string variable not substituted in multi-line string

Question

Here is the code/output all in one:

PS C:\code\misc> cat .\test.ps1; echo ----; .\test.ps1
$user="arun"
$password="1234*"
$jsonStr = @"
{
        "proxy": "http://$user:[email protected]:80",
        "https-proxy": "http://$user:[email protected]:80"
}
"@
del out.txt
echo $jsonStr >> out.txt
cat out.txt
----
{
        "proxy": "http://1234*@org.proxy.com:80",
        "https-proxy": "http://1234*@org.proxy.com:80"
}

Content of string variable $user is not substituted in $jsonStr.

What is the correct way to substitute it?

Matt · Accepted Answer · 2019-01-02 16:48:28Z

33

The colon is the scope character: $scope:$variable. PowerShell thinks you are invoking the variable $password from the scope $user. You might be able to get away with a subexpression.

$user="arun"
$password="1234*"
@"
{
        "proxy": "http://$($user):$($password)@org.proxy.com:80",
        "https-proxy": "http://$($user):$($password)@org.proxy.com:80"
}
"@

Or you could use the format operator

$user="arun"
$password="1234*"
@"
{{
        "proxy": "http://{0}:{1}@org.proxy.com:80",
        "https-proxy": "http://{0}:{1}@org.proxy.com:80"
}}
"@ -f $user, $password

Just make sure you escape curly braces when using the format operator.

You could also escape the colon with a backtick

$jsonStr = @"
{
        "proxy": "http://$user`:[email protected]:80",
        "https-proxy": "http://$user`:[email protected]:80"
}
"@

edited Jan 2, 2019 at 16:48

answered Feb 3, 2015 at 5:13

Matt

47k9 gold badges90 silver badges125 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

string variable not substituted in multi-line string

1 Answer 1

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related