Procedure Delete removes a substring from the specified source string.
PROCEDURE Delete(
VAR source : DYNARRAY[] of CHAR;
index : INTEGER;
count : INTEGER);def vs.Delete(source, index, count):
return source| Name | Type | Description |
|---|---|---|
| source | DYNARRAY[] of CHAR | Source string. |
| index | INTEGER | Start position in text string. |
| count | INTEGER | Length of substring. |
Per Raymond Mullin, on the VS list, there is a bug in Delete that prevents it from getting the last character if the string is a dynarray of char. The following test script should leave T = '', but it leaves T = '7'.
PROCEDURE DeleteTest;
VAR
T :DYNARRAY [] of CHAR;
BEGIN
T := '1234567';
Delete(T, 1, 7);
Message('T: ', T);
END;
RUN(DeleteTest);theStr:='A sample string';
Delete(theStr,3,7);
{deletes 'sample' from the string value}def DeleteTest():
T = '1234567'
T = vs.Delete(T, 1, 7)
vs.Message('T: ', T)
DeleteTest()prefix := '';
while (not ValidNumStr(Copy(inBearingStr, 1, 1), num)) & (Len(inBearingStr) > 0) do BEGIN
prefix := Concat(prefix, Copy(inBearingStr, 1, 1));
Delete(inBearingStr, 1, 1);
format := 'bearing';
END;
BEGIN
IF bConvKeyNote & GetKeyNoteData( textFoundH ) then BEGIN
if Copy( KNprefix, Len(KNprefix), 1) = ' ' then BEGIN
Delete(KNprefix, Len(KNprefix), 1);
if ( KNsuffix = '' ) & ( KNprefix = GetText( textFoundH ) ) then BEGIN
IF not ValidNumStr( GetText( textFoundH ), t_real ) THEN BEGIN
TextOrigin(0,0);
CreateText(Concat(' ', KNNoteNo));
BEGIN
{Returns the date string up to the third space character}
pos1 := Pos (' ', input);
dummy1 := Copy(input, 1, pos1);
Delete(input, 1, pos1);
pos1 := Pos (' ', input);
dummy1 := concat(dummy1, Copy(input, 1, pos1));
Delete(input, 1, pos1);
pos1 := Pos (' ', input);import vs
# Procedure Delete removes a substring from the specified source string.
source = 'Example'
index = 1
count = 5
result = vs.Delete(source, index, count)Availability: from All Versions