Custum input with text splitting

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Summary:

create a custom Input with text splitting for longer input

Custum input with text splitting

AntKou 25 Jan 2013 16:35

After that nobody answered my question:

How can the text displayed faster?

- is there another way to edit text on multiple lines?

2nd problem is that this routine doesn't work sometimes.

I used:

If C=3                 //C is the selected note
length(Str1 → A
If C=4
length(Str2 → A
If A>115             //5 times 23 characters (3pixel)
115 → A

0 → H
For(B,1,5
If H+23<A           //23 characters in a line (3pixel)
Then
If C=3
Text(B*8+14,2,sub(Str1,H+1,H+23   //drawing text
If C=4
Text(B*8+14,2,sub(Str2,H+1,H+23
Else
If C=3
Text(B*8+14,2,sub(Str1,H+1,A  //drawing last line text with less than 23 characters
If C=4
Text(B*8+14,2,sub(Str2,H+1,A
End
A-23 → A                               //characters left decreased
H+23 → H                               //lenght for start in the string increased
If A<0                                    //if there are no more characters break the loop
5 → B
End

Here is a screen:

EDIT: Inserted image — burr

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Last edited on 25 Jan 2013 21:40 by burr Show more

Re: Custum input with text splitting

Xeda Elnara 25 Jan 2013 18:58

If you want 23 characters per line and Str1 is the string to display:

length(Str1→B
Delvar A
length(Str1
While A≤B
Text(8A/23+22,2,sub(Str1,A+1,min(B-A-1,23
A+23→A
End

I have not tested that yet. Sorry if there are mistakes, but I have to go to class!

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Re: Custum input with text splitting

AntKou 25 Jan 2013 21:25

Thanks a lot!

I tested this:

length(Str1→B
DelVar A                                 //there was a 2nd length(Str1, but Ans is not used here
While A≤B
Text(8A/23+22,2,sub(Str1,A+1,min(B-A,23  //not min(B-A-1,23 because this hides the last character
A+23→A
End

Is there a possibility to display text more efficiently? Because the spaces only take 1 pxl in contrast to the capital letters and numbers.

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Re: Custum input with text splitting

burr 25 Jan 2013 21:37

You could add a separate If conditional for spaces…

Re: Custum input with text splitting

(account deleted) 25 Jan 2013 22:15

My idea is that you can use Pxl-Test( to test a common pixel used for all text…

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Re: Custum input with text splitting

AntKou 25 Jan 2013 22:37

Ifirst thought about that too but it would be easier to use a list with the number of spaces in the first, 2nd, 3rd, 4th, 5th line. Then you just need to code:

Text(8A/23+22,2,sub(Str1,A+1,int((L(SPACE)(1)*2)/3)+min(B-A,23

Every space takes 1 pxl, the capital letters and numbers 3 pxl. When a space is used, 2 pxl won't be displayed in the row, but a letter uses 3 pxl. The missing pxls together are divided into 3, because a letter uses 3 pxl. Round can't be used because it could give a greater value which would cause an error.

Problem: It is hard to make this list, is there an easier way to know how many spaces are in the first line? Even if there would be the list, Notes would crash if the string was edited outside Notes.

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Re: Custum input with text splitting

Xeda Elnara 26 Jan 2013 13:29

Yes, you can make use of the great inString( command :D I wasn't sure how you actually wanted to do the text wrapping (you could even make it wrap on spaces so that whole words aren't split across two lines). Anyways, I have a few ideas for you, but maybe the best approach would be to display the text one character at a time. Either that, or precompute the pixel length of each string. Ther are small assembly programs that will do that, but for BASIC:

Delvar CDelvar Dlength(Str1→B   ;C and D are your coordinates
For(A,1,B
sub(Str1,A,1→Str2        ;Get the next token. I am assuming they are 1 character tokens.
4                      ;We will assume the char is 4 pixels wide unless told otherwise
If Str2="             ;There is a space here
1
If inString(".:!i",Str2   ;these are tokens that are two pixels wide. That is a lowercase i, not the imaginary one.
2
If inString(",()[]",Str2  ;these are tokens that are 4 pixels wide
3
If inString("mw",Str2
5
If Str2="*
6
Ans→E
If D+Ans>90    ;max number of pixels you are allowing in the display window. I guessed 90?
Then
Delvar DC+1→C
End
Text(8C+14,D+2,Str2
D+E→D
End

EDIT: Also, to get a list of where all the spaces occur using inString(:

Delvar L1SetUpEditor L1
0
Repeat not(Ans
Ans→L1(1+dim(L1
inString(Str1," ",Ans+1
End

The first element will be a zero, everything after that will be the offset into the string to a space. What you can do to get the size of each word, then, is use ΔList(L1 and then the first element will be the length of the first word plus 1, the second element will be the length of the second word plus 1, and so on. You won't be given the length of the last word, though, unless there is a space after it.

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Last edited on 26 Jan 2013 13:39 by Xeda Elnara Show more