The suspected preexisting duplicate question is indeed a duplicate:
Given that + with an array as the LHS concatenates arrays, you must nest the RHS with the unary form of , (the array-construction operator) if it is an array that should be added as a single element:
# Sample input
$items = [pscustomobject] @{ Name = 'n1'; Value = 'v1'},
[pscustomobject] @{ Name = 'n2'; Value = 'v2'}
$ret = @() # create an empty *array*
foreach ($item in $items) {
$subret = $item.Name, $item.Value # use of "," implicitly creates an array
$ret += , $subret # unary "," creates a 1-item array
}
# Show result
$ret.Count; '---'; $ret[0]; '---'; $ret[1]
This yields:
2
---
n1
v1
---
n2
v2
The reason the use of [System.Collections.ArrayList] with its .Add() method worked too - a method that is generally preferable when building large arrays - is that .Add() only accepts a single object as the item to add, irrespective of whether that object is a scalar or an array:
# Sample input
$items = [pscustomobject] @{ Name = 'n1'; Value = 'v1'},
[pscustomobject] @{ Name = 'n2'; Value = 'v2'}
$ret = New-Object System.Collections.ArrayList # create an *array list*
foreach ($item in $items) {
$subret = $item.Name, $item.Value
# .Add() appends whatever object you pass it - even an array - as a *single* element.
# Note the need for $null = to suppress output of .Add()'s return value.
$null = $ret.Add($subret)
}
# Produce sample output
$ret.Count; '---'; $ret[0]; '---'; $ret[1]
The output is the same as above.
$ret += , $subretsolves the problem.