CODEWITHSUNDEEP
javajquery
Kunal Vashist
Kunal Vashist

Reputation: 2471

How to parse Json object in ajax

I am not able to parse tje Json object returned from the servlet in ajax,

I need to put json object values in there relative field

From my java code i am sending the below String in the form of JSON 

String  webVisitorDetails = "{"+"companyName : \""+webVisitor.getCompanyName()+ "\","+
                                                "address : \""+webVisitor.getProfessionalAddress()+ "\","+
                                                "city : \""+webVisitor.getCity()+ "\","+
                                                "zipCode : \""+webVisitor.getZipCode()+ "\","+
                                                "clientId : \""+webVisitor.getCustomerAccountNumber()+ "\"}";

In ajax

$.ajax({
    url: "ships",
    data: {
        email: email.toString()
    },
    success: function(data) {
        $.each(data, function(k, v) {
            console.log(k + " Value " + v);
            $("#city").text(v.city);
            $("#zipcode").text(v.getZipCode);
            $("#Adress").text(v.getProfessionalAddress);
        });
    },
    error: function(data) {
        console.log("error:", data);
    },
    type: "post",
    datatype:"json",
});

Upvotes: 1

Views: 1377

Answers (3)

Niks Jain
Niks Jain

Reputation: 1647

If it's creating a problem because of the single-inverted comma ('), then just do:

jQuery.parseJSON(data.replace(/'/g, '"'))

If that is the case then it should work for you...

Upvotes: 0

Wolfram
Wolfram

Reputation: 8052

Note that the jQuery setting is dataType with a capital T. To do the JSON parsing manually, use the parseJSON function. However, if you set the Content-Type of your servlet response to application/json, the datatype should be auto-detected.

After you fixed this: Does it work? What is the value of the data argument of your success handler?

console.debug(data);

As Neal already said, JSON parsing expects valid JSON strings starting with jQuery 1.4. You can validate your JSON jsonlint.com.

In jQuery 1.4 the JSON data is parsed in a strict manner; any malformed JSON is rejected and a parse error is thrown. (See json.org for more information on proper JSON formatting.)

To avoid the manual building of JSON strings, use something like the JSON-java processor (from iNan's comment) or other Java implementations listed on json.org.

Upvotes: 2

Naftali
Naftali

Reputation: 146310

You json string is incorrect

The keys must be surrounded by double quotes.

Read the requirements here

Upvotes: 0

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