CODEWITHSUNDEEP
xmlpowershellvariablesobjectupdates
AdilZ
AdilZ

Reputation: 1197

Powershell: saving xml changes in a new xml variable or object, keeping the original object unchanged

I have a xml variable that I get from Invoke-restmethod. Lets call it $object1

I want to make a new variable or object $object2 that is the same as $object1 but with a few changes to the values

Here is what I have tried:

$object2 = $object1

$object2.sweets.candy.Where({$_.lemondrop -gt 1}) | Foreach{$_.gumdrop = 20}

Now the problem with that is when I double check $object1 the value is changed there too... I dont want that

I just want the chage to remain in $object2

What am I doing wrong here?

Upvotes: 2

Views: 283

Answers (1)

mklement0
mklement0

Reputation: 438198

You need to clone $object1 in order to obtain an independent copy of it:

$object2 = $object1.Clone() # assumes that $object's type implements ICloneable

[System.Xml.XmlDocument] is a reference type (as opposed to a value type such as [int]), which means that:

$object2 = $object1

simply makes $object2 reference (point to) the very same object as $object1.

You can verify this as follows:

[object]::ReferenceEquals($object1, $object2)

Generally:

  • = with a value type instance creates a copy of the actual value, i.e., it implicitly clones.

  • = with a reference type instance creates a copy of the reference to the instance, i.e., it creates another reference to the very same object.

  • The same applies when you add elements to a (mutable) collection; e.g. (assume having created a list with $list = [System.Collections.Generic.List[object]] @()):

    • $var = 42; $list.Add($var) adds a copy of the value-type ([int]) instance to the list.
    • $var = [pscustomobject] @{ foo = 'bar' }; $list.Add($var) adds a reference to the reference-type ([pscustomobject]) instance to the list.

To determine if a given type is a value type or a reference type, use the type's Boolean .IsValueType property:

# Inspect a type directly.
PS> [System.Xml.XmlDocument].IsValueType
False    # -> a *reference* type

# Inspect a type via an instance thereof:
PS> $xml = [xml] '<xml/>'; $xml.GetType().IsValueType
False

Upvotes: 2

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