CODEWITHSUNDEEP
javascriptphpjqueryajaxjson
Vladimir Štus
Vladimir Štus

Reputation: 217

Parse JSON json_encoded database results

Here is my code in controller. I want grab this results into ajax success part. my question how to display results in ajax after using JSON.parse() function, can some one help me with loop; json code is something like this [{'id:1'},{info:name}] ??? 

function ajax(){

            $rpp= $_POST['rpp'];
            $last = $_POST['last'];
            $pn =  $_POST['pn'];


        if($pn<1){
            $pn=1;
        }
        elseif($pn>$last){
            $pn =$last;
        }

        $l = ($pn - 1) * $rpp;
        $this->db->limit($l, $rpp);
        $query = $this->db->get('pages');

        $data = array();

if ($query->num_rows() > 0) {
foreach($query->result() as $row) {
    $data[] = $row;
    }
}
$json = json_encode($data);

echo $json;
    }

ajax part

function request_page(pn)
{ 
    var rpp = <?php echo $rpp; ?>; // results per page
    var last = <?php echo $last; ?>; // last page number

    var results_box = document.getElementById("results_box");
    var pagination_controls = document.getElementById("pagination_controls");
    results_box.innerHTML = "loading results ...";
    $.ajax({
                type: "POST",
                url: "<?php echo site_url('search/ajax')?>",
                data: { 'rpp' : rpp , 'last' : last, 'pn' : pn},
                dataType: "text",
                success: function(msg){
                    alert(msg);
             // $.each($.parseJSON(msg), function() {
       // $('#results_box').html(this.id + " " + this.info);
   // });
                }


            });

Upvotes: 0

Views: 703

Answers (2)

neiker
neiker

Reputation: 8991

First remove that dataType form your $.ajax(), and then add a correct content-type header in php like:

header('Content-type: application/json');

This way, jQuery will parse your json correctly.

Now, you can do something like:

success: function(res){
    $.each(res, function(el) {
        console.log(el.id);
        console.log(el.info);
    });
}

Upvotes: 0

Axel Amthor
Axel Amthor

Reputation: 11106

just change

   dataType: "text",

to

   dataType: "json",

and jQuery is doing the parsing.

[EDIT]

supposed you have: [{'id':1,'info':'name'},{'id':2,'info':'nom'} ] (your json in your post is slightly unusable and not the result of an json_encode of an array) ...

   success: function(msg){
                var id = msg[0].id;
                var info = msg[0].info;
                ...

There's no error handling and no exception handling in this code which I regard as necessary!

Upvotes: 1

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