JAVA , Xml parsing

Question

I need some help reading an xml document.

I got a Class Person and i want create a List from that xml

the xml is something like:

<root>
<field1></field1>
<field2></field1>
<field3></field1>
<Persons>
<id></id>
<List>
<Person>
<Name>...</Name>
<LastName>...</LastName>
</Person>
<Person>
<Name>...</Name>
<LastName>...</LastName>
</Person>
<Person>
<Name>...</Name>
<LastName>...</LastName>
</Person>
</List
</Persons>
<field4></field1>
<field5></field1>
<field6></field1>

</root>

i'm using dom parser (org.w3c.dom)

Can anyone please sohw me what's the best way to get the Persons information ?

Thanks

score 1 · Answer 1 · answered Feb 16 '12 at 09:59

1

If you want only read info, you'd better (after loading DOM) use XPath on it. XPath is present in J2SE API. Write if you need special examples.

answered Feb 16 '12 at 09:59

Alumashka

11
2

score 1 · Answer 2 · answered Feb 16 '12 at 10:07

1

You have to use Simple API for XML (SAX). You may also use Streaming API for XML (StaX) (tutorial).

answered Feb 16 '12 at 10:07

KV Prajapati

92,042
19
143
183

score 1 · Answer 3 · answered Feb 16 '12 at 10:09

1

I prefer JAXB. Its also present in the J2SE API.

Write if you need help.

answered Feb 16 '12 at 10:09

ZehnVon12

3,676
3
18
23

thanks, im new to java, could u please show me a code example? Thanks for your help – ibm123 Feb 16 '12 at 16:12
Because your are new to java I add a new answer how to use regular expressions with this example. How about that? – ZehnVon12 Feb 17 '12 at 08:52
Thanks for the effort, I will review your answer – ibm123 Feb 19 '12 at 09:34

score 1 · Answer 4 · edited May 23 '17 at 10:34

I hate to just leave this here, but I answered a similar question here.

In Java you have quite a few options on actually parsing the XML - XPath will be the slowest but gives you a nice expression language to query the content with. DOM will be the second slowest but give you a tree-model in memory of your doc to walk. SAX will be faster, but requires you build the list as it parses through the doc on the fly and lastly STAX will be the fastest, but requires that you write some specific code to your format to build your list out.

Lastly, I would recommend a library I wrote called SJXP that gives you the performance of STAX with the ease of XPath... it is the perfect blend of the two.

You write rules like "/root/Persons/list/Person/Name" and give it your doc and it will fire every time it hits a name and call a user-provided callback for you, handing you the name it found.

You create a few rules for all the values you want and viola... you can create a START_TAG rule for the "/root/Persons/list/Person" open-tag, and create a new "Person p = new Person()" in your code, then as every sub-element hits, you just set the appropriate value on the person, something like this (as an example):

IRule linkRule = new DefaultRule(Type.CHARACTER, "/root/Persons/list/Person/Name") {
    @Override
    public void handleParsedCharacters(XMLParser parser, String text, Object userObject) {
        // Get the last person we added on open-tag.
        Person p = personList.get(personList.size() - 1);

        // <Name> tag was parsed, 'text' is our parsed Name. Set it.
        p.setName(text);
    }
}

The nice thing about SJXP is that the memory overhead is lower than the other parser approaches and performance higher (SAX will parse the elements on a match, STAX-based parsing doesn't parse the elements out of the stream until they are requested).

You will end up writing equally confusing code just to traverse your DOM and all the Node elements to build your list.

LASTLY, if you felt comfortable with XML->Object mapping, you could do what another person said and leverage JAXB. You will need to write a schema for your XML files, then it will generate Java objects for you that map perfect to them. Then you can just map your XML file directly to your Java object and call something like "persons.getList()" or whatever JAXB generates for you.

The memory overhead and performance will be on par with DOM parsing in that case (roughly).

JAXB does not require an XML schema. You can start from Java classes and add annotations to customize the XML representation: http://wiki.eclipse.org/EclipseLink/Examples/MOXy/GettingStarted/TheBasics — bdoughan, Feb 16 '12 at 16:49
Doh, forgot about the annots. Good correction @BlaiseDoughan. Last time I used JAXB was between it and Castor with some fairly complex schemas so I am a bit out of the loop with it. — Riyad Kalla, Feb 16 '12 at 21:40

score 0 · Answer 5 · answered Feb 16 '12 at 10:05

0

XPath is one of the solution,

if you do not want to use another library...

Than try defining the DTD and using the ID parameter, most of the parsers have getElementById(ID) funciton

answered Feb 16 '12 at 10:05

Darwly

344
1
5
22

score 0 · Answer 6 · answered Feb 17 '12 at 08:50

Another easy way is to use regular expressions:

Pattern pattern = Pattern.compile("<Person>.*?<Name>(.*?)</Name>.*?<LastName>(.*?)</LastName>.*?</Person>", Pattern.MULTILINE | Pattern.DOTALL);
Matcher matcher = pattern.matcher(xml);
while (matcher.find())
{
  String name = matcher.group(1);
  String lastName = matcher.group(2);
}

Store the name and lastName in your own Persons-Datastructure.

Define the Pattern.compile command as a constant outside your method because it needs time for initialization.

Please see http://docs.oracle.com/javase/6/docs/api/java/util/regex/Pattern.html

JAVA , Xml parsing

6 Answers6