string end

[cjkenworthy]

I've got a string:

(value1, value2, value3, value4,)

I want to trim the string at the end by removing the last comma. Which vbscript function would let me do this - to achieve:


(value1, value2, value3, value4)

 

[sfriday]

I could not find a simple utility but this will achieve the result required.

orgstring = "(value1, value2, value3, value4,)"
arrstring = Split(orgstring,",&quot
for i = 0 to ubound(arrstring) -1
If newstring = "" Then
newstring = arrstring(i)
Else
newstring = newstring & "," & arrstring(i)
End If
Next
newstring = newstring & &quot"
msgbox newstring
Regards
Steve Friday

 

[cjkenworthy]

The reaseon why I ask this is because I can't get the below code to enter the first 'IF':

(the code reads in the values of a dynamic number of text fields from a form into a sting strTERMNO)

dim formcounter, J

formcounter = Request.Form("hdnrecords&quot
response.write("Number of fields: " & formcounter) 'WORKS OK
response.write(&quot;<BR>&quot

for J=1 to formcounter

response.write(&quot;Counter J is: &quot; & J )
response.write(&quot;<BR>&quot

if J = formcounter then 'DOES NOT ENTER HERE
strTERMNO = strTERMNO & Request.Form(&quot;txttermno&quot; & J)
else
strTERMNO = strTERMNO & Request.Form(&quot;txttermno&quot; & J) & &quot;, &quot;
end if

next

I basically want it to use the 'else' for every time it reads txttermno , until the last one (where J = number of fields) e.g if it was teh second iteration and number ofr txttermno's was 2 then it would add it to the string, but not append another comma. Unfortunately it just uses the else every time, and a comma is always at the end.

?

 

[sfriday]

Changed your code a bit, it appears not to like the way the number is formated, by checking that j = Abs(formcounter) it seems to work. If you just change the line:

if J = Abs(formcounter) then

Hope it helps


dim formcounter, J

formcounter = &quot;3&quot;
msgbox &quot;Number of fields: &quot; & formcounter 'WORKS OK

For J= 1 to formcounter

msgbox &quot;Counter J is: &quot; & J

msgbox formcounter


if J = ABS(formcounter) then 'DOES NOT ENTER HERE
msgbox &quot;If&quot;
strTERMNO = strTERMNO & &quot;txttermno&quot; & J
else
msgbox &quot;else&quot;
strTERMNO = strTERMNO & &quot;txttermno&quot; & J & &quot;, &quot;
end if

next
msgbox strTERMNO Regards
Steve Friday

 

[sfriday]

Just one more note,

if you run Msgbox VarType(formcounter) you will probably get a return of 8 which means it is a string, you could convert this to an integer by doing CInt(formcounter) in which case your If statement could be

if J = CInt(formcounter) then


Regards
Steve Friday

 

[dilettante]

Turn it around.

If you put the &quot;, &quot; in front of each concatenated substring, you can use Mid(strTERMNO, 3) to easily chop off the leading &quot;, &quot; after you are done.

dim formcounter, J

formcounter = CInt(Request.Form(&quot;hdnrecords&quot)

for J=1 to formcounter
strTERMNO = strTERMNO & &quot;, &quot; & Request.Form(&quot;txttermno&quot; & J)
next

response.write(Mid(strTERMNO, 3))

 

[cjkenworthy]

I used the CInt option, thanks for the help.

Can for loop conditions only deal with INT's in the conditions?


Thanks again

 

[sfriday]

Not neccessarily, you could do a loop

Do until testvar = &quot;Found&quot;
blah blah blah
Loop

but it does appear that you should always check that your variable is an integer and not a string when performing a numerical loop check. I was not aware of this issue until I started playing with your code.

We all learn something new every day.





Regards
Steve Friday