Hello, Can you please let me know when I am using the command Split(“someone@example@com”, “@”,3)–> its spliting the string into 3 strings
whereas when Split(“someone@example@com”, “@”,0) → it should give me the output as "someone@example@com but it is showing indexOutofRangeException why?
Please read this.
Split(expression[,delimiter[,count[,compare]]])
| Parameter | Description |
|---|---|
| expression | Required. A string expression that contains substrings and delimiters |
| delimiter | Optional. A string character used to identify substring limits. Default is the space character |
| count | Optional. The number of substrings to be returned. -1 indicates that all substrings are returned |
| compare | Optional. Specifies the string comparison to use. |
Can have one of the following values:
- 0 = vbBinaryCompare - Perform a binary comparison
- 1 = vbTextCompare - Perform a textual comparison|
Hello thats right -1 as the limit/count returns me the all the available sibstrings separated by the delimiter, but in the case of 0 as the limit/count it should return null or the complete string? I am sorry can you please explain this a bit
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0 and 1 are compare options which are 4th options in the Split function.
you can use 0 number or “vbBinaryCompare” word in quotes.
you can use 1 number or “vbTextCompare” word in quotes as 4th option.
You have used only 3 options in your Split function and referring for 4th option.
I request you to please read full article about Split function. it will give full insight.
Hello I guess you got me wrong I am not referring to the compare functionality in the split function I am referring to the count functionality. I am passing 3 parameters in my split function (String,delimiter,Count) no compare has been used now, If I take count value as 1 it will return me a string substring that is the string itself no spiliting will happen but when I give count value as 0 it should give me an empty String array or null or the complete string in the output. I hope now you will get me when I doubt is?
This usually starts from 1 buddy
That’s why
—Even in the previous one as we have mention 3 it gave us three SubStrings out of the string
And so we need to mention from 1
As it is a Count and not a index
That’s the reason why you were getting that error
Hope this would help you
Cheers @Avijit_Pal
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Hello Palaniyappan I got you ,but it should have given me an empty string right like for the example if I take Split(“XX”,“X”) in this case we are 2 delimiters one after the other but it gives us a an empty string[*] because the split function is considering this as a empty string.
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yes ofcourse it would give us empty string obviously because we dont have any value aside to the delimiter i.e., on either sides of the delimiter
for example
but if i have any character aside to the delimiter only then we would get the value
here (2) next to split method is a index and usually index starts from 0 position but the count usually starts from 1 as said earlier
hope this would help you
Cheers @Avijit_Pal
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Thanks a lot @Palaniyappan and @rkelchuri 
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cheers @Avijit_Pal
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Hey @Palaniyappan a question that popped up if you use a for loop
like for item in Split(“@@”, “@”) it will print out 3 empty string but seems like when you used writeline activity where you used the method .ToString it concatenated all the empty string as 1 and the value 23 as another Is it so?
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no buddy i didnt concatenate
if the input is
str_input = “@@23”
then the output with the split method having delimiter will be in writeline like this
now string has @@23 so three places with @ as delimiter
|@|@|23 like this
0@1@2 - index position
–to get the value of second index position that is 23 from the string i used
Split(str_input,“@”)(2).tostring
cheers @Avijit_Pal
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Great thanks a lot @Palaniyappan, now its clear…
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