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c++ - 使用 Boost 更好的 XML 格式化?

转载 作者:行者123 更新时间:2023-11-30 03:59:02 25 4
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我正在使用 Boost Property Trees 将我的类实例导出为 XML 节点。它有效,但它只是将所有内容放在 1 行中。我希望它有缩进,例如:

<?xml version="1.0" encoding="utf-8"?>
<root>
<sensorconfigurations>
<configuration>
<name>SensorConfiguration1</name>
<sensorid>1</sensorid>
<signalindex>1</signalindex>
<mappingscheme>mappingscheme1</mappingscheme>
<soundpack>test1.wav</soundpack>
</configuration>
<configuration>
<name>SensorConfiguration2</name>
<sensorid>2</sensorid>
<signalindex>2</signalindex>
<mappingscheme>mappingscheme1</mappingscheme>
<soundpack>test2.wav</soundpack>
</configuration>
<configuration>
<name>SensorConfiguration3</name>
<sensorid>3</sensorid>
<signalindex>3</signalindex>
<mappingscheme>mappingscheme2</mappingscheme>
<soundpack>test3.wav</soundpack>
</configuration>
</sensorconfigurations>
</root>

这有可能吗?我是否缺少 write_xml 方法中的参数?

这是我的代码:

void SensorConfigurationBank::save()
{
using boost::property_tree::ptree;
ptree pt;
for(map<string, SensorConfiguration>:: iterator it = sensorConfigurations_.begin(); it != sensorConfigurations_.end(); ++it)
{
ptree myTree;
myTree.put("name", it->second.getName());
myTree.put("sensorid", it->second.getSensorID());
myTree.put("signalindex", it->second.getsignalIndex());
MappingScheme myScheme = it->second.getMScheme();
myTree.put("mappingscheme", myScheme.getId());
SoundPack mySound = it->second.getSound();
myTree.put("soundpack", mySound.filepath_);

pt.add_child("root.sensorconfigurations.configuration", myTree);
}
write_xml("SensorConfigurationBank2.xml", pt);
}

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