What could be a preg_replace pattern to change anything within quotes that is being assigned to $my_variable, regardless of length?
For example,
$my_variable = "whatever it is"
should be changed so that it reads:
$my_variable = "whatever new value"
I figured it out:
/\$my_variable=\"(.*?)\";/
Related
I am trying to extract whatever is between two strings. The first string is a known string, the second string could be from a list of strings.
For example,
We have the start string and the end strings. We want to get the text between these.
start = "start"
end = ["then", "stop", "other"]
Criteria
test = "start a task then do something else"
result = "a task"
test = "start a task stop doing something else"
result = "a task"
test = "start a task then stop"
result = "a task"
test = "start a task"
result = "a task"
I have looked at using a regex, and I got one which works for between two strings, I just cannot create one which words with a option of strings:
(?<=start\s).*(?=\sthen)
I have tried using this:
(?<=start\s).*(?=\sthen|\sstop|\sother)
but this will include 'then, stop or other' in the match like so:
"start a task then stop" will return "a task then"
I have also tried to do a 'match any character except the end list" in the capture group like so: (?<=start\s)((?!then|stop|other).*)(?=\sthen|\sstop|\sother) but this has the same effect as the one above.
I am using swift, so I am also wondering whether this can be achieved by finding the substring between two strings.
Thanks for any help!
You may use
(?<=start\s).*?(?=\s+(?:then|stop|other)|$)
See the regex demo. To search for whole words, add \b word boundary in proper places:
(?<=\bstart\s).*?(?=\s+(?:then|stop|other)\b|$)
See another regex demo
Details
(?<=start\s) - a positive lookbehind that matches a location immediately preceded with start string and a whitespace
.*? - any 0+ chars other than line break chars, as few as possible
(?=\s+(?:then|stop|other)|$) - a position in the string that is immediately followed with
\s+ - 1+ whitespaces
(?:then|stop|other) - one of the words
|$ - or end of string.
In sales column i have values with pound sign £1200. It is not readable by Data frame in scala, please help me for the same. i want column value in double, 1200. I am using below method but its not working.
def getRemovedDollarValue = udf(
(actualSales: String) => {
val actualSalesDouble = actualSales
.replace(",", "")
.replace("$", "")
.replace("\\u00A3","")
.replace("\\U00A3","")
.replaceAll("\\s", "_").trim().toDouble
java.lang.Double.parseDouble(actualSalesDouble.toString)
}
)
You need write: .replace("\u00A3","") instead of escaping .replace("\\u00A3","").
But I prefer just: .replace("£", "") - it is more readable.
I think the proposed solutions and comments all work but don't address the confusion behind why your code isn't working.
From the Pattern docs:
Thus the strings "\u2014" and "\\u2014", while not equal, compile into the same pattern, which matches the character with hexadecimal value 0x2014.
replace and replaceAll are both replacing all occurrences in a String, but only replaceAll is taking in a regular expression. You're passing in "\\u00A3" which will work as a pattern, but not a unicode literal due to the added backslash. As already suggested, either use replace with a unicode literal or the actual symbol, or change to replaceAll.
I am working on the exercises from https://www.scala-exercises.org/std_lib/formatting
For the following question, m answer seems incorrect but I do not know why.
val c = 'a' //unicode for a
val d = '\141' //octal for a
val e = '\"'
val f = '\\'
"%c".format(c) should be("a") //my answers
"%c".format(d) should be("a")
"%c".format(e) should be(")
"%c".format(f) should be(\)
your answer should be enclosed in quotes
"%c".format(e) should be("\"")
"%c".format(f) should be("\\")
because it wouldn't recognize string unless it's enclosed in quotes
Your last two lines are invalid Scala code and cannot be compiled:
// These are wrong
"%c".format(e) should be(")
"%c".format(f) should be(\)
The be() function needs to be passed a String, and neither of those calls are being passed a String. A String needs to start and end with a double-quote (there are some exceptions).
// In this case you started a String with a double-quote, but you are never
// closing the string with a second double-quote
"%c".format(e) should be(")
// In this case you are missing both double-quotes
"%c".format(f) should be(\)
In this case the code should be:
"%c".format(e) should be("\"")
"%c".format(f) should be("\\")
If you want a character to be treated literally in a String, you need to "escape" it with a backslash. So if you want to literally show a double-quote, you need to prefix it with a backslash:
\"
And as a String:
"\""
Similarily for a backslash:
\\
As a String:
"\\"
Using an IDE makes this easier to see. Using IntelliJ the String is green but the special non-literal characters are highlighted in orange.
Check quote signs.
https://www.tutorialspoint.com/scala/scala_strings.htm
https://docs.scala-lang.org/overviews/core/string-interpolation.html
https://learnxinyminutes.com/docs/scala/
You can run Scala code online and check yourself here:
https://scastie.scala-lang.org
https://ideone.com/
my #para_text = $mech->xpath('/html/body/form/table/tbody/tr[2]/td[3]/table/tbody/tr[2]/td/table/tbody/tr[3]/td/div/div/div', type => $mech->xpathResult('STRING_TYPE'));
#BELOW IS JUST TO MAKE SURE THE ABOVE CAPTURED THE CORRECT TEXT
print "Look here: #para_text";
$mech->click_button( id => "lnkHdrreplyall");
$mech->eval_in_page('document.getElementsByName("txtbdy")[0].value = "#para_text"');
In the last line of my code I need to put the contents of the #para_text array as the text to output into a text box on a website however from the "document" till the end of the line it needs to be surrounded by ' ' to work. Obviously this doesnt allow interpolation as that would require " " Any ideas on what to do?
To define a string that itself contains double quotes as well as interpolating variable values, you may use the alternative form of the double quote qq/ ... /, where you can choose the delimiter yourself and prevent the double quote " from being special
So you can write
$mech->eval_in_page(qq/document.getElementsByName("txtbdy")[0].value = "#para_text"/)
can you please tell me how to remove currency formatting from a variable (which is probably treated as a string).
How do I strip out currency formatting from a variable and convert it to a true number?
Thank you.
example
PS C:\Users\abc> $a=($464.00)
PS C:\Users\abc> "{0:N2}" -f $a
<- returns blank
However
PS C:\Users\abc> $a=-464
PS C:\Users\abc> "{0:C2}" -f $a
($464.00) <- this works
PowerShell, the programming language, does not "know" what money or currency is - everything PowerShell sees is a variable name ($464) and a property reference (.00) that doesn't exist, so $a ends up with no value.
If you have a string in the form: $00.00, what you can do programmatically is:
# Here is my currency amount
$mySalary = '$500.45'
# Remove anything that's not either a dot (`.`), a digit, or parentheses:
$mySalary = $mySalary -replace '[^\d\.\(\)]'
# Check if input string has parentheses around it
if($mySalary -match '^\(.*\)$')
{
# remove the parentheses and add a `-` instead
$mySalary = '-' + $mySalary.Trim('()')
}
So far so good, now we have the string 500.45 (or -500.45 if input was ($500.45)).
Now, there's a couple of things you can do to convert a string to a numerical type.
You could explicitly convert it to a [double] with the Parse() method:
$mySalaryNumber = [double]::Parse($mySalary)
Or you could rely on PowerShell performing an implicit conversion to an appropriate numerical type with a unary +:
$mySalaryNumber = +$mySalary