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I need to replace a character in a string with a space and the rplstr function in WebSmart will not accommodate this.

Product: WebSmart ILE Type: Frequently Asked Question

Over the years, a number of clients have had the need to use RPLSTR with "" or " " as the newval, but when you do this, you get an error:
RNX0100 Length or start position is out of range for the string operation.
In PW_RTS081 procedure  RPLSTR_
The solution appeared to be that we add support for blanks and/or spaces in the rplstr function.
However, when we researched this, there were some technical constraints (see the bottom if you're interested, a task has been logged for this).  
An easy way to do this is to use the SQL scalar replace function.
Remove spaces:
crtfld(alphafld, 25, 'A');
alphafld= " S p a c e s ";
sqlexec("set :alphafld= replace(:alphafld, ' ', '')";
// alphafld contains "Spaces"
Replace with spaces:
crtfld(alphafld, 25, 'A');
alphafld= "Replace_With_Spaces";
sqlexec("set :alphafld= replace(:alphafld, '_', ' ')";
//  alphafld contains "Replace With Spaces"
Earlier Solution:
Create a uparser function rplstr2, by copying rplstr from PARSER with the trim() removed from both Old and New parms:
         PROT='ALPHA rplstr2(ALPHA &string, ALPHA &oldval, ALPHA &newval)';
    rplstr_(~&string!:~&oldval!:~ &newval!)';
Earliest Solution/Workaround
You can also try this in your code:
func rplmethis()
 crtfld(mystring, 64, "A", 0, "string");
 crtfld(temp1, 64, "A", 0, "string");
 crtfld(pos1, 2, "N", 0, "counter");
 crtfld(pos2, 2, "N", 0, "position");
 crtfld(quote, 1, "A", 0, "quote");
 crtfld(stringlen, 2, "N", 0, "length of string");
 crtfld(temp2, 64, "A", 0, "string");
 pos1 = 1;
 quote = "[";
 mystring = "Pickles are ve[ry green do[n't you know";
 stringlen = len(mystring);
 pos2 = scanstr(quote, mystring,pos1);
 if(pos2 > 0)
  temp1 = substr(mystring, pos1, pos2-1) + "";
  while (pos2 > 0)
   pos1 = pos2 + 1;
   pos2 = scanstr(quote, mystring,pos1);
    if(pos2 <> 0)
     temp2 = substr(mystring, pos1, pos2 - pos1);
     temp1 = trim(temp1) + "" + trim(temp2);
    temp1 = trim(temp1) + "" + substr(mystring, pos1, stringlen - pos1 + 1); 
   pos2 = scanstr(quote, mystring,pos1);         
Tech Notes
The reason the RPLSTR function errors out when trying to replace with " " is because the call to rplstr_ trims the parms (in the PARSER member of PW_PSRC).  Therefore the " " is trimmed and thus ends up with length 0. A possible solution to this would be to change the RPLSTR PML function to have additional parms to indicate if the oldval and/or newval should be trimmed, i.e.
resstr = rplstr(strval, oldval, trim(*YES, *NO), newval, trim(*YES, *NO));
That way, you could pass in " " with *NO for the newval, and it won't trim it, and the RPLSTR function wouldn't error out.
MDH 20060127: A better solution will be to create a new function that trims, because at this point we can't change how an existing function works.
KJH 20101007. FS Task 6968 has been logged in relation to this.

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