Constructor not matched for class

Question

I am trying to deserialize an xml file, But I am having errors on deserializing it using the code below:

try {
    Strategy strategy = new CycleStrategy("id", "ref");
    Serializer serializer = new Persister(strategy);
    File source = new File("ActionToLettersAndBorrowers.xml");
    ProcessEngineObject op = serializer.read(ProcessEngineObject.class, source);

    System.out.println(op.getName());

} catch (Exception e) {
    e.printStackTrace();
}

Am i missing something out? I got the idea from simplexml website.

Please, make sure that the class you are trying to deserialize has a default public constructor. — Laurent De Cant, Apr 18 '18 at 12:37
[Related](https://stackoverflow.com/q/47407587/335858). Unfortunately, the other question has no answer, only a work-around. — Sergey Kalinichenko, Apr 18 '18 at 12:39
@LaurentDeCant i have a constructor. public Email(String to, String from, String subject, String body) { this.to = to; this.from = from; this.subject = subject; this.body = body; } — miggyyy, Apr 18 '18 at 14:54
@GBlodgett org.simpleframework.xml.core.PersistenceException: Constructor not matched for class nhmfc.filenet.xml.Email — miggyyy, Apr 18 '18 at 14:55
I tried to convert the constructor into a public void method, and it works. I don't know if simplexml cannot read constructors? — miggyyy, Apr 18 '18 at 15:36
SimpleXml like most of XML framework needs a no-arg constructor for deserialization. That was your error, and by removing your constructor, the default one was applied. Possible duplicate of [this error](https://stackoverflow.com/a/7471452/4629012) — Pagbo, Apr 19 '18 at 12:23

score 2 · Answer 1 · answered Feb 25 '20 at 13:13

In Kotlin you just need to predefine your properties. Example:

@Root(name = "Response")
data class LinkResponse @JvmOverloads constructor(
@field:Element(name = "Field1")
var bucket: String = "",
@field:Element(name = "Field2")
var key: String = "",
@field:Element(name = "Field3")
var etag: String = "",
@field:Element(name = "Field4")
var location: String = ""
)

score 0 · Answer 2 · answered Jun 26 '19 at 21:45

0

You can define two constructors, like:

class Strategy {
...
    Strategy(){...}
    Strategy(String id, String ref){...}
}

Another approach could be defining a no-arg constructor and then set the properties you need.

answered Jun 26 '19 at 21:45

marcolav

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Constructor not matched for class

2 Answers2