How can I divide a string with dots as delimiters in Groovy?
If I have a string like "22112018", how do I convert it to "22.11.2018"?
EDIT:
I wasn't really sure how to formulate the question. I wanted to 'split' the string but split() method doesn't do what I need (doesn't mean the same).
This answer in comments (by #ernest_k) was good enough for what I needed:
text = "22112018"
"${text[0..1]}.${text[2..3]}.${text[4..7]}"
However, it was not an "answer" in the SO way, so I'm accepting the answer by #tim_yates (also works and is probably a more precise and robust solution).
I assume this is a date...
You could do:
Date.parse('ddMMyyyy', '22112018').format('dd.MM.yyyy')
instead of just grabbing characters
Related
how should I define a function that takes in a string and returns the string in upper and lowercases according to even and odd indexing?
def myfunc(string):
for some in string:
if string.index%2==0:
I have written this much but I do not know what should I type now.
please help.
Here you can find the string methods Specially look at upper() and lower() which will be your friend here.
In TI-BASIC, the + operation is overloaded for string concatenation (in this, if nothing else, TI-BASIC joins the rest of the world).
However, any attempt to concatenate involving an empty string raises a Dimension Mismatch error:
"Fizz"+"Buzz"
FizzBuzz
"Fizz"+""
Error
""+"Buzz"
Error
""+""
Error
Why does this occur, and is there an elegant workaround? I've been using a starting space and truncating the string when necessary (doesn't always work well) or using a loop to add characters one at a time (slow).
The best way depends on what you are doing.
If you have a string (in this case, Str1) that you need to concatenate with another (Str2), and you don't know if it is empty, then this is a good general-case solution:
Str2
If length(Str1
Str1+Str2
If you need to loop and add a stuff to the string each time, then this is your best solution:
Before the loop:
" →Str1
In the loop:
Str1+<stuff_that_isn't_an_empty_string>→Str1
After the loop:
sub(Str1,2,length(Str1)-1→Str1
There are other situations, too, and if you have a specific situation, then you should post a simplified version of the relevant code.
Hope this helps!
It is very unfortunate that TI-Basic doesn't support empty strings. If you are starting with an empty string and adding chars, you have to do something like this:
"?
For(I,1,3
Prompt Str1
Ans+Str1
End
sub(Ans,2,length(Ans)-1
Another useful trick is that if you have a string that you are eventually going to evaluate using expr(, you can do "("+Str1+")"→Str1 and then freely do search and replace on the string. This is a necessary workaround since you can't search and replace any text involving the first or last character in a string.
I am currently looking for a way in which I can add a underscore
in a string generated as such:
>>> bin(1)[2:].zfill(8)
'00000001'
I want the string to be '0_0_0_0_0_0_0_1'
How do i add the underscore between each character?
This can be achieved in multiple ways and has already been answered a lot of times here. Usually people seem to do to some kind of replace operation but there are other ways. Here's how to use the String join() method.
s = "00000001"
"_".join(s)
I want to remove all the characters from a string expect whatever character is between a certain set of characters. So for example I have the input of Grade:2/2014-2015 and I want the output of just the grade, 2.
I'm thinking that I need to use the FIND function to grab whatever is between the : and the / , this also needs to work with double characters such 10 however I believe that it would work so long as the defining values with the FIND function are correct.
Unfortunately I am totally lost on this when using the FIND function however if there is another function that would work better I could probably figure it out myself if I knew what function.
It's not particularly elegant but =MID(A1,FIND(":",A1)+1,FIND("/",A1) - FIND(":",A1) - 1) would work.
MID takes start and length,FIND returns the index of a given character.
Edit:
As pointed out, "Grade:" is fixed length so the following would work just as well:
=MID(A1,7,FIND("/",A1) - 7)
You could use LEFT() to remove "Grade:"
And then use and then use LEFTB() to remove the year.
Look at this link here. This is the way I would go about it.
=SUBSTITUTE(SUBSTITUTE(C4, "Grade:", ""), "/2014-2015", "")
where C4 is the name of your cell.
I'm getting a garbled JSON string from a HTTP request, so I'm looking for a temp solution to select the JSON string only.
The request.params() returns this:
[{"insured_initials":"Tt","insured_surname":"Test"}=, _=1329793147757,
callback=jQuery1707229194729661704_1329793018352
I would like everything from the start of the '{' to the end of the '}'.
I found lots of examples of doing similar things with other languages, but the purpose of this is not to only solve the problem, but also to learn Scala. Will someone please show me how to select that {....} part?
Regexps should do the trick:
"\\{.*\\}".r.findFirstIn("your json string here")
As Jens said, a regular expression usually suffices for this. However, the syntax is a bit different:
"""\{.*\}""".r
creates an object of scala.util.matching.Regex, which provides the typical query methods you may want to do on a regular expression.
In your case, you are simply interested in the first occurrence in a sequence, which is done via findFirstIn:
scala> """\{.*\}""".r.findFirstIn("""[{"insured_initials":"Tt","insured_surname":"Test"}=, _=1329793147757,callback=jQuery1707229194729661704_1329793018352""")
res1: Option[String] = Some({"insured_initials":"Tt","insured_surname":"Test"})
Note that it returns on Option type, which you can easily use in a match to find out if the regexp was found successfully or not.
Edit: A final point to watch out for is that the regular expressions normally do not match over linebreaks, so if your JSON is not fully contained in the first line, you may want to think about eliminating the linebreaks first.