CODEWITHSUNDEEP
powershellazure-devops
Nilay
Nilay

Reputation: 65

How to pass a variable having a dollar sign to another variable in PowerShell?

I am using an On-Prem Server to run VSTS Build/Releases. Currently, I am trying to pass a variable from VSTS: $(password) to my script. Suppose the value of this variable $(password) is 'stringwith$sign'`

This variable $(password) needs to be injected into a string in my script:

$string = "I need to inject my password string from VSTS here "$(password)""

The String should look like this:

$string = I need to inject my password string from VSTS here "stringwith$sign"

How do I achieve this? The build/release will fail if I simply add it as $(password) since it thinks $sign in "stringwith$sign" is a variable. I cannot even use '' quotes since my variable $(password) needs to be inserted in $string.

Upvotes: 4

Views: 5590

Answers (3)

Marina Liu
Marina Liu

Reputation: 38106

You just need to use the format ${env:password} instead of $(password) to get the variable password's value.

Such as if you add a PowerShell task with below script:

$string="I need to inject my password string from VSTS here ${env:test}" 
echo $string

Then it will show I need to inject my password string from VSTS here stringwith$sign in the build log.

Upvotes: 1

4c74356b41
4c74356b41

Reputation: 72171

You can also use format operator:

$myVar = 'okay$dollar'
'My string with my var {0} inside' -f $myVar

or Get-Variable:

$myVar = 'okay$dollar'
"My string with my var $(Get-Variable myvar | select -expandproperty value) inside"

Upvotes: 0

DeanOC
DeanOC

Reputation: 7282

Without seeing any sample code, it's a bit hard to tell how your script works.

But basically, if you are setting a string literal that contains special characters, you can stop them from being parsed by using single quotes instead of double-quotes. For example, if you execute

$password = "stringwith$sign"
$password

Then the value of password is stringwith.

This is because powershell has parsed the string and treated $sign as being the name of a variable and has attempted to insert the value of $sign. But as $sign hasn't been declared, the default value of empty string is used.

However, if you used single quotes, i.e.

$password = 'stringwith$sign'
$password

Then the value of password is stringwith$sign.

Subsequently, setting 

$string = "I need to inject my password string from VSTS here ""$password"""

gives $string the value of

I need to inject my password string from VSTS here "stringwith$sign"

Upvotes: 4

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