divide a text to array

[Silvio.Falconi]

if I have a string sample :

Local cMessage:= "12345678901234567890123456789012345678901234567890123456789012345678901234567890"

I wish insert into a array records 20 cr sample

aadd( atextlinea,"12345678901234567890")

for a sample if there are 21 cr on cmessage the atextlinea must be 1

but I not Know the len of the cmessage and I wish create the atextlinea array

some can help me

[pieter]

Hello Silvio.Falconi,

Maybe I can help you. What do you mean with cr?

Pieter

[Silvio.Falconi]

Characters

[pieter]

Oké,

"but I not Know the len of the cmessage..."
With LEN() you can calculated the length of a string
Local cMessage:= "12345678901234567890123456789012345678901234567890123456789012345678901234567890"
MsgInfo(LEN( cMessage ) ) //give 80

"for a sample if there are 21 cr on cmessage the atextlinea must be 1"
I am not sure if I understand this. Does it mean that if cmessage has 21 characters, the length of atexlinea array has to be 1? With the 21th character from cmessage on the first position of array atextlinea?

Pieter

[Silvio.Falconi]

yes I know len function
But U perhaps not understood what I wish.
I explain you
I have a string of n character (I not Know the Len of this string)
I wish create a simply array with a string of 20 character each
sample :
aadd( atext,"12345678901234567890")
aadd( atext,"12345678901234567890")
aadd( atext,"12345678901234567890")

the strings must be 20 charactes
for a sample : I hace a string
len( cstring) := 41
I must create 3 record ( 2 of 20 cr and 1 of 1 cr)
but this one must be of 1 cr + space(19)
allway 20 cr

[Silvio.Falconi]

for a sample :

if len(cMessage)==40
aadd(atext,Left( cmessage, 20 ))
aadd(atext,right(cMessage,20))
else
cTxt := Left( cmessage, 20 )
cTxt := cTxt + Space( 20 - Len( cTxt)
endif

but not run ok because I can have a big string

[joseluisysturiz]

yes I know len function
But U perhaps not understood what I wish.
I explain you
I have a string of n character (I not Know the Len of this string)
I wish create a simply array with a string of 20 character each
sample :
aadd( atext,"12345678901234567890")
aadd( atext,"12345678901234567890")
aadd( atext,"12345678901234567890")

the strings must be 20 charactes
for a sample : I hace a string
len( cstring) := 41
I must create 3 record ( 2 of 20 cr and 1 of 1 cr)
but this one must be of 1 cr + space(19)
allway 20 cr

Creo que si cada vez que tomes los primeros 20 usas una VAR auxiliar donde vas restando los ya usados, esa VAR te controlara el tamaño restante, cuando te queden 21 el Len(VarAux) sera igual a 21, luego Len(Var_Aux) sera 1, luego creo podrias usar la funcion PAD para el relleno de espacios, es la idea que se me viene a mente...saludos...

[Carlos Mora]

Code: Select all | Expand

aText:= Array()
FOR i:= 1 TO Len( cMessage ) STEP 20
   aAdd( aText, PadR( SubStr( cMessage, i, 20 ), 20)  ) // Edited 
NEXT
 

[pieter]

Yes, I was also doing something with substr(), the code Carlos Mora has send works almost, but if I run this code, I see that there still is a problem at the end.
I will think about it further.

Pieter

[nageswaragunupudi]

Code: Select all | Expand

   local cMessage:= "12345678901234567890123456789012345678901234567890123456789012345678901234567890"
   local i,n
   local aTextLinea  := {}

   cMessage  := PadR( cMessage, ( n := Ceiling( Len( cMessage ) / 20 ) * 20 ) )
   for i := 1 to n step 20
      AAdd( aTextLinea, SubStr( cMessage, i, 20 ) )
   next
 

[Carlos Mora]

Yes, I was also doing something with substr(), the code Carlos Mora has send works almost, but if I run this code, I see that there still is a problem at the end.
I will think about it further.

Pieter

IF last element with less than 20 chars, Justadd sth like

Code: Select all | Expand

If Len( aTail( aText ) ) != 20
  aText[ Len(aText) ] := PadR(   aText[ Len(aText) ], 20 )
ENDIF

Or look at the EDITED source

[Silvio.Falconi]

thanks to all ...I'm trying your samples

[pieter]

Sorry, the code of Carlos seems to work now. (I don't know what happend before exactly).

Pieter

[pieter]

Code from Carlos (I only changed aText:= Array() to atext := { })

Local atext := { }
Local cMessage:= "abcdefghijklmnopqrstuvwxyz1234567890abcdefg"
MsgInfo(LEN( cMessage ) )// gives 43
FOR i:= 1 TO Len( cMessage ) STEP 20
aadd( aText, SubStr( cMessage, i, 20 ) )
NEXT
MsgInfo(aText[1])
MsgInfo(aText[2])
MsgInfo(aText[3])

[Silvio.Falconi]

OK IT RUN OK THANKS