How to address "Select-String -pattern" return object via pipeline without using foreach loop?

Question

Let's say I have a file with a following content:

1,first_string,somevalue
2,second_string,someothervalue
n,n_nd_string,somemorevalue

I need to Get-Content of this file, get the last string and get the number before the "," symbol (n in this case). (I'll just increment it and append n+1 string to this file, but it does not matter right now). I want all this stuff be done with pipeline cascade

I have come to this solution so far:

[int]$number = Get-Content .\1.mpcpl | Select-Object -Last 1 | Select-String -Pattern '^(\d)' | ForEach-Object {$_.matches.value}

It actually works, but I wonder if there are any ways of addressing Select-String -Pattern '^(\d)' return object without using the foreach loop? Beacause I know that the return collection in my case will only consist of a 1 element (I'm selecting a single last string of a file and I get only one match)

You might need a -match operator to get the first match only. Or -replace: -replace '^(\d).*', '$1' — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Aug 28, 2019 at 19:31
hm, but how exactly do I apply -match or -replace operators in a pipeline? for example, after I get the last string of a file. — Ivan
– Ivan, Commented Aug 28, 2019 at 19:42

Wiktor Stribiżew · Accepted Answer · 2019-08-28 20:41:10Z

1

You may use

$num = [int]((Get-Content .\1.mpcpl | Select-Object -Last 1) -replace '^(\d).*', '$1')

Notes

(Get-Content .\1.mpcpl | Select-Object -Last 1) - reads the file and gets the last line
(... -replace '^(\d).*', '$1') - gets the first digit from the returned line (NOTE: if the line does not start with a digit, it will fail as the output will be the whole line)
[int] casts the string value to an int.

Another way can be getting a match and then retrieving it from the default $matches variable:

[IO.File]::ReadAllText($filepath) -match '(?m)^(\d).*\z' | Out-Null
[int]$matches[1]

The (?m)^(\d).*\z pattern gets the digit at the start of the last line into Group 1, hence $matches[1].

edited Aug 28, 2019 at 20:41

answered Aug 28, 2019 at 20:12

Wiktor Stribiżew

631k41 gold badges502 silver badges633 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Ivan Over a year ago

I like the parentheses solution. Thanks

js2010 · Accepted Answer · 2019-08-28 20:34:22Z

0

It looks like a csv to me...

import-csv 1.mpcpl -header field1,field2,field3 | select -last 1 | select -expand field1

output:

answered Aug 28, 2019 at 20:34

js2010

28.4k6 gold badges82 silver badges88 bronze badges

1 Comment

Ivan Over a year ago

a good one, but this particular file may look like a csv file but it not necessarily should be it, so I decided to get more general solution for different data structures with one common feature -- line number at the beginning of each line.

Collectives™ on Stack Overflow

How to address "Select-String -pattern" return object via pipeline without using foreach loop?

2 Answers 2

1 Comment

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related