Been scratching my head on this one...
I'd like to remove .com and capitalize S and T from: "sometext.com"
So output would be Some Text
Thank you in advance
For most of this you can use the replace() member of the String object.
The syntax is:
$string = $string.replace('what you want replaced', 'what you will replace it with')
Replace can be used to erase things by using blank quotes '' for the second argument. That's how you can get rid of .com
$string = $string.replace('.com','')
It can also be used to insert things. You can insert a space between some and text like this:
$string = $string.replace('et', 'e t')
Note that using replace does NOT change the original variable. The command below will print "that" to your screen, but the value of $string will still be "this"
$string = 'this'
$string.replace('this', 'that')
You have to set the variable to the new value with =
$string = "this"
$string = $string.replace("this", "that")
This command will change the value of $string to that.
The tricky part here comes in changing the first t to capital T without changing the last t. With strings, replace() replaces every instance of the text.
$string = "text"
$string = $string.replace('t', 'T')
This will set $string to TexT. To get around this, you can use Regex. Regex is a complex topic. Here just know that Regex objects look like strings, but their replace method works a little differently. You can add a number as a third argument to specify how many items to replace
$string = "aaaaaa"
[Regex]$reggie = 'a'
$string = $reggie.replace($string,'a',3)
This code sets $string to AAAaaa.
So here's the final code to change sometext.com to Some Text.
$string = 'sometext.com'
#Use replace() to remove text.
$string = $string.Replace('.com','')
#Use replace() to change text
$string = $string.Replace('s','S')
#Use replace() to insert text.
$string = $string.Replace('et', 'e t')
#Use a Regex object to replace the first instance of a string.
[regex]$pattern = 't'
$string = $pattern.Replace($string, 'T', 1)
What you're trying to achieve isn't well-defined, but here's a concise PowerShell Core solution:
PsCore> 'sometext.com' -replace '\.com$' -replace '^s|t(?!$)', { $_.Value.ToUpper() }
SomeText
-replace '\.com$' removes a literal trailing .com from your input string.
-replace '^s|t(?!$), { ... } matches an s char. at the start (^), and a t that is not (!) at the end ($); (?!...) is a so-called negative look-ahead assertion that looks ahead in the input string without including what it finds in the overall match.
Script block { $_.Value.ToUpper() } is called for each match, and converts the match to uppercase.
-replace (a.k.a -ireplace) is case-INsensitive by default; use -creplace for case-SENSITIVE replacements.
For more information about PowerShell's -replace operator see this answer.
Passing a script block ({ ... }) to dynamically determine the replacement string isn't supported in Windows PowerShell, so a Windows PowerShell solution requires direct use of the .NET [regex] class:
WinPs> [regex]::Replace('sometext.com' -replace '\.com$', '^s|t(?!$)', { param($m) $m.Value.ToUpper() })
SomeText
Related
I have a method that has an if statement that catches if it finds a special character. What I want to do now if find the position of the special characters and replace it with _A
Some Examples
test# becomes test_A
I#hope#someone#knows#the#answer# becomes I_Ahope_Asomeone_Aknows_Athe_Aanswer_A
or if it has more than one special character
You?didnt#understand{my?Question# becomes You_Adidnt_Aunderstand_Amy_AQuestion_A
Would I have to loop through the whole string and when I reach that character change it to _A or is there a quicker way of doing this?
# is just a character like any other, you can use the -replace operator:
PS C:\>'I#hope#someone#knows#the#answer#' -replace '#','_A'
I_Ahope_Asomeone_Aknows_Athe_Aanswer_A
Regex is magic, you can define all the special cases you like (braces will have to be escaped):
PS C:\>'You?didnt#understand{my?Question#' -replace '[#?\{]','_A'
You_Adidnt_Aunderstand_Amy_AQuestion_A
So your function could look something like this:
function Replace-SpecialChars {
param($InputString)
$SpecialChars = '[#?\{\[\(\)\]\}]'
$Replacement = '_A'
$InputString -replace $SpecialChars,$Replacement
}
Replace-SpecialChars -InputString 'You?didnt#write{a]very[good?Question#'
If you are unsure of which characters to escape, have the regex class do it for you!
function Replace-SpecialChars {
param(
[string]$InputString,
[string]$Replacement = "_A",
[string]$SpecialChars = "#?()[]{}"
)
$rePattern = ($SpecialChars.ToCharArray() |ForEach-Object { [regex]::Escape($_) }) -join "|"
$InputString -replace $rePattern,$Replacement
}
Alternatively, you can use the .NET string method Replace():
'You?didnt#understand{my?Question#'.Replace('#','_A').Replace('?','_A').Replace('{','_A')
But I feel the regex way is more concise
Here's the scenario -- One step of the process involves fixing city names when the data is obviously misspelled, along with some basic conversions like "MTN" to "Mountain" and so forth. I've built a variable containing several substitution strings, and I'm trying to apply that set of subs on one of the input fields later down the line.
my $citysub = <<'EOF';
s/DEQUEEN/DE QUEEN/;
s/ELDORADO/EL DORADO/;
... # there are about 100 such substitution strings
EOF
...
while ($line <INFILE>)
{
...
#field = split(/","/,$line); # it's a comma-delimited file with quoted strings; this is spltting exactly like I intend; at the end, I'll piece it back together properly
...
# the 9th field and 12th field are city names, i.e., $field[8] and $field[12]
$field[8] =~ $citysub; # this is what I'm wanting to do, but it doesn't work!
# since that doesn't work, I'm using the following, but it's much slower, obviiously
$field[8] = `echo $field[8]|sed -e "$citysub"`; # external calls to system commands
So, what's the proper syntax to insert a multi-line substitution string and apply it toward a single array value?
my %citysub = ( "DEQUEEN" => "DE QUEEN", "ELDORADO" => "EL DORADO" );
for my $find ( keys %citysub ) {
my $replace = $citysub{ $find };
$field[8] =~ s/$find/$replace/g;
}
Explanation: Create a hash of "thing to match" => "thing to replace with". then loop over that hash and run s/// with the thing to match and the thing to replace with.
In Powershell given the following string
$string = "this is a sample of 'my' text $PSP.what do you think"
how do I use the -replace function to convert the string to
this is a sample of 'my' text Hello.what do you think
I obviously need to escape the string somehow, Also $PSP is not a declared variable in my script
I need to change all mentions of $PSP for some other string
Use the backtick character (above the tab key):
$string = "this is a sample of 'my' text `$PSP.what do you think"
To replace the dollar sign using the -replace operator, escape it with backslash:
"this is a sample of 'my' text `$PSP.what do you think" -replace '\$PSP', 'hello'
Or use the string.replace method:
$string = "this is a sample of 'my' text `$PSP.what do you think"
$string.Replace('$PSP','Hello)'
this is a sample of 'my' text Hello.what do you think
Unless you modify your original string (e.g. by escaping the $), this is isn't (really) possible.
Your $string doesn't really contain a $PSP, as it is replaced by nothing in the assignment statement.
$string = "this is a sample of 'my' text $PSP.what do you think"
$string -eq "this is a sample of 'my' text .what do you think"
evaluates to:
True
This comes up as the first answer in google even though it is really old, so I will add my slight variation.
In my case I was reading in a file and replacing a string with $s in it.
The short version of my file is:
<version>$version$<version>
In the case where one is actiong on a (file) stream, variables are not autoreplaced so there is no need to escape the $ in the file.
In the replacement pattern you can avoid the interpretation of the variable using ' instead of ".
My final command looked like:
(gc $fileName) | % { $_.replace('$version$', "$BuildNumber") } | sc $fileName
This is a file read(get-content) piped through the replace and back in to the file with a set-content.
You should try
$string = $string.Replace("\$PSP", "Hello")
or
$string = $string.Replace("\$PSP", $the_new_value)
or to be more generic use Regex
$string = [regex]::Replace($string, "\$\w+", "Hello")
In perl suppose I have a string like 'hello\tworld\n', and what I want is:
'hello world
'
That is, "hello", then a literal tab character, then "world", then a literal newline. Or equivalently, "hello\tworld\n" (note the double quotes).
In other words, is there a function for taking a string with escape sequences and returning an equivalent string with all the escape sequences interpolated? I don't want to interpolate variables or anything else, just escape sequences like \x, where x is a letter.
Sounds like a problem that someone else would have solved already. I've never used the module, but it looks useful:
use String::Escape qw(unbackslash);
my $s = unbackslash('hello\tworld\n');
You can do it with 'eval':
my $string = 'hello\tworld\n';
my $decoded_string = eval "\"$string\"";
Note that there are security issues tied to that approach if you don't have 100% control of the input string.
Edit: If you want to ONLY interpolate \x substitutions (and not the general case of 'anything Perl would interpolate in a quoted string') you could do this:
my $string = 'hello\tworld\n';
$string =~ s#([^\\A-Za-z_0-9])#\\$1#gs;
my $decoded_string = eval "\"$string\"";
That does almost the same thing as quotemeta - but exempts '\' characters from being escaped.
Edit2: This still isn't 100% safe because if the last character is a '\' - it will 'leak' past the end of the string though...
Personally, if I wanted to be 100% safe I would make a hash with the subs I specifically wanted and use a regex substitution instead of an eval:
my %sub_strings = (
'\n' => "\n",
'\t' => "\t",
'\r' => "\r",
);
$string =~ s/(\\n|\\t|\\n)/$sub_strings{$1}/gs;
If I had:
$foo= "12."bar bar bar"|three";
how would I insert in the text ".." after the text 12. in the variable?
Perl allows you to choose your own quote delimiters. If you find you need to use a double quote inside of an interpolating string (i.e. "") or single quote inside of a non-interpolating string (i.e. '') you can use a quote operator to specify a different character to act as the delimiter for the string. Delimiters come in two forms: bracketed and unbracketed. Bracketed delimiters have different beginning and ending characters: [], {}, (), [], and <>. All other characters* are available as unbracketed delimiters.
So your example could be written as
$foo = qq(12."bar bar bar"|three);
Inserting text after "12." can be done many ways (TIMTOWDI). A common solution is to use a substitution to match the text you want to replace.
$foo =~ s/^(12[.])/$1../;
the ^ means match at the start of the sting, the () means capture this text to the variable $1, the 12 just matches the string "12", and the [] mean match any one of the characters inside the brackets. The brackets are being used because . has special meaning in regexes in general, but not inside a character class (the []). Another option to the character class is to escape the special meaning of . with \, but many people find that to be uglier than the character class.
$foo =~ s/^(12\.)/$1../;
Another way to insert text into a string is to assign the value to a call to substr. This highlights one of Perl's fairly unique features: many of its functions can act as lvalues. That is they can be treated like variables.
substr($foo, 3, 0) = "..";
If you did not already know where "12." exists in the string you could use index to find where it starts, length to find out how long "12." is, and then use that information with substr.
Here is a fully functional Perl script that contains the code above.
#!/usr/bin/perl
use strict;
use warnings;
my $foo = my $bar = qq(12."bar bar bar"|three);
$foo =~ s/(12[.])/$1../;
my $i = index($bar, "12.") + length "12.";
substr($bar, $i, 0) = "..";
print "foo is $foo\nbar is $bar\n";
* all characters except whitespace characters (space, tab, carriage return, line feed, vertical tab, and formfeed) that is
If you want to use double quotes in a string in Perl you have two main options:
$foo = "12.\"bar bar bar\"|three";
or:
$foo = '12."bar bar bar"|three';
The first option escapes the quotes inside the string with backslash.
The second option uses single quotes. This means the double quotes are treated as part of the string. However, in single quotes everything is literal so $var or #array isn't treated as a variable. For example:
$myvar = 123;
$mystring = '"$myvar"';
print $mystring;
> "$myvar"
But:
$myvar = 123;
$mystring = "\"$myvar\"";
print $mystring;
> "123"
There are also a large number of other Quote-like Operators you could use instead.
$foo = "12.\"bar bar bar\"|three";
$foo =~s/12\./12\.\.\./;
print $foo; # results in 12...\"bar bar bar\"|three"