[RESOLVED] blank line at end in richtextbox

I trying to add sentences on a new line to the end of text in a richtextbox
Like this:
with rtb
.selstart = len(.text)
.seltext = vbcrlf & "some new text"
end select
This works okay if there is no vbcrlf already in the text i am adding to.
If the text i am adding to has a vbcrlf at the end a blank line is added
before my new text.
How can i find if there is a vbcrlf and how to delete it ?
.

vb Code:

if right(.text,2)=vbCrLf then .seltext = "newtext" else .seltext = vbCrLf & "newtext"

That should work, offhand

Quote:

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Originally Posted by smUX

vb Code:

if right(.text,2)=vbCrLf then .seltext = "newtext" else .seltext = vbCrLf & "newtext"

That should work, offhand

---

Than you
That would probably work i just need a way to select just the last line in the rtb
Every line above the last line will have a vbcrlf

using strrev(.text,vbCrLf) will give you the point at which the list vbCrLf is, but you'd then have to do other stuff to check and see if the last line is empty or not...the suggestion I gave might be the best option...that or...

vb Code:

if right(.text,2)=vbCrLf then .text=left(.text,len(.text)-2)

Quote:

---

Originally Posted by smUX

using strrev(.text,vbCrLf) will give you the point at which the list vbCrLf is, but you'd then have to do other stuff to check and see if the last line is empty or not...the suggestion I gave might be the best option...that or...

vb Code:

if right(.text,2)=vbCrLf then .text=left(.text,len(.text)-2)

---

seems vb does not even see the vbcrlf i can see the 2 little squares in bread but
vb does not. I might have code against the rtf code directly.
Thought this was going to be easy

Do me a favour and run this line of code...paste the results from the debug window here so I have an idea of what the problem is :-)

vb Code:

debug.print asc(right(.text,1)), asc(mid(.text,len(.text)-2,1))

I think the second part is right, not 100% sure...it should return the asc codes for the linefeed first then the carriage return...10 then 13

Quote:

---

Originally Posted by smUX

Do me a favour and run this line of code...paste the results from the debug window here so I have an idea of what the problem is :-)

vb Code:

debug.print asc(right(.text,1)), asc(mid(.text,len(.text)-2,1))

I think the second part is right, not 100% sure...it should return the asc codes for the linefeed first then the carriage return...10 then 13

---

debug with vbcrlf
10 33
without vbcrlf
33 101

33 = !
101 = e
10 you know, linefeed

So you don't have a vbCrLf at the end, you have a Lf :-)

Hope that helps you a little

Edit: Whoops...just made sure of that, and code *was* wrong

vb Code:

debug.print asc(right(.text,1)), asc(mid(.text,len(.text)-1,1))

Quote:

---

Originally Posted by smUX

33 = !
101 = e
10 you know, linefeed

So you don't have a vbCrLf at the end, you have a Lf :-)

Hope that helps you a little

Edit: Whoops...just made sure of that, and code *was* wrong

vb Code:

debug.print asc(right(.text,1)), asc(mid(.text,len(.text)-1,1))

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I think i am in business i was coding against .selrtf rather than .text
Thanks a lot