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我对 exiv2 进行了一些修改，试图在 EXIF、IPTC 和 XMP 之间扩展它的日期/时间转换功能。http://dev.exiv2.org/issues/864

IPTC 日期和 IPTC 时间存储在不同的字段中，所以首先我尝试解析 IPTC 日期，稍后在我的代码中我尝试解析时间。

我是编程的初级初学者，所以这看起来是一个非常基本的问题，但这里有。该代码被添加到 cnvIptcValue 函数内的 exiv2 文件 convert.cpp 中。

我从命令行得到的确切错误：

$ exiv2 -eiX iptc.jpg 
Warning: Failed to convert Iptc.Application2.DateCreated to Xmp.photoshop.DateCreated, unable to parse '1944-09-08'
Warning: Failed to convert Iptc.Application2.DigitizationDate to Xmp.xmp.CreateDate, unable to parse '2015-03-06'

起初我认为日期转换代码失败：

if (std::string(from) == "Iptc.Application2.DateCreated" || std::string(from) == "Iptc.Application2.DigitizationDate")  {

int year, month, day, hour, min, sec, tzHour, tzMinute;
char tzSign;
char buf[30];


    if (sscanf(value.c_str(), "%d-%d-%d", &year, &month, &day) != 3) {

    EXV_WARNING << "Failed to convert " << from << " to " << to
                << ", unable to parse '" << value << "'\n";
    return;
}

但进一步的测试表明，失败的是时间转换代码：

    const char* iptcTime = 0;  
if (std::string(from) == "Iptc.Application2.DateCreated") {
    iptcTime = "Iptc.Application2.TimeCreated";
}
else if (std::string(from) == "Iptc.Application2.DigitizationDate") {
    iptcTime = "Iptc.Application2.DigitizationTime";
}
if (iptcTime) {              
    if (sscanf(value.c_str(), "%d:%d:%d%1c%d:%02d", &hour, &min, &sec, &tzSign, &tzHour, &tzMinute) != 6) {

        EXV_WARNING << "Failed to convert " << from << " to " << to
                    << ", unable to parse '" << value << "'\n";

        return;
    }

所以时间代码试图解析日期字段。通过添加几行代码将其更改为时间字段：

    Exiv2::IptcData::iterator iptctime_pos = iptcData_->findKey(IptcKey(from));
    std::string value = iptctime_pos->toString();

我还扩展了错误消息代码以查看时间字段中的内容。现在的结果是：

Warning: Failed to convert Iptc.Application2.TimeCreated to Xmp.photoshop.DateCreated, unable to parse '11:11:14+00:00'
Year Month Day:          1944 9 8
Hour Minute Second:      32767 36913992 1978972555
Zone Sign Hour Minute:   "U+007F"  -184143665 6457088
Warning: Failed to convert Iptc.Application2.DigitizationTime to Xmp.xmp.CreateDate, unable to parse '16:26:47+00:00'
Year Month Day:          2015 3 6
Hour Minute Second:      32767 36909480 36909144
Zone Sign Hour Minute:   "U+007F"  -184143665 6457088

它正确解析日期字段，但不正确解析时间。为什么是这样？是不是解析错了？

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1 回答 1

0

解决了！需要进行两项更改。

首先是确保扫描与之前设置的相同的密钥。我的试错尝试让我从“iptctime”而不是“value”进行扫描，这导致了解析错误。

std::string value = iptctime_pos->toString();

if (sscanf(value.c_str(), "%d:%d:%d%1c%d:%d", &hour, &min, &sec, &tzSign, &tzHour, &tzMinute) != 6) {

其次是添加一个else if防止 DateTime 属性被通用分配代码重写的功能。

完整的工作结果是：

if (std::string(from) == "Iptc.Application2.DateCreated" || std::string(from) == "Iptc.Application2.DigitizationDate")  {

    int year, month, day, hour, min, sec, tzHour, tzMinute;
    char tzSign;
    std::string iptctime;
    char buf[30];


    (sscanf(value.c_str(), "%d-%d-%d", &year, &month, &day)

    const char* iptcTime = 0;  
    if (std::string(from) == "Iptc.Application2.DateCreated") {
        iptcTime = "Iptc.Application2.TimeCreated";
    }
    else if (std::string(from) == "Iptc.Application2.DigitizationDate") {
        iptcTime = "Iptc.Application2.DigitizationTime";
    }
    if (iptcTime) {              
        Exiv2::IptcData::iterator iptctime_pos = iptcData_->findKey(IptcKey(iptcTime));
        if (iptctime_pos == iptcData_->end()) return;
            if (iptctime_pos->key() == iptcTime) {
                std::string value = iptctime_pos->toString();

                sscanf(value.c_str(), "%d:%d:%d%1c%d:%d", &hour, &min, &sec, &tzSign, &tzHour, &tzMinute)
        }
    }
    snprintf(buf, sizeof(buf), "%04d-%02d-%02dT%02d:%02d:%02d%1c%02d:%02d",
         year, month, day, hour, min, sec, tzSign, tzHour, tzMinute);
    buf[sizeof(buf) - 1] = 0;
    (*xmpData_)[to] = buf;
}

else if (std::string(from) != "Iptc.Application2.DateCreated" || std::string(from) != "Iptc.Application2.DigitizationDate")  {

        (*xmpData_)[to] = value;
}

进行更多测试，我会将其上传到 exiv2 存储库，以在即将发布的 exiv2 0.25 版本中提供！

PS：感谢@tivn 的帮助。

于 2015-04-13T01:51:55.957 回答