I am trying to write a Powershell code to identify a string with a specific character from a filename from multiple files.
An example of a filename
20190902091031_202401192_50760_54206_6401.pdf
$Variable = $Filename.Substring(15,9)
Results:
202401192 (this is what I am after)
However in some instances the filename will be like below
20190902091031_20240119_50760_54206_6401.pdf
$Variable = $Filename.Substring(15,9)
Results:
20240119_ (this is NOT what I am after)
I am trying to find a code to identify the 9th character,
IF the 9th character = "_"
THEN Set
$Variable = $Filename.Substring(15,8)
Results:
20240119
All credit to TheMadTechnician who beat me to the punch with this answer.
To expand on the technique a bit, use the split method or operator to split a string every time a certain character shows up. Your data is separated by the underscore character, so is a perfect example of using this technique. By using either of the following:
$FileName.Split('_')
$FileName -split '_'
You can turn your long string into an array of shorter strings, each containing one of the parts of your original string. Since you want the 2nd one, you use the array descriptor [1] (0 is 1st) and you're done.
Good luck
Related
Context
I am designing a code that runs a bunch of calculations, and outputs figures. At the end of the code, I want to save everything in a nice way, so my take on this is to go to a user specified Output directory, create a new folder and then run the save process.
Question(s)
My question is twofold:
I want my folder name to be unique. I was thinking about getting the current date and time and creating a unique name from this and the input filename. This works but it generates folder names that are a bit cryptic. Is there some good practice / convention I have not heard of to do that?
When I get the datetime string (tn = datestr(now);), it looks like that:
tn =
'07-Jul-2022 09:28:54'
To convert it to a nice filename, i replace the '-',' ' and ':' characters by underscores and append it to a shorter version of the input filename chosen by the user. I do that using strrep:
tn = strrep(tn,'-','_');
tn = strrep(tn,' ','_');
tn = strrep(tn,':','_');
This is fine but it bugs me to have to use 3 lines of code to do so. Is there a nice one liner to do that? More generally, is there a way to look for every non letter or number character in a string and replace it with a given character? I bet that's what regexp is there for but frankly I can't quite get a hold on how regexps work.
Your point (1) is opinion based so you might get a variety of answers, but I think a common convention is to at least start the name with a reverse-order date string so that sorting alphabetically is the same as sorting chronologically (i.e. yymmddHHMMSS).
To answer your main question directly, you can use the built-in makeValidName utility which is designed for making valid variable names, but works for making similarly "plain" file names.
str = '07-Jul-2022 09:28:54';
str = matlab.lang.makeValidName(str)
% str = 'x07_Jul_202209_28_54'
Because a valid variable can't start with a number, it prefixes an x - you could avoid this by manually prefixing something more descriptive first.
This option is a bit more simple than working out the regex, although that would be another option which isn't too nasty here using regexprep and replacing non-alphanumeric chars with an underscore:
str = regexprep( str, '\W', '_' ); % \W (capital W) matches all non-alphanumeric chars
% str = '07_Jul_2022_09_28_54'
To answer indirectly with a different approach, a nice trick with datestr which gets around this issue and addresses point (1) in one hit is to use the following syntax:
str = datestr( now(), 30 );
% str = '20220707T094214'
The 30 input (from the docs) gives you an ISO standardised string to the nearest second in reverse-order:
'yyyymmddTHHMMSS' (ISO 8601)
(note the T in the middle isn't a placeholder for some time measurement, it remains a literal letter T to split the date and time parts).
I normally use your folder naming approach with a meaningful prefix, replacing ':' by something else:
folder_name = ['results_' strrep(datestr(now), ':', '.')];
As for your second question, you can use isstrprop:
folder_name(~isstrprop(folder_name, 'alphanum')) = '_';
Or if you want more control on the allowed characters you can use good old ismember:
folder_name(~ismember(folder_name, ['0':'9' 'a':'z' 'A':'Z'])) = '_';
I have a line saved as $variable1, for example:
file"yourtxthere/i/master.ext" autostart:true, width
How would I go about writing the correct syntax using Regex in Powershell to grab everything within the quotes. Are regular expressions the best way to do this in Powershell?
Though regular expressions are powerful, this is a simple case. Thus a simple solution is based on string object's split() method.
$variable1 = 'file"yourtxthere/i/master.ext" autostart:true, width'
# Splitting requires at least one "
if ($variable1.IndexOf('"') -gt 0) {
$variable1.Split('"')[1]
}
Splitting the string will result an array of three elements:
file
yourtxthere/i/master.ext
autostart:true, width
As Powershell's arrays begin with index zero, the desired data is at index location 1. In case the string doesn't contain a double quote, the split won't work. This is checked with the statement
if ($variable1.IndexOf('"') -gt 0)
Without one, splitting a quoteless string would return just a one-celled array and trying to access index 1 would result in an error message:
$variable1.Split('!')[1]
Index was outside the bounds of the array.
I'm not 100% sure I understand the question but assuming you had that stored in $variable1 as a string like so:
$variable1 = 'file"yourtxthere/i/master.ext" autostart:true, width'
Then you could extract just the quoted text by splitting the string using the double quotes as a delimiter like this:
$variable1.Split('"')[1]
That splits the string into an array with three parts. The first part is everything before the first double quotes and the second everything in between the first and second quote marks, and the third everything after the second quotation marks. The [1] tells it to output the second part.
I think this picture might illustrate it a little better.
There are other ways to split that up and in fact the variable isn't necessary at all if you do something like this:
('file"yourtxthere/i/master.ext" autostart:true, width').split('"')[1]
That will work fine so long as there are never any extra double quotes before the first occurrence.
Problem
I am working on a script that has a user provide a specific IP address and I want to mask this IP in some fashion so that it isn't stored in the logs. My problem is, that I can easily do this when I know what the first three values of the IP typically are; however, I want to avoid storing/hard coding those values into the code to if at all possible. I also want to be able to replace the values even if the first three are unknown to me.
Examples:
10.11.12.50 would display as XX.XX.XX.50
10.12.11.23 would also display as XX.XX.XX.23
I have looked up partial string replacements, but none of the questions or problems that I found came close to doing this. I have tried doing things like:
# This ended up replacing all of the numbers
$tempString = $str -replace '[0-9]', 'X'
I know that I am partway there, but I aiming to only replace only the first 3 sets of digits so, basically every digit that is before a '.', but I haven't been able to achieve this.
Question
Is what I'm trying to do possible to achieve with PowerShell? Is there a best practice way of achieving this?
Here's an example of how you can accomplish this:
Get-Content 'File.txt' |
ForEach-Object { $_ = $_ -replace '\d{1,3}\.\d{1,3}\.\d{1,3}','xx.xx.xx' }
This example matches a digit 1-3 times, a literal period, and continues that pattern so it'll capture anything from 0-999.0-999.0-999 and replace with xx.xx.xx
TheIncorrigible1's helpful answer is an exact way of solving the problem (replacement only happens if 3 consecutive .-separated groups of 1-3 digits are matched.)
A looser, but shorter solution that replaces everything but the last .-prefixed digit group:
PS> '10.11.12.50' -replace '.+(?=\.\d+$)', 'XX.XX.XX'
XX.XX.XX.50
(?=\.\d+$) is a (positive) lookahead assertion ((?=...)) that matches the enclosed subexpression (a literal . followed by 1 or more digits (\d) at the end of the string ($)), but doesn't capture it as part of the overall match.
The net effect is that only what .+ captured - everything before the lookahead assertion's match - is replaced with 'XX.XX.XX'.
Applied to the above example input string, 10.11.12.50:
(?=\.\d+$) matches the .-prefixed digit group at the end, .50.
.+ matches everything before .50, which is 10.11.12.
Since the (?=...) part isn't captured, it is therefore not included in what is replaced, so it is only substring 10.11.12 that is replaced, namely with XX.XX.XX, yielding XX.XX.XX.50 as a result.
i am facing issue while converting unicode data into national characters.
When i convert the Unicode data into national using national-of function, some junk character like # is appended after the string.
E.g
Ws-unicode pic X(200)
Ws-national pic N(600)
--let the value in Ws-Unicode is これらの変更は. getting from java end.
move function national-of ( Ws-unicode ,1208 ) to Ws-national.
--after converting value is like これらの変更は #.
i do not want the extra # character added after conversion.
please help me to find out the possible solution, i have tried to replace N'#' with space using inspect clause.
it worked well but failed in some specific scenario like if we have # in input from user end. in that case genuine # also converted to space.
Below is a snippet of code I used to convert EBCDIC to UTF. Before I was capturing string lengths, I was also getting # symbols:
STRING
FUNCTION DISPLAY-OF (
FUNCTION NATIONAL-OF (
WS-EBCDIC-STRING(1:WS-XML-EBCDIC-LENGTH)
WS-EBCDIC-CCSID
)
WS-UTF8-CCSID
)
DELIMITED BY SIZE
INTO WS-UTF8-STRING
WITH POINTER WS-XML-UTF8-LENGTH
END-STRING
SUBTRACT 1 FROM WS-XML-UTF8-LENGTH
What this code does is string the UTF8 representation of the EBCIDIC string into another variable. The WITH POINTER clause will capture the new length of the string + 1 (+ 1 because the pointer is positioned to the next position after the string ended).
Using this method, you should be able to know exactly how long second string is and use that string with the exact length.
That should remove the unwanted #s.
EDIT:
One thing I forgot to mention, in my case, the # signs were actually EBCDIC low values when viewing the actual hex on the mainframe
Use inspect with reverse and stop after first occurence of #
So I'm reading multiple text files in Matlab that have, in their first columns, a column of "times". These times are either in the format 'MM:SS.milliseconds' (sorry if that's not the proper way to express it) where for example the string '29:59.9' would be (29*60)+(59)+(.9) = 1799.9 seconds, or in the format of straight seconds.milliseconds, where '29.9' would mean 29.9 seconds. The format is the same for a single file, but varies across different files. Since I would like the times to be in the second format, I would like to check if the format of the strings match the first format. If it doesn't match, then convert it, otherwise, continue. The code below is my code to convert, so my question is how do I approach checking the format of the string? In otherwords, I need some condition for an if statement to check if the format is wrong.
%% Modify the textdata to convert time to seconds
timearray = textdata(2:end, 1);
if (timearray(1, 1) %{has format 'MM.SS.millisecond}%)
datev = datevec(timearray);
newtime = (datev(:, 5)*60) + (datev(:, 6));
elseif(timearray(1, 1) %{has format 'SS.millisecond}%)
newtime = timearray;
You can use regular expressions to help you out. Regular expressions are methods of specifying how to search for particular patterns in strings. As such, you want to find if a string follows the formats of either:
xx:xx.x
or:
xx.x
The regular expression syntax for each of these is defined as the following:
^[0-9]+:[0-9]+\.[0-9]+
^[0-9]+\.[0-9]+
Let's step through how each of these work.
For the first one, the ^[0-9]+ means that the string should start with any number (^[0-9]) and the + means that there should be at least one number. As such, 1, 2, ... 10, ... 20, ... etc. is valid syntax for this beginning. After the number should be separated by a :, followed by another sequence of numbers of at least one or more. After, there is a . that separates them, then this is followed by another sequence of numbers. Notice how I used \. to specify the . character. Using . by itself means that the character is a wildcard. This is obviously not what you want, so if you want to specify the actual . character, you need to prepend a \ to the ..
For the second one, it's almost the same as the first one. However, there is no : delimiter, and we only have the . to work with.
To invoke regular expressions, use the regexp command in MATLAB. It is done using:
ind = regexp(str, expression);
str represents the string you want to check, and expression is a regular expression that we talked about above. You need to make sure you encapsulate your expression using single quotes. The regular expression is taken in as a string. ind would this return the starting index of your string of where the match was found. As such, when we search for a particular format, ind should either be 1 indicating that we found this search at the beginning of the string, or it returns empty ([]) if it didn't find a match. Here's a reproducible example for you:
B = {'29:59.9', '29.9', '45:56.8', '24.5'};
for k = 1 : numel(B)
if (regexp(B{k}, '^[0-9]+:[0-9]+\.[0-9]+') == 1)
disp('I''m the first case!');
elseif (regexp(B{k}, '^[0-9]+\.[0-9]+') == 1)
disp('I''m the second case!');
end
end
As such, the code should print out I'm the first case! if it follows the format of the first case, and it should print I'm the second case! if it follows the format of the second case. As such, by running this code, we get:
I'm the first case!
I'm the second case!
I'm the first case!
I'm the second case!
Without knowing how your strings are formatted, I can't do the rest of it for you, but this should be a good start for you.