I want to grab the first char of a var string and the first char of the following caracter
Example:
$var1 = "Jean-Martin"
I want a way to grab the first letter "J" then I want to take the first char following the "-" (dash) which is "M".
Something like this?
$initial1 = $var1[0]
$initial2 = $var1.Split('-')[1][0]
Strings in Powershell use the System.String class from the .Net framework. As such, they are indexable to retrieve individual characters and have many methods available such as the Split method used above.
See the documentation here.
$var1 = "Jean-Martin"
To get the first character:
$var1[0]
To get the first character after the dash:
$characterToSeek = '-'
$var1[$var1.IndexOf($characterToSeek)+1]
Another option using regex:
PS> $var1 -replace '^(.)[^-]+-(.).+$','$1$2'
JM
Related
I have a text file sample.txt containing
computer
computer.pc = 1
pc
i want only number 1, where i want to assign that value to a variable
$number = Get -content "sample.txt"
You can extract the number by using the Regex Match method.
Example code to do this:
$number = ([regex]::Match((Get-content "sample.txt"), "\d+")).Value
The pattern \d+ means to match one or more decimal digits and using the Match method will return the first match found.
See Quantifiers in Regular Expressions for additional information regarding the quantifiers available.
I have the following Powershell variable
$var = "AB-0045"
I would like to increase the number in the string to become "AB-0046".
I can do:
$newNumber = [int]$var.Substring($var.length -4,4) + 1
Which will give me the desired number 46, but then I have to append that 46 as a string to a new string "AB-00".
Is there a better way to do that?
Now that you have the integer, you'll have to convert back to string formatted in the way you'd like and concatenate.
I'd recommend adding to "AB-" rather than "AB-00" in case your number goes over 100.
To pad leading zeros, you can use the -f operator.
e.g. "{0:d4}" -f 45
You'll still need to get the integer first (45 in the example) from your original string.
I tested with regex class Replace() method and string class Split() method with string formatter. Split() seems faster provided your string is always in the same format. The Replace() method does not care what happens before the last 4 numbers:
# Replace Method
[regex]::Replace($var,'\d{4}$',{([int]$args[0].Value+1).ToString('0000')})
# Split method
$a,[int]$b = $var.split('-'); "{0}-{1:0000}" -f $a,++$b
I need to create a String from double the use String.Trim() to remove the full stop, but it doesn't remove it. I think there is also a way to do this numerically but I'd like to do it with the string. Is there a reason it won't remove it? The output from the code is 5.5
$MyDouble = 5.5
[String]$MyDouble2 = $MyDouble
$MyDouble2.Trim(".")
$MyDouble2
String.Trim() only trims from the beginning and end of strings, so it has no effect in your command, because the . only occurs inside your input string.
If you truly want to remove just the . and keep the post-decimal-point digits, use the -replace operator:
$MyDouble2 -replace '\.' # -> '55'
Note:
* -replace takes a regex (regular expression) as the search operand, hence the need to escape regex metacharacter . as \.
* The above is short for $MyDouble2 -replace '\.', ''. Since the replacement string is the empty string in this case, it can be omitted.
If you only want to extract the integer portion, use either 4c74356b41's .Split()-based answer, or adapt the regex passed to -replace to match everything from the . through the end of the string.
$MyDouble2 -replace '\..*' # -> '5'
#Matt mentions the following alternatives:
For removing the . only: Using String.Replace() to perform literal substring replacement (note how . therefore does not need \-escaping, as it did with -replace, and that specifying the replacement string is mandatory):
$MyDouble2.Replace('.', '') # -> '55'
For removing the fractional part of the number (extracting the integer part only), using a numerical operation directly on $MyDouble (as opposed to via the string representation stored in $MyDouble2), via Math.Floor():
[math]::Floor($MyDouble) # -> 5 (still a [double])
Looking at some documentation for .Trim([char[]]) you will see that
Removes all leading and trailing occurrences of a set of characters specified in an array from the current String object.
That does not cover the middle of strings, so using the .Replace() method would accomplish that.
I think there is also a way to do this numerically but I'd like to do it with the string.
Just wanted to mention that converting numbers to strings to then drop decimals via string manipulation is a poor approach. Assuming your example is what you are actually trying to do, I suggest using a static method from the [math] class instead.
$MyDouble = 5.5
[math]::Floor($MyDouble)
$MyDouble = 5.5
[String]$MyDouble2 = $MyDouble
$MyDouble2.Replace(".", "")
Well, why would it trim not the last (or first) character? It wouldn't, what you need (probably) is:
$MyDouble = 5.5
[String]$MyDouble2 = $MyDouble
$MyDouble2.Split(".")[0]
$MyDouble = 5.5
[String]$MyDouble2 = $MyDouble
$res=$MyDouble2 -split "\."
$res[0..($res.Count-1)] -join ""
I want to strip a string of all new lines and commas (and place it into an array), so I created this:
let results = text.componentsSeparatedByCharactersInSet(NSCharacterSet(charactersInString: ",\n"))
However, the newlines are still existing in my array (the commas are being removed). What's the correct way of adding newline to the NSCharacterSet? Or, how to add comma to NSCharacterSet.newLineCharacterSet.
Thanks.
Here is janky solution, but still looking for a more elegant one.
var results = text.componentsSeparatedByCharactersInSet(NSCharacterSet(charactersInString: ","))
text = results.joinWithSeparator(" ")
results = text.componentsSeparatedByCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet())
(one-line) SOLUTION:
var results = text.componentsSeparatedByCharactersInSet(NSCharacterSet(charactersInString: " ,\u{000A}\u{000B}\u{000C}\u{000D}\u{0085}"))
Explanation is below.
You can unite two NSCharacterSet by first using an NSMutableCharacterSet, for example:
let charset = NSMutableCharacterSet(charactersInString: ",")
charset.formUnionWithCharacterSet(NSCharacterSet.newlineCharacterSet())
let results = text.componentsSeparatedByCharactersInSet(charset)
So MartinR brought to my attention that there are more line feeds than just "\n".
I looked at the values used in NSCharacterSet.newlineCharacterSet and added them all, giving me:
var results = text.componentsSeparatedByCharactersInSet(NSCharacterSet(charactersInString: " ,\u{000A}\u{000B}\u{000C}\u{000D}\u{0085}"))
This got rid of all the whitespace, commas, and new lines. Interestingly - when I used all the newline values separately to see if I could figure out which newline was being used in my case, none of them worked. But when used all together, it strips my new lines.
This may be a very simple task for many but I could not find anything appropriate for me.
I have a file name: filenm_A006.2011.269.10.47.G25_2010
I want to separate all its parts (separated by . and _) to use them separately. How can I do it with simple matlab commands?
Kind Regards,
Mushi
I recommend regexp:
fname = 'filenm_A006.2011.269.10.47.G25_2010';
parts = regexp(fname, '[^_.]+', 'match');
parts =
'filenm' 'A006' '2011' '269' '10' '47' 'G25' '2010'
You can now refer to parts{1} through parts{8} for the pieces. Explanation: the regexp pattern [^_.] means all characters not equal to _ or ., and the + means you want groups of at least 1 character. Then 'match' asks the regexp function to return a cell array of the strings of all the matches of that pattern. There are other regexp modes; for example, the indices of each piece of the file.
Use the command
strsplit.
cellArrayOfParts = strsplit(fileName,{'.' '_'});
You can use strsplit to split it:
strsplit('filenm_A006.2011.269.10.47.G25_2010',{'_','.'})
ans =
'filenm' 'A006' '2011' '269' '10' '47' 'G25' '2010'
Another option is to use regexp, like Peter suggested.