I am creating a script converting a csv file in an another format.
To do so, i need my numbers to have a fixed format to respect column size : 00000000000000000,00 (20 characters, 2 digits after comma)
I have tried to format the number with -f and the method $value.toString("#################.##") without success
Here is an example Input :
4000000
45817,43
400000
570425,02
15864155,69
1068635,69
128586256,9
8901900,04
29393,88
126858346,88
1190011,46
2358411,95
139594,82
13929,74
11516,85
55742,78
96722,57
21408,86
717,01
54930,49
391,13
2118,64
Any hints are welcome :)
Thank you !
tl;dr:
Use 0 instead of # in the format string:
PS> $value = 128586256.9; $value.ToString('00000000000000000000.00')
00000000000128586256.90
Note:
Alternatively, you could construct the format string as an expression:
$value.ToString('0' * 20 + '.00')
The resulting string reflects the current culture with respect to the decimal mark; e.g., with fr-FR (French) in effect, , rather than . would be used; you can pass a specific [cultureinfo] object as the second argument to control what culture is used for formatting; see the docs.
As in your question, I'm assuming that $value already contains a number, which implies that you've already converted the CSV column values - which are invariably strings - to numbers.
To convert a string culture-sensitively to a number, use [double]::Parse('1,2'), for instance (this method too has an overload that allows specifying what culture to use).
Caveat: By contrast, a PowerShell cast (e.g. [double] '1.2') is by design always culture-invariant and only recognizes . as the decimal mark, irrespective of the culture currently in effect.
zerocukor287 has provided the crucial pointer:
To unconditionally represent a digit in a formatted string and default to 0 in the absence of an available digit, use 0, the zero placeholder in a .NET custom numeric format string
By contrast, #, the digit placeholder, represents only digits actually present in the input number.
To illustrate the difference:
PS> (9.1).ToString('.##')
9.1 # only 1 decimal place available, nothing is output for the missing 2nd
PS> (9.1).ToString('.00')
9.10 # only 1 decimal place available, 0 is output for the missing 2nd
Since your input uses commas as decimal point, you can split on the comma and format the whole number and the decimal part separately.
Something like this:
$csv = #'
Item;Price
Item1;4000000
Item2;45817,43
Item3;400000
Item4;570425,02
Item5;15864155,69
Item6;1068635,69
Item7;128586256,9
Item8;8901900,04
Item9;29393,88
Item10;126858346,88
Item11;1190011,46
Item12;2358411,95
Item13;139594,82
Item14;13929,74
Item15;11516,85
Item16;55742,78
Item17;96722,57
Item18;21408,86
Item19;717,01
Item20;54930,49
Item21;391,13
Item22;2118,64
'# | ConvertFrom-Csv -Delimiter ';'
foreach ($item in $csv) {
$num,$dec = $item.Price -split ','
$item.Price = '{0:D20},{1:D2}' -f [int64]$num, [int]$dec
}
# show on screen
$csv
# output to (new) csv file
$csv | Export-Csv -Path 'D:\Test\formatted.csv' -Delimiter ';'
Output in screen:
Item Price
---- -----
Item1 00000000000004000000,00
Item2 00000000000000045817,43
Item3 00000000000000400000,00
Item4 00000000000000570425,02
Item5 00000000000015864155,69
Item6 00000000000001068635,69
Item7 00000000000128586256,09
Item8 00000000000008901900,04
Item9 00000000000000029393,88
Item10 00000000000126858346,88
Item11 00000000000001190011,46
Item12 00000000000002358411,95
Item13 00000000000000139594,82
Item14 00000000000000013929,74
Item15 00000000000000011516,85
Item16 00000000000000055742,78
Item17 00000000000000096722,57
Item18 00000000000000021408,86
Item19 00000000000000000717,01
Item20 00000000000000054930,49
Item21 00000000000000000391,13
Item22 00000000000000002118,64
I do things like this all the time, usually for generating computernames. That custom numeric format string reference will come in handy. If you want a literal period, you have to backslash it.
1..5 | % tostring 00000000000000000000.00
00000000000000000001.00
00000000000000000002.00
00000000000000000003.00
00000000000000000004.00
00000000000000000005.00
Adding commas to long numbers:
psdrive c | % free | % tostring '0,0' # or '#,#'
18,272,501,760
"Per mille" character ‰ :
.00354 | % tostring '#0.##‰'
3.54‰
I'm using PowerShell and running a tool to extract Lenovo hardware RAID controller info to identify the controller number for use later on in another command line (this is part of a SCCM Server Build Task Sequence). The tool outputs a lot of data and I'm trying to isolate just what I need from the output.
I've been able to isolate what I need, but I'm thinking there has to be a more efficient way so looking for optimizations. I'm still learning when it comes to working with strings.
The line output from the tool that I'm looking for looks like this:
0 0 0 252:0 17 DRIVE Onln N 557.861 GB dsbl N N dflt -
I'm trying to get the 3 characters to the left of the :0 (the 252 but on other models this could be 65 or some other 2 or 3 digit number)
My existing code is:
$ControllerInfo = cmd /c '<path>\storcli64.exe /c0 show'
$forEach ($line in $ControllerInfo) {
if ($line -like '*:0 *') {
$ControllerNum = $line.split(':')[0] # Get everything left of :
$ControllerNum = $ControllerNum.Substring($ControllerNum.Length -3) # Get last 3 chars of string
$ControllerNum = $ControllerNum.Replace(' ', '') # Remove blanks
Write-Host $ControllerNum
break #stop looping through output
}
}
The above works but I'm wondering if there's a way to combine the three lines that start with $ControllerNum = so I can have just have a single $ControllerNum = (commands) line to set the variable instead of doing it in 3 lines. Basically want to combine the Split, Substring and Replace commands into a single line.
Thanks!
Here's another option:
$ControllerNum = ([regex]'(\d{2,3}):0').Match($line).Groups[1].Value
Used on your sample 0 0 0 252:0 17 DRIVE Onln N 557.861 GB dsbl N N dflt -
the result in $ControllerNum wil be 252
If you want just the last digits before the first :, without any whitespace, you can do that with one or two regex expressions:
$line -replace '^.*\b(\d+):.*$','$1'
Regex explanation:
^ # start of string
.* # any number of any characters
\b # word boundary
( # start capture group
\d+ # 1 or more strings
) # end capture group
: # a literal colon (:)
.* # any number of any characters
$ # end of string
replacement:
$1 # Value captured in the capture group above
So basically what I'm trying to achieve is to get a MAC address from a text file and increment the value by one.
Been bashing my head against the Google/StackOverflow wall for a couple of hours, think there's a concept I'm just not getting.
PowerShell:
$Last_MAC_Address = (Get-Content -LiteralPath "\\UNC\Path\Last MAC Address.txt")
Write-Host ($Last_MAC_Address)
# Output: 00155DE10B73
$Next_MAC_Address = (($Last_MAC_Address | Format-Hex) + 1)
This is a 3 step process, and although PetSerAl answered it in the comments as a one liner, I'll break it down slightly for posterity (and use a different class).
The first step is to get the Hex number as a decimal (mathematical base 10, not type).
The Second step is the incrementation of the decimal.
And the final step is converting it back to hexadecimal.
broken down and not a one liner this will accomplish the task at hand:
$asDecimal = [System.Convert]::ToInt64("00155DE10B73", 16)
$asDecimal++
$asHex = [System.Convert]::ToString($asDecimal, 16)
Another option is to prefix the value with 0x and cast it to an int64:
$Next_MAC_Address = ([int64]"0x$Last_MAC_Address"+1).ToString('X12')
You could also use the format operator (-f) instead of the ToString() method:
$Next_MAC_Address = '{0:X12}' -f ([int64]"0x$Last_MAC_Address"+1)
There is, however, one thing that may be worth noting. MAC addresses aren't just random 6-byte numbers without any inner structure. They actually consist of two parts. The first 3 bytes form the Organizationally Unique Identifier (OUI), a vendor-specific prefix (00-15-5D is one of the OUIs belonging to Microsoft). Only the last 3 bytes are a random number, a unique identifier for each card from the vendor identified by the OUI.
Taking that into consideration you may want to split the MAC address accordingly, e.g. like this:
$oui, $nid = $Last_MAC_Address -split '(?<=^[0-9a-f]{6})(?=[0-9a-f]{6}$)'
or like this:
$oui = $Last_MAC_Address.Substring(0, 6)
$nid = $Last_MAC_Address.Substring(6, 6)
and increment only the NIC identifier, and only if it wouldn't overflow:
if ($nid -ne 'ffffff') {
$Next_MAC_Address = "{0}{1:X6}" -f $oui, ([int64]"0x$nid"+1)
} else {
Write-Error 'MAC address overflow.'
}
I have an .nc file I'm reading in matlab, and getting info out of the time variable.
the code looks like this
>> ncreadatt(model_list{3},'T','units')
ans =
'months since 1850-01-01'
what I want to do is get just the '1850' out of the answer.
Regular expression is a very powerful tool to parse and manipulate strings.
Matlab has regexp command:
line = 'months since 1850-01-01';
res = regexp( line, '\s(\d+)-', 'tokens', 'once');
year = str2double(res{1})
And the results is:
year =
1850
The regular expression used '\s(\d+)-' means:
\s - look for a single white space character (the space before 1850).
'(\d+)' - look for one or more digit ('\d+'), the parentheses means that all charcters matching here will be saved as a "token".
'-' - look for a single '-' after the digits.
You can play with it on ideone.
I am working with PowerShell to create a renaming script for a number of files in a directory.
Two questions here:
I have a string variable $strPrefix = "ACV-100-" and an integer counter $intInc = 000001 and I wish to increment the counter $intInc 1 -> 2 and then concatenate the two and store it in a variable $strCPrefix in the following format: ACV-100-000002.
I believe the $intInc will need to be cast in order to convert it once incrementing is complete but I am unsure how to do this.
Secondly, I have found that the script will display 000001 as 1, 000101 as 101 and so on... I need the full 6 digits to be displayed as this will form a file name. How do I keep or pad the numbers before I process the concatenation?
$intInc = 1
$strCprefix = $strprefix + "{0:000000}" -f $intInc # give ACV-100-000001
hope can help
This should do it:
$intInc = 1
...
$filename = $strPrefix + ('0' * (6 - $intInc.ToSTring().Length)) + ($intInc++).ToString()