I have an application where in i need to save the data input by a user in a form in an XML file at a specified location and i need to perform this using Java . I am relatively very new to XML handling in java. I would like some suggestions as to how to start the task .
我有一个应用程序,我需要将用户输入的数据保存在指定位置的XML文件中,我需要使用Java执行此操作。我对java中的XML处理相对比较新。我想就如何开始这项任务提出一些建议。
Any code snippets and links will be helpful ...
任何代码片段和链接都会有所帮助......
Thank You
谢谢
4 个解决方案
#1
17
There is very good framework JAXB for this also there is Simple
有一个非常好的框架JAXB,这也有简单
But I have used this XStream
但我使用过这个XStream
Person joe = new Person("Joe", "Walnes");
joe.setPhone(new PhoneNumber(123, "1234-456"));
joe.setFax(new PhoneNumber(123, "9999-999"));
Now, to convert it to XML, all you have to do is make a simple call to XStream:
现在,要将其转换为XML,您所要做的就是对XStream进行简单的调用:
String xml = xstream.toXML(joe);
The resulting XML looks like this:
生成的XML如下所示:
<person>
<firstname>Joe</firstname>
<lastname>Walnes</lastname>
<phone>
<code>123</code>
<number>1234-456</number>
</phone>
<fax>
<code>123</code>
<number>9999-999</number>
</fax>
</person>
Also See
另见
- JAXB
- JAXB
- where-i-can-find-a-detailed-comparison-of-java-xml-frameworks
- 其中-I-可以找到的-A-详细对比-的-Java的XML的框架
#2
1
There are many open source libraries, but I would simply use JAXB, the standard. Although I have to say the XStream library suggested by other answerers looks very promising, too!
有许多开源库,但我只是使用标准的JAXB。虽然我不得不说其他回答者建议的XStream库看起来也很有前途!
#3
#4
0
Consider using xstream (http://x-stream.github.io/). XStream's API is very simple:
考虑使用xstream(http://x-stream.github.io/)。 XStream的API非常简单:
YourObjectGraph yourData=buildYourData();
XStream xstream=new XStream();
String yourXML=xstream.toXml(yourData);
// do something with your shiny XML
Importing is just as easy:
导入同样简单:
YourObjectGraph yourData=(YourObjectGraph)xstream.fromXml(yourXml);
#1
17
There is very good framework JAXB for this also there is Simple
有一个非常好的框架JAXB,这也有简单
But I have used this XStream
但我使用过这个XStream
Person joe = new Person("Joe", "Walnes");
joe.setPhone(new PhoneNumber(123, "1234-456"));
joe.setFax(new PhoneNumber(123, "9999-999"));
Now, to convert it to XML, all you have to do is make a simple call to XStream:
现在,要将其转换为XML,您所要做的就是对XStream进行简单的调用:
String xml = xstream.toXML(joe);
The resulting XML looks like this:
生成的XML如下所示:
<person>
<firstname>Joe</firstname>
<lastname>Walnes</lastname>
<phone>
<code>123</code>
<number>1234-456</number>
</phone>
<fax>
<code>123</code>
<number>9999-999</number>
</fax>
</person>
Also See
另见
- JAXB
- JAXB
- where-i-can-find-a-detailed-comparison-of-java-xml-frameworks
- 其中-I-可以找到的-A-详细对比-的-Java的XML的框架
#2
1
There are many open source libraries, but I would simply use JAXB, the standard. Although I have to say the XStream library suggested by other answerers looks very promising, too!
有许多开源库,但我只是使用标准的JAXB。虽然我不得不说其他回答者建议的XStream库看起来也很有前途!
#3
1
I would start by looking at the XStream library. It's very simple to convert POJOs (plain old java objects) to and from XML. I have a blog post detailing some of the gotchas.
我将从查看XStream库开始。将POJO(普通的旧Java对象)与XML进行转换非常简单。我有一篇博文,详细介绍了一些问题。
#4
0
Consider using xstream (http://x-stream.github.io/). XStream's API is very simple:
考虑使用xstream(http://x-stream.github.io/)。 XStream的API非常简单:
YourObjectGraph yourData=buildYourData();
XStream xstream=new XStream();
String yourXML=xstream.toXml(yourData);
// do something with your shiny XML
Importing is just as easy:
导入同样简单:
YourObjectGraph yourData=(YourObjectGraph)xstream.fromXml(yourXml);