CODEWITHSUNDEEP
powershellpowershell-2.0powershell-3.0
user1863886
user1863886

Reputation:

Alternating string

I am trying to use regular expressions to match against a string that starts with 7 numbers, then has a "K" inbetween it, and then 3 numbers again. For example: 1234567K890.

I currently have $_a -match '^\d{7}K\d{3}'. However, this does not work for my purposes. Does anyone have a solution?

Upvotes: 1

Views: 109

Answers (2)

Mitul
Mitul

Reputation: 9854

PS C:\> "1234567K890" -match "\d{7}(k)\d{3}"

This \d{7} matches 7 digits then (k) matches letter k and \d{3} matches last three characters.

Upvotes: 2

Anthony Neace
Anthony Neace

Reputation: 26023

Tested this, works for your example and some others:

$string = "1234567K890"
$string -match '^[0-9]{7}(k)[0-9]{3}$'"

It matches against exactly 7 numbers, then against K (casing does not matter), then against exactly 3 numbers. The characters at the beginning and the end of the string restrict against whitespace at the beginning and end of the string -- if you want whitespace to be allowed, you can just remove them.

Here's a powershell regex reference, which may help in the future.

Upvotes: 2

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