Very good explanation, thank you.
Am 22.05.2017 um 23:33 schrieb d4t4f...@gmail.com:
> Matti
>
> Your string has 3 substrings when split with SPLIT (STRING, "#"):
>
> Index 0: "Hello"
> Index 1: ""
> Index 2: "12345"
>
> It is documented that separators are single characters, not strings; you can
A variant of the previous suggestions, to ignored (5) voids :-)
Public Sub Main()
Dim ss As String[]
Dim s As String
ss = Split("hello#12345", "#")
For Each s In ss
If Not IsNull(s) Then Print s;
Next
End
Regards
Gianluigi
2017-05-23 8:38 GMT+02:00 Fabien Bodard :
> A=Split
A=Split (replace ("##yyy##1234","##",chr (27)),chr (27))
Or do not use double characters as separator .
Le 22 mai 2017 23:35, a écrit :
> Matti
>
> Your string has 3 substrings when split with SPLIT (STRING, "#"):
>
> Index 0: "Hello"
> Index 1: ""
> Index 2: "12345"
>
> It is documented
Matti
Your string has 3 substrings when split with SPLIT (STRING, "#"):
Index 0: "Hello"
Index 1: ""
Index 2: "12345"
It is documented that separators are single characters, not strings; you can
specify a string because any of its individual characters may be used as
separator.
If you also sp
Ok, thank you, Tobi.
I'll have to start thinking again (but not tonight anymore).
Matti
Am 22.05.2017 um 22:17 schrieb Tobias Boege:
> On Mon, 22 May 2017, Matti wrote:
>> If I have a string sStr="hello##12345" and want to split it:
>>
>> Dim aSplit as String[]
>> aSplit = Split(sStr, "##")
>> Pri
On Mon, 22 May 2017, Matti wrote:
> If I have a string sStr="hello##12345" and want to split it:
>
> Dim aSplit as String[]
> aSplit = Split(sStr, "##")
> Print aSplit[0], aSplit[1]
>
> Returns always only "hello". Maybe "12345" is excluded because being an
> integer?
>
It is not an integer in
