" var3="$value" var4="$value" var5="$value" var6="$value" var7="$value" var8="$value" var9="$value" var10="$value"

declare var1..10="$value"

declare varT,z,3="$value"

share|improve this answer

edited May 15 at 8:08

answered May 15 at 7:30

laenkeiolaenkeio

4617

answered May 15 at 7:30

laenkeiolaenkeio

4617

answered May 15 at 7:30

laenkeiolaenkeio

4617

6

let is doing an arithmetic evaluation. In bash, this is equivalent to (( ... )). This is why your code only works for integers.

Using an array instead of specially named variables:

var=( "$value" "$value" "$value" "$value" "$value" "$value" "$value" "$value" "$value" "$value" )

printf 'var[5]=%sn' "$var[5]"

or an associative array,

declare -A var

var=( ["T"]=$value ["z"]=$value [3]=$value )

printf 'var[T]=%sn' "$var["T"]"

You could also do a loop:

for varname in var0 var1 varT varfoo; do
 declare -n var="$varname"
 var=$value
done

This loop loops over names of variables. In the loop body, a name reference variable is created that references the current loop variable. When the value of the name reference variable is set, the named variable is set.

share|improve this answer

edited May 15 at 8:07

answered May 15 at 7:50

Kusalananda♦Kusalananda

147k18278462

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What does the -n option do? I get an error with it. Ah... namerefs (>= 4.3, IIRC?)
  
  – muru
  May 15 at 8:08
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @muru Yes, declare -n creates a name reference variable. You would have to run bash 4.3+ to use name references.
  
  – Kusalananda♦
  May 15 at 8:09
  
  

  
  
  
  

add a comment | 

6

let is doing an arithmetic evaluation. In bash, this is equivalent to (( ... )). This is why your code only works for integers.

Using an array instead of specially named variables:

var=( "$value" "$value" "$value" "$value" "$value" "$value" "$value" "$value" "$value" "$value" )

printf 'var[5]=%sn' "$var[5]"

or an associative array,

declare -A var

var=( ["T"]=$value ["z"]=$value [3]=$value )

printf 'var[T]=%sn' "$var["T"]"

You could also do a loop:

for varname in var0 var1 varT varfoo; do
 declare -n var="$varname"
 var=$value
done

This loop loops over names of variables. In the loop body, a name reference variable is created that references the current loop variable. When the value of the name reference variable is set, the named variable is set.

share|improve this answer

edited May 15 at 8:07

answered May 15 at 7:50

Kusalananda♦Kusalananda

147k18278462

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What does the -n option do? I get an error with it. Ah... namerefs (>= 4.3, IIRC?)
  
  – muru
  May 15 at 8:08
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @muru Yes, declare -n creates a name reference variable. You would have to run bash 4.3+ to use name references.
  
  – Kusalananda♦
  May 15 at 8:09
  
  

  
  
  
  

add a comment | 

let is doing an arithmetic evaluation. In bash, this is equivalent to (( ... )). This is why your code only works for integers.

Using an array instead of specially named variables:

var=( "$value" "$value" "$value" "$value" "$value" "$value" "$value" "$value" "$value" "$value" )

printf 'var[5]=%sn' "$var[5]"

or an associative array,

declare -A var

var=( ["T"]=$value ["z"]=$value [3]=$value )

printf 'var[T]=%sn' "$var["T"]"

You could also do a loop:

for varname in var0 var1 varT varfoo; do
 declare -n var="$varname"
 var=$value
done

This loop loops over names of variables. In the loop body, a name reference variable is created that references the current loop variable. When the value of the name reference variable is set, the named variable is set.

share|improve this answer

edited May 15 at 8:07

answered May 15 at 7:50

Kusalananda♦Kusalananda

147k18278462

var=( "$value" "$value" "$value" "$value" "$value" "$value" "$value" "$value" "$value" "$value" )

printf 'var[5]=%sn' "$var[5]"

declare -A var

var=( ["T"]=$value ["z"]=$value [3]=$value )

printf 'var[T]=%sn' "$var["T"]"

for varname in var0 var1 varT varfoo; do
 declare -n var="$varname"
 var=$value
done

share|improve this answer

edited May 15 at 8:07

answered May 15 at 7:50

Kusalananda♦Kusalananda

147k18278462

answered May 15 at 7:50

Kusalananda♦Kusalananda

147k18278462

answered May 15 at 7:50

Kusalananda♦Kusalananda

147k18278462

3

value=balabala
eval var1..10=$value
echo $var1..10

share|improve this answer

edited May 15 at 12:16

Stéphane Chazelas

319k57603969

answered May 15 at 7:39

dedowsdidedowsdi

2993

add a comment | 

3

value=balabala
eval var1..10=$value
echo $var1..10

share|improve this answer

edited May 15 at 12:16

Stéphane Chazelas

319k57603969

answered May 15 at 7:39

dedowsdidedowsdi

2993

add a comment | 

value=balabala
eval var1..10=$value
echo $var1..10

share|improve this answer

edited May 15 at 12:16

Stéphane Chazelas

319k57603969

answered May 15 at 7:39

dedowsdidedowsdi

2993

value=balabala
eval var1..10=$value
echo $var1..10

share|improve this answer

edited May 15 at 12:16

Stéphane Chazelas

319k57603969

answered May 15 at 7:39

dedowsdidedowsdi

2993

edited May 15 at 12:16

Stéphane Chazelas

319k57603969

edited May 15 at 12:16

Stéphane Chazelas

319k57603969

answered May 15 at 7:39

dedowsdidedowsdi

2993

answered May 15 at 7:39

dedowsdidedowsdi

2993

2

Value='-%% this is a test %%-'
for VarName in myVar1 myVar2 myOtherVar; do printf -v "$VarName" %s "$Value"; done

share|improve this answer

edited May 15 at 12:20

Stéphane Chazelas

319k57603969

answered May 15 at 11:08

BuygrushBuygrush

211

New contributor

Buygrush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

2

Value='-%% this is a test %%-'
for VarName in myVar1 myVar2 myOtherVar; do printf -v "$VarName" %s "$Value"; done

share|improve this answer

edited May 15 at 12:20

Stéphane Chazelas

319k57603969

answered May 15 at 11:08

BuygrushBuygrush

211

New contributor

Buygrush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

Value='-%% this is a test %%-'
for VarName in myVar1 myVar2 myOtherVar; do printf -v "$VarName" %s "$Value"; done

share|improve this answer

edited May 15 at 12:20

Stéphane Chazelas

319k57603969

answered May 15 at 11:08

BuygrushBuygrush

211

New contributor

Buygrush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Value='-%% this is a test %%-'
for VarName in myVar1 myVar2 myOtherVar; do printf -v "$VarName" %s "$Value"; done

share|improve this answer

edited May 15 at 12:20

Stéphane Chazelas

319k57603969

answered May 15 at 11:08

BuygrushBuygrush

211

New contributor

Buygrush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited May 15 at 12:20

Stéphane Chazelas

319k57603969

edited May 15 at 12:20

Stéphane Chazelas

319k57603969

answered May 15 at 11:08

BuygrushBuygrush

211

New contributor

Buygrush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered May 15 at 11:08

BuygrushBuygrush

211