Powershell - reading, creating and manipulating a csv file

Question

I am trying to write a script that would create and maintain a csv spreadsheet. It would be ran on a schedule and if if didn't find any .csv file it should create it from scratch.

This is the way I go about it:

# === Paths
$ipsFileLocation = "C:\stuff\ips.txt"
$dbLocation = "C:\stuff\db.csv"

# === Actual script starts here

Write-Host "Importing the CSV File..."
Try 
{ 
    $db = Import-Csv $dbLocation
} 
Catch [System.IO.FileNotFoundException] 
{ 
    Write-Host "The database file is not found! Creating a new one..."
    $db = @()
}

$ipList = get-content $ipsFileLocation

Update-Ips $db $ipList

Export-Csv -InputObject $db -Path $dbLocation -NoTypeInformation

# === Helper methods
function Update-Ips($db, $ips) 
{
    foreach ($ip in $ips)
    {   
        $foundRow = $db | Where { $_.IP -eq $ip } #check if the particular ip exists in database
        if ($foundRow -eq $null)
        {
            #no ip found in database -> Create new entry
            $newRecord = "" | select IP,Host,Responds,LastLogon
            $newRecord.IP = $ip
            $db += $newRecord
        }
    }
}

The ips.txt file is just a text file containing IP addresses, each in a new line.

The script runs, however the export-csv method leaves me with a .csv file that looks like this:

"Count","Length","LongLength","Rank","SyncRoot","IsReadOnly","IsFixedSize","IsSynchronized" "0","0","0","1","System.Object[]","False","True","False"

Not quite what I imagined.

Also, right at the end I would expect the $db object to looks like this: desired result

Instead, I get something that doesn't look like a table at all:

result that i get

Could someone please tell me that am I doing wrong and what is the difference between my desired result and the result that I actually get from running my script?

Is that the entire script? i see a reference to $csv which would be empty if not populated somewhere else — Matt
– Matt, Commented Jul 19, 2014 at 14:09
Good catch Matt, thanks. Fixed now, but the problem persists and my question stands. — Rafał Saltarski
– Rafał Saltarski, Commented Jul 19, 2014 at 14:48
As far as what the output of $db looks like, well that's just list view verse table view. It doesn't affect anything other than how you see it in the console. — Benjamin Hubbard
– Benjamin Hubbard, Commented Jul 19, 2014 at 15:56

Frode F. · Accepted Answer · 2014-07-19 15:49:12Z

3

You're trying to export the array itself (-InputObject $db) to CSV. That's why you get the properties of the array (count etc.).

Replace

Export-Csv -InputObject $db -Path $dbLocation -NoTypeInformation

with

$db | Export-Csv -Path $dbLocation -NoTypeInformation

answered Jul 19, 2014 at 15:49

Frode F.

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