在delphi中从TMemo中搜索和删除

Question

I have a code, to search and delete from memo lines. but there is a little problem that, for example, memo lines contains 1, 11, 12, 13, 21, 22 etc. when e=1 it removes all lines which contains 1. I need to delete only defined search (e=1)

  for i := 0 to memo3.lines.count-1 do
  begin
    if (pos(IntToStr(e), Memo3.Lines[i]) > 0) then begin
      Memo3.Lines.Delete(i);  
      Memo3.Lines.Delimiter := '-';
      Memo3.Lines.StrictDelimiter := True;
      t:= Memo3.Lines.DelimitedText;
      Label5.Caption:=t;
    end;

Answer 1

Checking only for the string

If you only want to find the string in isolation, then you'll have to look if before and after the string are no digits:

NumStr := IntToStr(e);
Str := Memo3.Lines[i];
NumPos := Pos(NumStr, Str);
if NumPos > 0 then
begin
  if (NumPos > 1) and IsDigit(Str[NumPos - 1])) or
     (NumPos < Length(Str)) and IsDigit(Str[NumPos + 1])) then
    Continue; // i.e. skip deleting etc.

Alternatively, you try to find spaces, tabs, etc. around the NumStr you find and only delete if you find that the number is a single "word" in that string.

One number per line?

Now if your TMemo only contains exactly one single number per line , then things are much easier, and you don't need Pos() at all:

NumStr := IntToStr(e);
for i := Memo3.Lines.Count - 1 downto 0 do
begin
  if NumStr = Memo3.Lines[i] then
  begin
    Memo3.Lines.Delete(i);
    ...
  end;
end;

Note that I didn't repeat the call to IntToStr() for each iteration of the loop. I Just did this once and assigned the result of the call to NumStr . Function calls take time.

Deleting in a loop

If you want to delete from an indexed list of items, like the Lines property of a TMemo , then, in order not to skip any lines, always loop backwards, like I do above. Because if you delete line i , all lines after i will move one index down, so when you go to line i+1 , that will be the second to next line , not the next line ( because that got index i , after the deletion ).

But when you loop backwards, the previous line will be i-1 , and that index did not change.

Simple example:

Original situation:

index: text
0:     A
1:     B
2:     C  <-- delete!
3:     D
4:     E

After deletion:

0:     A
1:     B
2:     D <-- now at index 2, was at index 3
3:     E <-- now at index 3, was at index 4

If you delete C at index 2, then D and E go down one index, and now D is at index 2. But the upward loop increments i to 3, so now you inspect line 3, and D is never checked . But if you go downward, then i becomes 1, and that didn't change index and still contains B .

Answer 2

When changing the lines in the memo by deleting you cannot loop from 0 to count - 1 because the index changes everytime you delete a line

This loop will delete all lines that contain the value in e

for i := Memo3.Lines.Count - 1 downto 0 do
begin
  if pos(e, Memo3.Lines[i]) > 0 then
  begin
    Memo3.Lines.Delete(i);
  end;
end;

I have however no clue what the other lines of code in your sample should do, maybe you can elaborate what you are trying to achieve with that.

EDIT
If you want to delete only rows that match the value in e then use this loop

Value := IntToStr(e);
for i := Memo1.Lines.Count - 1 downto 0 do
begin
  if Value = Trim(Memo1.Lines[i]) then
  begin
    Memo1.Lines.Delete(i);
  end;
end;