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replace " " with %
Posted: 23 Oct 2018 11:14
by Jannis636
Why i can not replace an empty string (" ") with another charactor? stringName =replace(stringName, " ", "%"); does not work?!
Re: replace " " with %
Posted: 23 Oct 2018 11:32
by Desmanto
I just try this
Code: Select all
stringName = "";
a = replace(stringName, "", "%");
Put condition debug dialog after it and {a} is %
It can be that your stringName is not really empty. You can try to use encodeURL(stringName) to see if it is really empty.
Re: replace " " with %
Posted: 24 Oct 2018 13:28
by Jannis636
I solved it with
replaceAll(someText, "\\s", "%")