How Could I Display Data In An Div

Question

0.00/5 (No votes)

See more:

Hi,
I want to display the title and url from the array, it displays the value but as [object object] when i am debugging, i want it to display the text value and not only as objects.

here is my code

C#

function successHandler(data) {

        var jsonObject = JSON.parse(data.body);
        var dataArr = $.makeArray(jsonObject.d.results);
        var resultArr = $.map(dataArr, function (item) {



            return { label: item.Title, value: item.Url };


        });


UPDATE
========================

Unable to get property length?

function successHandler(data) {
   
        var jsonObject = JSON.parse(data.body);
        var dataArr = $.makeArray(jsonObject.d.results);
        var resultArr = $.map(dataArr, function (item) {  
            
            $.each(resultArr, function (key, value) {

                $("#result").text(value[key].label + " " + value[key].value);
               
            });
         
            
          
            
        });

        var resultDisplay = document.getElementById("result");
        resultDisplay.innerHtml = resultArr;


}

New UPDATE
====
Error: Unable to get property 'length' of undefined or null reference

C#

function successHandler(data) {

    var jsonObject = JSON.parse(data.body);
    var dataArr = $.makeArray(jsonObject.d.results);
    var resultArr = $.map(dataArr, function () {
        $.each(resultArr, function (key, value) {


        })
        var str = '';
        str += (value.label + " " + value.value);
        $("#result").text(str);



    });


}

Posted 12-Jan-15 21:43pm

Kurac1

Updated 12-Jan-15 23:51pm

v3

Add a Solution

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

SujayC · Answer 1 · 2015-01-12T21:56:00

Solution 1

what you are returning from map function is an object (simple JSON), instead try using resultArr[0].label & resultArr[0].value

Ex:

JavaScript

resultDisplay.innerHtml = resultArr[0].label + " " + resultArr[0].value;

Posted 12-Jan-15 21:56pm

SujayC

Comments

Kurac1 13-Jan-15 4:06am

Doing like that it only displays data one time second time it wonk work

SujayC 13-Jan-15 5:15am

$("#result").text(value[key].label + " " + value[key].value);
try changing the above line of code to
var str = ''; //this statement should be out of each block
str += value.label + " " + value.value);
$("#result").text(str);
value in the each loop will itself be individual json object from the array. Further

Further, there seems to be some syntax/code formatting error in your updated snippet, hope you have corrected that as well.

Kurac1 13-Jan-15 4:09am

I have updated

Kurac1 13-Jan-15 5:50am

i will update again, it dont work

SujayC 13-Jan-15 11:05am

sorry if i wasn't clear in my previous comment what i meant was the following

function successHandler(data) {
var jsonObject = JSON.parse(data.body);
var dataArr = $.makeArray(jsonObject.d.results);
var resultArr = $.map(dataArr, function (item) {
return { 'label' : item.Title, 'value' : item.Url }
});
var str = '';
$.each(resultArr, function (key, value) {
str += (value.label + " " + value.value);
});
$("#result").text(str);
}