Forum Discussion
QuickBaseCoachD
8 years agoQrew Captain
There is no replace function.
But you can do this.
var text String = [My Field];
var text Delimiters = " !@#$%^&*";
List("",
Part($String,1,$Delimiters),
Part($String,2,$Delimiters),
Part($String,3,$Delimiters),
Part($String,4,$Delimiters),
Part($String,5,$Delimiters),
Part($String,6,$Delimiters),
Part($String,7,$Delimiters),
Part($String,8,$Delimiters),
Part($String,9,$Delimiters))
But you can do this.
var text String = [My Field];
var text Delimiters = " !@#$%^&*";
List("",
Part($String,1,$Delimiters),
Part($String,2,$Delimiters),
Part($String,3,$Delimiters),
Part($String,4,$Delimiters),
Part($String,5,$Delimiters),
Part($String,6,$Delimiters),
Part($String,7,$Delimiters),
Part($String,8,$Delimiters),
Part($String,9,$Delimiters))