Powershell Echo Statement + Variable In One line

Question

I'm running a script, and I want it to print a "statement + variable + statement" at the end [when successful]. I've tried a few thing but it always returns as 3 separate lines, instead of one. The Echo "" before and after is just to make it easier to read when printed by spacing it out, I've tried it with and without and I get the same result.

$filename = "foo.csv"

    echo ""
    echo "The file" $filename "has been processed."
    echo ""

I get this:

The file
foo.csv
has been processed.

Possible duplicate of How to concatenate strings and variables in PowerShell? — BioGenX
– BioGenX, Commented Mar 17, 2017 at 15:52

henrycarteruk · Accepted Answer · 2017-03-17 16:03:37Z

42

If you use double quotes you can reference the variable directly in the string as they allow variable expansion, while single quotes do not allow this.

$filename = "foo.csv"
Write-Output "The file $filename has been processed."

-> The file foo.csv has been processed.

Also, echo is actually just an alias for Write-Output, so I've used the full name.

answered Mar 17, 2017 at 16:03

henrycarteruk

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4 Comments

physlexic Over a year ago

So simple. I got thrown off because it gets grayed out. I thought it was being converted to plain text. Thanks.

henrycarteruk Over a year ago

I'd certainly recommend reading up on double vs single quotes, such a small thing can have a very big effect :)

Gorodeckij Dimitrij Over a year ago

this not work properly with $args[0] if they are few - its print it all args :/

henrycarteruk Over a year ago

@GorodeckijDimitrij yes, arrays and strings need to be handled differently.

4b0 · Accepted Answer · 2019-12-03 11:09:11Z

14

echo ("The file "+ $filename +" has been processed.")

edited Dec 3, 2019 at 11:09

4b0

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answered Dec 3, 2019 at 11:07

tom

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1 Comment

Ankit Kante Over a year ago

Please add a description to ur answer

BioGenX · Accepted Answer · 2017-03-17 15:55:25Z

7

In powershell, you can use Write-Host as follows:

$filename = "foo.csv"
Write-Host 'The file' $filename 'has been processed.'

-> The file foo.csv has been processed.

answered Mar 17, 2017 at 15:55

BioGenX

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1 Comment

Dr Rob Lang Over a year ago

When writing out an object then Write-Host will give you the object's type, not its contents. Better to useWrite-Output

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