在Delphi 5中是否有任何方法可以将字符串转换为TDateTime，您可以在其中指定要使用的实际格式？
我正在研究一个工作处理器，它接受来自不同工作站的任务。这些任务有一系列参数，其中一些是日期，但是（不幸的是，我的控制之外）它们作为字符串传递。由于作业可以来自不同的工作站，因此用于将日期格式化为字符串的实际日期时间格式可能（当然，实际的执行）也不同。
谷歌搜索，我发现的唯一快速解决方案是偷偷改变ShortDateFormat变量，然后将其恢复到原始值。由于ShortDateFormat是一个全局变量，并且我在线程环境中工作，唯一可行的方法是同步每次访问它，这是完全不可接受的（也是可撤销的）。
我可以将SysUtils单元中的库代码复制到我自己的方法中，并调整它们以使用指定的格式而不是全局变量，但我只是想知道是否有更充分的东西我错过了
亲切的问候，并提前感谢，
威廉
更新
更简洁地说：
我需要类似StrToDate（或StrToDateTime）的内容，并添加了选项，指定将字符串转换为TDateTime时应使用的确切格式。
答案 0 :(得分：28)
请改用VarToDateTime。它支持字符串中的更多日期格式并自动转换它们。
var
DateVal: TDateTime;
begin
DateVal := VarToDateTime('23 Sep 2010');
ShowMessage(DateToStr(DateVal));
end;
我看到你正在使用Delphi 5.某些版本的Delphi需要将variants添加到uses子句中;大多数更高版本为您添加它。我不记得Delphi 5属于哪个类别。
答案 1 :(得分：8)
我为FreePascal的dateutils单元创建了这样的例程，如果需要移植，它应该很容易移植。
代码：
（代码是文件末尾的最后一个（巨大的）程序）
文件：
http://www.freepascal.org/docs-html/rtl/dateutils/scandatetime.html
请注意，它不是formatdatetime的完全反转，它有一些扩展名：
FormatDateTime的反转不是100％反向，只是因为可以放置例如格式字符串中的时间标记两次，scandatetime不知道选择哪个时间。
像hn这样的字符串不能安全地反转。例如。 1：2（1分钟后2分钟）发送12，解析为12:00然后 错过了“n”部分的字符。
扩展
（我相信这些评论在某种程度上已经过时了一些亚洲支持，但我不确定）
答案 2 :(得分：7)
Delphi的更高版本可以为字符串转换函数提供额外的TFormatSettings参数。 TFormatSettings是一个包含各种格式全局变量（ShortDateFormat，LongDateFormat等）的结构。因此，您可以以线程安全的方式覆盖这些值，甚至可以单次调用。
我不记得引入了哪个版本的Delphi，但我很确定它是在Delphi 5之后。
所以是的，据我所知，您需要同步对ShortDateFormat的每次访问，或者使用不同的函数。
答案 3 :(得分：6)
这是函数及其两个帮助程序，我编写了使用 完全 日期时间格式解析字符串。而且由于Stackoverflow也是代码维基：这里是所有代码：
class function TDateTimeUtils.TryStrToDateExact(const S, DateFormat: string; PivotYear: Integer;
out Value: TDateTime): Boolean;
var
Month, Day, Year: Integer;
Tokens: TStringDynArray;
CurrentToken: string;
i, n: Integer;
Partial: string;
MaxValue: Integer;
nCurrentYear: Integer;
function GetCurrentYear: Word;
var
y, m, d: Word;
begin
DecodeDate(Now, y, m, d);
Result := y;
end;
begin
Result := False;
{
M/dd/yy
Valid pictures codes are
d Day of the month as digits without leading zeros for single-digit days.
dd Day of the month as digits with leading zeros for single-digit days.
ddd Abbreviated day of the week as specified by a LOCALE_SABBREVDAYNAME* value, for example, "Mon" in English (United States).
Windows Vista and later: If a short version of the day of the week is required, your application should use the LOCALE_SSHORTESTDAYNAME* constants.
dddd Day of the week as specified by a LOCALE_SDAYNAME* value.
M Month as digits without leading zeros for single-digit months.
MM Month as digits with leading zeros for single-digit months.
MMM Abbreviated month as specified by a LOCALE_SABBREVMONTHNAME* value, for example, "Nov" in English (United States).
MMMM Month as specified by a LOCALE_SMONTHNAME* value, for example, "November" for English (United States), and "Noviembre" for Spanish (Spain).
y Year represented only by the last digit.
yy Year represented only by the last two digits. A leading zero is added for single-digit years.
yyyy Year represented by a full four or five digits, depending on the calendar used. Thai Buddhist and Korean calendars have five-digit years. The "yyyy" pattern shows five digits for these two calendars, and four digits for all other supported calendars. Calendars that have single-digit or two-digit years, such as for the Japanese Emperor era, are represented differently. A single-digit year is represented with a leading zero, for example, "03". A two-digit year is represented with two digits, for example, "13". No additional leading zeros are displayed.
yyyyy Behaves identically to "yyyy".
g, gg Period/era string formatted as specified by the CAL_SERASTRING value.
The "g" and "gg" format pictures in a date string are ignored if there is no associated era or period string.
PivotYear
The maximum year that a 1 or 2 digit year is assumed to be.
The Microsoft de-factor standard for y2k is 2029. Any value greater
than 29 is assumed to be 1930 or higher.
e.g. 2029:
1930, ..., 2000, 2001,..., 2029
If the PivotYear is between 0 and 99, then PivotYear is assumed to be
a date range in the future. e.g. (assuming this is currently 2010):
Pivot Range
0 1911..2010 (no future years)
1 1912..2011
...
98 2009..2108
99 2010..2099 (no past years)
0 ==> no years in the future
99 ==> no years in the past
}
if Length(S) = 0 then
Exit;
if Length(DateFormat) = 0 then
Exit;
Month := -1;
Day := -1;
Year := -1;
Tokens := TDateTimeUtils.TokenizeFormat(DateFormat);
n := 1; //input string index
for i := Low(Tokens) to High(Tokens) do
begin
CurrentToken := Tokens[i];
if CurrentToken = 'MMMM' then
begin
//Long month names, we don't support yet (you're free to write it)
Exit;
end
else if CurrentToken = 'MMM' then
begin
//Short month names, we don't support yet (you're free to write it)
Exit;
end
else if CurrentToken = 'MM' then
begin
//Month, with leading zero if needed
if not ReadDigitString(S, n, 2{MinDigits}, 2{MaxDigits}, 1{MinValue}, 12{MaxValue}, {var}Month) then Exit;
end
else if CurrentToken = 'M' then
begin
//months
if not ReadDigitString(S, n, 1{MinDigits}, 2{MaxDigits}, 1{MinValue}, 12{MaxValue}, {var}Month) then Exit;
end
else if CurrentToken = 'dddd' then
begin
Exit; //Long day names, we don't support yet (you're free to write it)
end
else if CurrentToken = 'ddd' then
begin
Exit; //Short day names, we don't support yet (you're free to write it);
end
else if CurrentToken = 'dd' then
begin
//If we know what month it is, and even better if we know what year it is, limit the number of valid days to that
if (Month >= 1) and (Month <= 12) then
begin
if Year > 0 then
MaxValue := MonthDays[IsLeapYear(Year), Month]
else
MaxValue := MonthDays[True, Month]; //we don't know the year, assume it's a leap year to be more generous
end
else
MaxValue := 31; //we don't know the month, so assume it's the largest
if not ReadDigitString(S, n, 2{MinDigits}, 2{MaxDigits}, 1{MinValue}, MaxValue{MaxValue}, {var}Day) then Exit;
end
else if CurrentToken = 'd' then
begin
//days
//If we know what month it is, and even better if we know what year it is, limit the number of valid days to that
if (Month >= 1) and (Month <= 12) then
begin
if Year > 0 then
MaxValue := MonthDays[IsLeapYear(Year), Month]
else
MaxValue := MonthDays[True, Month]; //we don't know the year, assume it's a leap year to be more generous
end
else
MaxValue := 31; //we don't know the month, so assume it's the largest
if not ReadDigitString(S, n, 1{MinDigits}, 2{MaxDigits}, 1{MinValue}, MaxValue{MaxValue}, {var}Day) then Exit;
end
else if (CurrentToken = 'yyyy') or (CurrentToken = 'yyyyy') then
begin
//Year represented by a full four or five digits, depending on the calendar used.
{
Thai Buddhist and Korean calendars have five-digit years.
The "yyyy" pattern shows five digits for these two calendars,
and four digits for all other supported calendars.
Calendars that have single-digit or two-digit years, such as for
the Japanese Emperor era, are represented differently.
A single-digit year is represented with a leading zero, for
example, "03". A two-digit year is represented with two digits,
for example, "13". No additional leading zeros are displayed.
}
if not ReadDigitString(S, n, 4{MinDigits}, 4{MaxDigits}, 0{MinValue}, 9999{MaxValue}, {var}Year) then Exit;
end
else if CurrentToken = 'yyy' then
begin
//i'm not sure what this would look like, so i'll ignore it
Exit;
end
else if CurrentToken = 'yy' then
begin
//Year represented only by the last two digits. A leading zero is added for single-digit years.
if not ReadDigitString(S, n, 2{MinDigits}, 2{MaxDigits}, 0{MinValue}, 99{MaxValue}, {var}Year) then Exit;
nCurrentYear := GetCurrentYear;
Year := (nCurrentYear div 100 * 100)+Year;
if (PivotYear < 100) and (PivotYear >= 0) then
begin
//assume pivotyear is a delta from this year, not an absolute value
PivotYear := nCurrentYear+PivotYear;
end;
//Check the pivot year value
if Year > PivotYear then
Year := Year - 100;
end
else if CurrentToken = 'y' then
begin
//Year represented only by the last digit.
if not ReadDigitString(S, n, 1{MinDigits}, 1{MaxDigits}, 0{MinValue}, 9{MaxValue}, {var}Year) then Exit;
nCurrentYear := GetCurrentYear;
Year := (nCurrentYear div 10 * 10)+Year;
if (PivotYear < 100) and (PivotYear >= 0) then
begin
//assume pivotyear is a delta from this year, not an absolute value
PivotYear := nCurrentYear+PivotYear;
end;
//Check the pivot year value
if Year > PivotYear then
Year := Year - 100;
end
else
begin
//The input string should contains CurrentToken starting at n
Partial := Copy(S, n, Length(CurrentToken));
Inc(n, Length(CurrentToken));
if Partial <> CurrentToken then
Exit;
end;
end;
//If there's still stuff left over in the string, then it's not valid
if n <> Length(s)+1 then
begin
Result := False;
Exit;
end;
if Day > MonthDays[IsLeapYear(Year), Month] then
begin
Result := False;
Exit;
end;
try
Value := EncodeDate(Year, Month, Day);
except
Result := False;
Exit;
end;
Result := True;
end;
class function TDateTimeUtils.TokenizeFormat(fmt: string): TStringDynArray;
var
i: Integer;
partial: string;
function IsDateFormatPicture(ch: AnsiChar): Boolean;
begin
case ch of
'M','d','y': Result := True;
else Result := False;
end;
end;
begin
SetLength(Result, 0);
if Length(fmt) = 0 then
Exit;
//format is only one character long? If so then that's the tokenized entry
if Length(fmt)=1 then
begin
SetLength(Result, 1);
Result[0] := fmt;
end;
partial := fmt[1];
i := 2;
while i <= Length(fmt) do
begin
//If the characters in partial are a format picture, and the character in fmt is not the same picture code then write partial to result, and reset partial
if IsDateFormatPicture(partial[1]) then
begin
//if the current fmt character is different than the running partial picture
if (partial[1] <> fmt[i]) then
begin
//Move the current partial to the output
//and start a new partial
SetLength(Result, Length(Result)+1);
Result[High(Result)] := partial;
Partial := fmt[i];
end
else
begin
//the current fmt character is more of the same format picture in partial
//Add it to the partial
Partial := Partial + fmt[i];
end;
end
else
begin
//The running partial is not a format picture.
//If the current fmt character is a picture code, then write out the partial and start a new partial
if IsDateFormatPicture(fmt[i]) then
begin
//Move the current partial to the output
//and start a new partial
SetLength(Result, Length(Result)+1);
Result[High(Result)] := partial;
Partial := fmt[i];
end
else
begin
//The current fmt character is another non-picture code. Add it to the running partial
Partial := Partial + fmt[i];
end;
end;
Inc(i);
Continue;
end;
//If we have a running partial, then add it to the output
if partial <> '' then
begin
SetLength(Result, Length(Result)+1);
Result[High(Result)] := partial;
end;
end;
class function TDateTimeUtils.ReadDigitString(const S: string; var Pos: Integer;
MinDigits, MaxDigits: Integer; MinValue, MaxValue: Integer;
var Number: Integer): Boolean;
var
Digits: Integer;
Value: Integer;
Partial: string;
CandidateNumber: Integer;
CandidateDigits: Integer;
begin
Result := False;
CandidateNumber := -1;
CandidateDigits := 0;
Digits := MinDigits;
while Digits <= MaxDigits do
begin
Partial := Copy(S, Pos, Digits);
if Length(Partial) < Digits then
begin
//we couldn't get all we wanted. We're done; use whatever we've gotten already
Break;
end;
//Check that it's still a number
if not TryStrToInt(Partial, Value) then
Break;
//Check that it's not too big - meaning that getting anymore wouldn't work
if (Value > MaxValue) then
Break;
if (Value >= MinValue) then
begin
//Hmm, looks good. Keep it as our best possibility
CandidateNumber := Value;
CandidateDigits := Digits;
end;
Inc(Digits); //try to be greedy, grabbing even *MORE* digits
end;
if (CandidateNumber >= 0) or (CandidateDigits > 0) then
begin
Inc(Pos, CandidateDigits);
Number := CandidateNumber;
Result := True;
end;
end;
答案 4 :(得分：4)
如果你想知道在后来的Delphi中如何解决这个问题，你可以在这里看一下稍微更现代的东西（看起来像Delphi 6） sysutils.pas ：
http://anygen.googlecome.com/.../SysUtils.pas
查看带有StrToDateTime参数的TFormatSettings的重载版本。
function StrToDateTime(const S: string;
const FormatSettings: TFormatSettings): TDateTime; overload;
答案 5 :(得分：1)
我不确定你想要什么。我不再使用Delphi 5了，但我很确定StrToDateTime函数存在于其中。使用它，您可以使用格式设置将字符串转换为TDateTime。然后，您可以使用FormatDateTime将此类TDateTime转换为任何格式，使您可以使用任何您希望的日期格式。
答案 6 :(得分：1)
使用RegExpr库（https://github.com/masterandrey/TRegExpr）
var
RE: TRegExpr;
begin
RE := TRegExpr.Create;
try
RE.Expression := '^(\d\d\d\d)/(\d\d)/(\d\d)T(\d\d):(\d\d):(\d\d)$';
if RE.Exec( Value ) then
begin
try
Result := EncodeDate( StrToInt( RE.Match[1] ),
StrToInt( RE.Match[2] ),
StrToInt( RE.Match[3] ) ) +
EncodeTime( StrToInt( RE.Match[4] ),
StrToInt( RE.Match[5] ),
StrToInt( RE.Match[6] ),
0 )
except
raise EConvertError.Create( 'Invalid date-time: ' + Value )
end
end
else
raise EConvertError.Create( 'Bad format: ' + Value )
finally
RE.Free
end
end;
答案 7 :(得分：0)
我会绕过它。在我看来，你有两个选择，你自己提到了
TDateTime。我很想知道你将如何确定04/05/2010的格式。
program DateTimeConvert;
{$APPTYPE CONSOLE}
uses
SysUtils;
function GetPart(const part, input, format: string): string;
var
I: Integer;
begin
for I := 1 to Length(format) do
if Uppercase(format[I]) = Uppercase(part) then
Result := Result + input[I];
end;
function GetDay(const input, format: string): string;
begin
Result := GetPart('d', input, format);
if Length(Result) = 1 then Result := SysUtils.Format('0%0:s', [Result]);
end;
function GetMonth(const input, format: string): string;
begin
Result := GetPart('m', input, format);
if Length(Result) = 1 then Result := SysUtils.Format('0%0:s', [Result]);
end;
function GetYear(const input, format: string): string;
begin
Result := GetPart('y', input, format);
end;
function ConvertToMyLocalSettings(const input, format: string): string;
begin
Result := SysUtils.Format('%0:s/%1:s/%2:s', [GetDay(input, format), GetMonth(input, format), GetYear(input, format)]);
end;
begin
Writeln(ConvertToMyLocalSettings('05/04/2010', 'dd/mm/yyyy'));
Writeln(ConvertToMyLocalSettings('05-04-2010', 'dd-mm-yyyy'));
Writeln(ConvertToMyLocalSettings('5-4-2010', 'd-m-yyyy'));
Writeln(ConvertToMyLocalSettings('4-5-2010', 'm-d-yyyy'));
Writeln(ConvertToMyLocalSettings('4-05-2010', 'M-dd-yyyy'));
Writeln(ConvertToMyLocalSettings('05/04/2010', 'dd/MM/yyyy'));
Readln;
end.