Powershell – How to you use an object’s property in a double-quoted string

powershellstring-interpolation

I have the following code:

$DatabaseSettings = @();
$NewDatabaseSetting = "" | select DatabaseName, DataFile, LogFile, LiveBackupPath;
$NewDatabaseSetting.DatabaseName = "LiveEmployees_PD";
$NewDatabaseSetting.DataFile = "LiveEmployees_PD_Data";
$NewDatabaseSetting.LogFile = "LiveEmployees_PD_Log";
$NewDatabaseSetting.LiveBackupPath = '\\LiveServer\LiveEmployeesBackups';
$DatabaseSettings += $NewDatabaseSetting;

When I try to use one of the properties in a string execute command:

& "$SQlBackupExePath\SQLBackupC.exe" -I $InstanceName -SQL `
  "RESTORE DATABASE $DatabaseSettings[0].DatabaseName FROM DISK = '$tempPath\$LatestFullBackupFile' WITH NORECOVERY, REPLACE, MOVE '$DataFileName' TO '$DataFilegroupFolder\$DataFileName.mdf', MOVE '$LogFileName' TO '$LogFilegroupFolder\$LogFileName.ldf'"

It tries to just use the value of $DatabaseSettings rather than the value of $DatabaseSettings[0].DatabaseName, which is not valid.
My workaround is to have it copied into a new variable.

How can I access the object's property directly in a double-quoted string?

Best Solution

When you enclose a variable name in a double-quoted string it will be replaced by that variable's value:

$foo = 2
"$foo"

becomes

"2"

If you don't want that you have to use single quotes:

$foo = 2
'$foo'

However, if you want to access properties, or use indexes on variables in a double-quoted string, you have to enclose that subexpression in $():

$foo = 1,2,3
"$foo[1]"     # yields "1 2 3[1]"
"$($foo[1])"  # yields "2"

$bar = "abc"
"$bar.Length"    # yields "abc.Length"
"$($bar.Length)" # yields "3"

PowerShell only expands variables in those cases, nothing more. To force evaluation of more complex expressions, including indexes, properties or even complete calculations, you have to enclose those in the subexpression operator $( ) which causes the expression inside to be evaluated and embedded in the string.