Check if char string is integer
- Thread starter nate100
- Start date
Code:
DEF VAR vNumber AS INT.
DEF VAR vCharString AS CHAR INITIAL "134343".
ASSIGN vNumber = INT(vCharString) NO-ERROR.
IF ERROR-STATUS:ERROR THEN
MESSAGE "Character String is not an integer." VIEW-AS ALERT-BOX INFORMATION.
ELSE
MESSAGE "Character String is an integer." VIEW-AS ALERT-BOX INFORMATION.KnutHandsome
Member
This approach has a couple of obscure gotchas, eg.
try it with vCharString = '-'.
Best way is to parse and check each character, though you may be able to use the above method if you trim out certain characters first.
Code:
function myinteger returns integer (str as char):
define variable str2 as character no-undo.
define variable i as integer no-undo.
str= trim(str).
do i = 1 to length(str):
str2 = substring(str,i,1).
if str2 = '-' then do:
if i <> 1 then
return ?.
else
next.
end.
if lookup(str2, '0,1,2,3,4,5,6,7,8,9,0') = 0 then
return ?.
end.
return integer(str).
end function.If function returns ? then input is not integer.
If none of the pieces of code given above work ... you can probably use the ASCII value .....
ASC('0') = 48
ASC('1) = 49
--
--
--
--
--
--
--
ASC('9') = 57.
This is a fool proof way to check whether the characters in your string are all integers or if there is a special character in there . Because Error-Status('-') will also cause an error.
Cheers
Sunil.
ASC('0') = 48
ASC('1) = 49
--
--
--
--
--
--
--
ASC('9') = 57.
This is a fool proof way to check whether the characters in your string are all integers or if there is a special character in there . Because Error-Status('-') will also cause an error.
Cheers
Sunil.
KnutHandsome
Member
Nice, but you arguably need a trim.
MESSAGE isinteger(' 90')
VIEW-AS ALERT-BOX INFO BUTTONS OK.
And what about
MESSAGE isinteger(?)
VIEW-AS ALERT-BOX INFO BUTTONS OK.
KnutHandsome
Member
MESSAGE isinteger(' 90')
VIEW-AS ALERT-BOX INFO BUTTONS OK.
It depend if leading space sholud be interpreted as correct integer or not.
MESSAGE isinteger(?)
VIEW-AS ALERT-BOX INFO BUTTONS OK.
Yes , you right, unknown value shold be treated.
KnutHandsome
Member
That is why I rather weaselly included 'arguably' in my post, but I suspect a user entering ' 90' in an input field would be suprised if it was rejected as a number.
Though some difficult types might argue that ? is a valid member of the Progress integer set.:read:
Not me though. Well, not today.
KnutHandsome
Member
Using cibulkas method, how about the unreadable:
I was trying to get it down to a geeky one line, but couldn't bypass errors 76 and 77, hence the variable and assign. Perhaps someone else can...
What was that I was saying in another thread about time well spent?
Code:
FUNCTION isInteger RETURNS LOGICAL( cInteger AS CHARACTER) :
DEF VAR lTest AS LOGICAL NO-UNDO.
ASSIGN lTest = (cInteger NE ?) AND
(STRING(INTEGER(cInteger)) EQ TRIM(cInteger)) NO-ERROR.
RETURN lTest.
END FUNCTION.I was trying to get it down to a geeky one line, but couldn't bypass errors 76 and 77, hence the variable and assign. Perhaps someone else can...
What was that I was saying in another thread about time well spent?
I've add LEFT-TRIM
FUNCTION isInteger RETURNS LOGICAL( cInteger AS CHARACTER) :
DEF VAR lTest AS LOGICAL NO-UNDO.
ASSIGN lTest = (cInteger NE ?) AND
(STRING(INTEGER(cInteger)) EQ LEFT-TRIM(TRIM(cInteger), "0":U)) NO-ERROR.
RETURN lTest.
END FUNCTION.
because of
MESSAGE isInteger(" 0123")
VIEW-AS ALERT-BOX INFO BUTTONS OK.
KnutHandsome
Member
KnutHandsome
Member
Code:
function myinteger returns integer (str as char):
define variable str2 as character no-undo.
define variable i as integer no-undo.
str= trim(str).
if str = '-' or str = '' then
return ?.
do i = 1 to length(str):
str2 = substring(str,i,1).
if str2 = '-' then do:
if i <> 1 then
return ?.
else
next.
end.
if lookup(str2, '0,1,2,3,4,5,6,7,8,9,0') = 0 then
return ?.
end.
return integer(str).
end function.Now it works with myinteger('') and myinteger(?).
KnutHandsome
Member
You have a redundant '0' in your lookup.
