CODEWITHSUNDEEP
jsonajaxxmlservlets
Fiordarancio
Fiordarancio

Reputation: 73

Got "XML Parsing Error: not well-formed" when I expect JSON

I am running a simple website that communicates with a TestServlet using ajax and jquery. The servlet gives a JSON object using doGet(), as follows:

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    String json = "{\"name\": \"jsontest\",\"type\":\"jsonobject\"}";
    response.getWriter().append(json);
}

When a button is pressed on the page, an ajax request is issued as follows:

$.ajax({
 type: "GET",
 url: "test",
 data: "",
 dataType: "json",
 success: function(reply) { window.alert("Success\n"+reply); },
 error: function(err) { window.alert(err); }

});

The request obtains success, but reply is null and I get from web console: 

XML Parsing Error: not well-formed
Location: http://localhost:8585/web/test
Line Number 1, Column 1: {"name": "jsontest","type":"jsonobject"}

I tried to specify mimeType: "application\json" but I got the same behavior. Instead, when I don't specify dataType or I put contentType: "json" I succeed to read correctly the JSON string but still I get the XML Parsing Error.

Can anyone explain why I get a XML Parsing Error while I am supposing to exchange JSON?

NOTE: I'm using Firefox and Tomcat9.

Thank you in advance.

Upvotes: 2

Views: 3220

Answers (1)

Fiordarancio
Fiordarancio

Reputation: 73

Simply, I found out that while I was expecting json content in the request through dataType, the servlet was not specifying that its response was coded with that type too. Adding:

response.setContentType("application/json");

solved the problem.

Upvotes: 2

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