I am looking for a method in velocity where I can put a check on certain part of string and carry on with subsequent methods. I tried with the following:
But it throws an error that 'contains' function is unrecognized.
##Check whether a certain string is a sub part of other string
#set($Name="")
#set($SomeName="ABC DEF"
#set($Name=$SomeName)
#if($Name.contains("DE"))
Correct!
#end
At first glance i found that there is a mistake bracket not close
#set($SomeName="ABC DEF"
should be changed to
#set($SomeName="ABC DEF")
Other than that i don't think there is a problem with contains function . i have tested it is working fine with me .
Related
**Having a problem when using String Interpolation in the filter pictured. It is telling me Too many positional arguments: 1 allowed, but 2 found
I know what that means, it wants me to remove the String Interpolation that I have in there. I have tried all the bracket combinations I can think of and nothing takes away this error. My String is just a plain string and works fine if I put it directly in the code at this spot but won't work with the interpolation.
Any suggestions on how I can fix this?
Thankyou in Advance For your help **s!
e.ingCatCode == "passedIngCatCode $ingSelSearch"
Use "" quotes and the $ variable inside the quotes.
This is the part of the code I have copied to see the output,
def check(string,sub_str):
if(string.find(sub_str)==-1):
print('no')
else:
print('yes)
# driver code for testing the above function
string='geeks for geeks'
sub_str='geeks'
I specifically wanted to understand how this expression works :
if(string.find(sub_str)==-1): . Also this code is for finding substrings in a given strings can some one tell if this is the optimal way, I know it is tutorial code but I have an easier way to find the substrings. Just wanted to know if that would make passing test cases easier hence the above code. Anyways thanks y'all for your answers.
The method find() returns the index of the string you are looking for. The string in front of find() is the one in which you are looking for the second string.
SentenceThatIsCompletelySearched.find(ForThisPartHere)
If the string you are looking for is present it will return the index (a number on which position of the sentence the string has been found).
If the string is not inside the sentence then find() will return -1 (a number).
So in your case you are checking if sub_str is inside string and if it is not present (return of -1) you will print "no". If it is you will print "yes".
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 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.