抄来的(与我的方法不一样,我是写了一个换行函数)
来宾发言:解决fastreport 中文换行乱码(2996字节) 来宾姓名:东方蜘蛛
来宾发言:
解决中文换行乱码 ，在fr_class 查找 WrapLine 过程，替换为下面的过程，就
可以解决。
//const s:string
procedure WrapLine(const s: wideString);
//解决中文换行乱码 ZWZ
var
i, cur, beg, last, LoopPos: Integer;
WasBreak, CRLF: Boolean;
begin
CRLF := False;
LoopPos := 0;
for i := 1 to Length(s)do
if (s =#10) or (s=#13) then
// if s in [#10, #13] then
begin
CRLF := True;
break;
end;
last := 1;
beg := 1;
if not CRLF and ((Length(s) 〈= 1) or (WCanvas.TextWidth(s) 〈= maxwidth)) then
OutLine(s + #1)
else
begin
cur := 1;
while cur 〈= Length(s)do
begin
if (s[cur] =#10) or (s[cur]=#13) then
// if s[cur] in [#10, #13] then
begin
OutLine(Copy(s, beg, cur - beg) + #1);
while (cur 〈 Length(s)) and
((s[cur] =#10) or (s[cur]=#13))
// (s[cur] in [#10, #13])
do
Inc(cur);
beg := cur;
last := beg;
if (s[cur] =#10) or (s[cur]=#13) then
// if s[cur] in [#13, #10] then
Exit else
continue;
end;
if s[cur] 〈> ' ' then
if WCanvas.TextWidth(Copy(s, beg, cur - beg + 1)) > maxwidth then
begin
WasBreak := False;
if (Flags and flWordBreak) 〈> 0 then
begin
i := cur;
while (i 〈= Length(s)) and
not ((s = ' ') or (s = ' ') or (s = '.') or (s = ',') or(s = '-'))do
//not (s in spaces)do
Inc(i);
b := BreakWord(Copy(s, last + 1, i - last - 1));
if Length(b) > 0 then
begin
i := 1;
cur := last;
while (i 〈= Length(b)) and
(WCanvas.TextWidth(Copy(s, beg, last - beg + 1 + Ord(b)) + '-') 〈= maxwidth)do
begin
WasBreak := True;
cur := last + Ord(b);
Inc(i);
end;
last := cur;
end;
end
else
if last = beg then
last := cur;
if WasBreak then
OutLine(Copy(s, beg, last - beg + 1) + '-')
else
if s[last] = ' ' then
OutLine(Copy(s, beg, last - beg)) else
begin
OutLine(Copy(s, beg, last - beg));
Dec(last);
end;
if ((Flags and flWordBreak) 〈> 0) and not WasBreak and (last = cur - 1) then
if LoopPos = cur then
begin
beg := cur + 1;
cur := Length(s);
break;
end
else
LoopPos := cur;
beg := last + 1;
last := beg;
end;
// if s[cur] in spaces then
last := cur;
if s[cur] = ' ' then
last := cur;
Inc(cur);
end;
if beg 〈> cur then
OutLine(Copy(s, beg, cur - beg + 1) + #1);
end;
end;
][