Getting Result From JSON, JQUERY Not functioning

Question

I have some problem. I have created some app that consuming JSON array.

The syntax is using Jquery 1.9

The logic is : getting values from textbox (valobj variable below) then write to toprightsec div area.

$("#btnsearch").click(function() {
        valobj = $('#search_box').val();
        $.getJSON("search.php", { q : valobj }, function(data,result){
            //show result from database
            $('toprightsec').append("Title" + data.title)
                            .append("Intro" + data.intro_text);
            //end show result
        }, JSON);
    });

JSON array is from PHP result. some example

{"content":[{"title":"Test Post 100","intro_text":"Intro Test"}]}

But, it's not working. any helps?

thanks in advance.

UPDATE

$("#btnsearch").click(function() {
    valobj = $('#search_box').val();
    $.getJSON("search.php", { q : valobj }, function(data,result){
        //show result from database
        $('.toprightsec').append("Title" + data.content[0].title)
                        .append("Intro" + data.content[0].intro_text);
        console.log(data)
        //end show result
    }, JSON);

UPDATE 2

$(document).ready(function () {

    $("#btnsearch").click(function() {
    valobj = $('#search_box').val();
    $.getJSON("search.php", { q : valobj }, function(data,result){
        //show result from database
        $.each(data.content, function() {
            $('.toprightsec').append("Title" + data.content[0].title)
                        .append("Intro" + data.content[0].intro_text);
        });

        //end show result
    }, JSON);
});

hi @x4rf41 it's not working. The section content still the same. nothing changed — randytan
– randytan, Commented Jun 17, 2013 at 10:21
Try using console.log(data) to look at the structure of the JSON in JS. — casraf
– casraf, Commented Jun 17, 2013 at 10:21

Code Lღver · Accepted Answer · 2013-06-17 10:35:30Z

2

Use the following code:

 $('.toprightsec').append("Title" + data.content[0].title)
                            .append("Intro" + data.content[0].intro_text);

instead of:

 $('toprightsec').append("Title" + data.title)
                            .append("Intro" + data.intro_text);

If toprightsec is id then use the # and if class then use the .. here you have not specify that.

edited Jun 17, 2013 at 10:35

answered Jun 17, 2013 at 10:21

Code Lღver

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11 Comments

x4rf41 Over a year ago

maybe his selector is wrong too. what would this $('toprightsec') select ?

Code Lღver Over a year ago

@x4rf41 I have mention about it in the answer. hope this will be understandable to OP.

randytan Over a year ago

hi @x4rf41 the toprightsec is a class. please see the update section

Code Lღver Over a year ago

@randytan then use the . before the toprightsec and is there only one tag or more than one with this class name?

randytan Over a year ago

only one class: <td width="25px"><div class="toprightsec">Content for class "toprightsec" Goes Here</div></td>

|

elrado · Accepted Answer · 2013-06-17 10:22:39Z

0

You are using json array,so you have to access single array element and not whole array:

json_array.title //WRONG 
json_array[0].title //OK

Try something like:

$("#btnsearch").click(function() {
        valobj = $('#search_box').val();
        $.getJSON("search.php", { q : valobj }, function(data,result){
            //show result from database
            if (data.length){
                var _props = data.pop().content;
                $('toprightsec').append("Title" + _props.title)
                                .append("Intro" + _props.intro_text);
            }
            //end show result
        }, JSON);
    });

answered Jun 17, 2013 at 10:22

elrado

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2 Comments

x4rf41 Over a year ago

you are forgetting about the content array

x4rf41 Over a year ago

there is another one :D var _props = data.pop().content; should be var _props = data.content.pop();

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