How to escape string variables when calling ExpandString in PowerShell

Question

I have a PowerShell script to take a JSON file as a parameter and convert it to object. In the JSON string (as a template) there may have some variables which should be replaced with the actual values during runtime. For example, $var_1, $var_x and $var_y in this JSON string:

{
    "key1": "My name is ${var_1}",
    "key2":
    {
        "key3": "Another variable ${var_x}",
        "key4": "${var_y}"
    }
}

The names of these variables are unknown to the script. I use ExpandString to do the job:

[string]$jsonText = Get-Content $inputFile;

if ([string]::IsNullOrWhiteSpace($jsonText)) {
    throw "No content in $inputFile"
}
$temp = $ExecutionContext.InvokeCommand.ExpandString($jsonText);
$jsonObj = ConvertFrom-Json $temp;

The problem is, when a variable contains double quotation mark, then calling ConvertFrom-Json will fail because the quotation mark is not escaped. For example, "My name is ${var_1}" may become "My name is a "new" name".

Is there a way to escape these string variables when calling ExpandString?

mklement0 · Accepted Answer · 2018-05-30 04:05:34Z

You can use the following technique, taking advantage of being able to embed entire statements in your template via $(...) the subexpression operator; therefore, it should also be noted that you should generally only use this technique with template strings that you trust, given that arbitrary commands could be executed.

$var_1 = '"new" name'

$template = @'
{ "key1": "My name is $((Get-Variable var_1).Value -replace '"', '\"')" }
'@

$ExecutionContext.InvokeCommand.ExpandString($template)

The above yields:

{ "key1": "My name is \"new\" name" }

which is valid JSON.

Note the use of Get-Variable to access the value of variable ${var_1} indirectly, so that its embedded " chars. can be safely escaped as \", as required by JSON.

Since the above is a little cumbersome for multiple references, you could define an auxiliary function:

$var_1 = '"new" name'

# Aux. function that takes a variable name and returns its value with
# " chars. \-escaped.
function esc ($varName) { (Get-Variable $varName).Value -replace '"', '\"' }    

# Now you can reference the variables using $(esc <var-name):
$template = @'
{ "key1": "My name is $(esc var_1)" }
'@

$ExecutionContext.InvokeCommand.ExpandString($template)

If you don't control the template string, more work is needed:

$var_1 = '"new" name'

# Aux. function that takes a variable name and returns its value with
# " chars. \-escaped.
function esc ($varName) { (Get-Variable $varName).Value -replace '"', '\"' }

# The original template with ${var}-style references.
$template = @'
{
  "key1": "My name is ${var_1}",
}
'@

# Modify the template to replace ${var} references with $(esc var).
$modifiedTemplate = $template -replace '\$\{(\w+)\}', '$$(esc $1)'

# Now expand the modified template.
$ExecutionContext.InvokeCommand.ExpandString($modifiedTemplate)

Note that the template transformation assumes: ^{Tip of the hat to briantist for his input.}

that there are no escaped variable references to be treated as literals in the input template (e.g., `${var}).
that all variable references have the form ${varName}, where varName is assumed to be composed of letters, digits, and underscores only (matching one or more \w instances).

A more robust solution that covers all edge cases may require using PowerShell's parsing API, which requires significantly more effort, however.

Chenry Lee · Accepted Answer · 2018-05-30 05:20:36Z

0

Is it allowed in that format? If the variable is quoted with "", it's a string, and won't add the marks to .json. e.g.:

$var_1 = "Json Sample"
$json_piece = @"
{
    "key1": "My name is $var_1",
    "key2":
    {
        "key3": "Another variable $var_2",
        "key4": "$var_3"
    }
}
"@

It outputs ($var_2 and $var_3 are not defined):

{
    "key1": "My name is Json Sample",
    "key2":
    {
        "key3": "Another variable ",
        "key4": ""
    }
}

If you add " to a json file like this:

{
    "key1": "My name is "Json Sample"",
    "key2":
    {
        "key3": "Another variable ",
        "key4": ""
    }
}

You will make an error: Excepted comma. It means if the variable contains "", you must escape them at first, if not, the .json file is unexpected.

P.S. Validation I validated it on FREEFORMATTER.COM.
It's my code:

{
  "Key": "value is "value"",
  "key2": "other value"
}

And I get 2 errors:

The JSON input is NOT valid according to RFC 4627 (JSON specification). Unexpected token v at position 20.

The JSON input is NOT valid in JavaScript, missing } after property list (At line #2), (At position #24)

edited May 30, 2018 at 5:20

answered May 30, 2018 at 3:42

Chenry Lee

3802 silver badges10 bronze badges

5 Comments

briantist Over a year ago

The point is that here it's like $var_1='"JsonSample"', so once the replacement happens, the resulting JSON does become invalid. They don't know what value the variable will actually contain at run-time.

Chenry Lee Over a year ago

If you use " in your .json file, you will get errors. " must be escaped before passing to the .json file. I just tried that.

Chenry Lee Over a year ago

I added the validation to my answer.

briantist Over a year ago

Perhaps I'm misunderstanding, but you seem to be restating the problem posed in the question.

Chenry Lee Over a year ago

Hmm, I found that, the point of this question is how to escape double quotation mark from the variable. @mklement0 made a good answer.

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