Jquery Json request and callback to html table

Question

I habe built a cascading drop down that functions very well. when the form is submitted, the Id is passed to second script. I can´t seem to get the callback function to ...function properly. Don´t know what wrong.

note I updated the code with the suggested changes, but something still goes wrong.

The php:

<?php
if (!empty($_GET['id'])) {
$id = $_GET['id'];

try {


$objDb = new PDO('mysql:host=localhost;dbname=blankett', 'root', 'root');
$objDb->exec('SET CHARACTER SET utf8');

$sql = "SELECT * 
    FROM `forms`
    WHERE `id` = '$id'";
$statement = $objDb->prepare($sql);
$list = $statement->fetchAll(PDO::FETCH_ASSOC);



if (!empty($list)) {

   $out = array();

  foreach ($list as $row ) {
  $out[] = '<tr><td><a href="'.$row['link_form'].'">'.$row['name_form'].'</a></td> <td>'.$row['date_added'].'</td></tr>';
   }

  echo json_encode(array('error' => false, 'list' => $out));
} else {
  echo json_encode(array('error' => true));
}


} catch(PDOException $e) {
echo json_encode(array('error' => true));
}

}else {
echo json_encode(array('error' => true));
}


?>

The Jquery ajax call.

$('#blankett_form').submit(function() {
var id = $(this).find('.update:last').val();
    if (id == '') {
        alert('Välj land och region.'); //glöm inte bort att ändra beroende på land.
    } else {
        var table = '<table class="table table-hover table-bordered"><thead><tr><th>blanketter.</th><th>datum tillagt.</th></tr></thead><tbody></tbody></table>'
        $('#formsubmit').empty().append(table)
        $ajax({
                url: 'func/blankett_func2.php',
                data: {'id':id},
                dataType: 'JSON',
                success: function(data)
                {
                    $.each(data.list, function(index, value){
                    $('#formsubmit tbody').append(value);
                    });
                }
          });
return false
});

I think there is still a problem with the script since it does not load. What could be the reason? — StenW
– StenW, Commented Mar 24, 2013 at 0:36

adeneo · Accepted Answer · 2013-03-24 00:28:25Z

1

Replace :

$each(data.out, function(){
    var trow = out;
    $('#formsubmit tbody').append(trow);
});

With :

$each(data.form, function(i, val){
    $('#formsubmit tbody').append(val);
});

as there is no out variable inside the each function, and it looks like you're naming the key form in the returned array ?

In the PHP you are redeclaring the array inside the loop, needs to be:

$out = array();

foreach ($list as $row ) {
     $out[] = '<tr><td><a href="'.$row['link_form'].'">'.$row['name_form'].'</a></td> <td>'.$row['date_added'].'</td></tr>';
}

and there's no need to pass an empty array to execute():

$statement->execute();

edited Mar 24, 2013 at 0:28

answered Mar 24, 2013 at 0:11

adeneo

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2 Comments

StenW Over a year ago

You are right. I don´t know if it is because I am tired, but I can´t seem to get it to work. I guess I should not use form, because it is reserved work in Jquery?

StenW Over a year ago

Any idea how I should rewrite the php?

Musa · Accepted Answer · 2013-03-24 00:19:57Z

1

You arent sending the data properly, you have to send the id key and the id value e.g.

data: {'id':id},

Also your json output is of the form {"error": false, "form": "html string"} so to access the html string you have to use data.form instead of data.out. So try

$.each(data.form, function(index, value){
    $('#formsubmit tbody').append(value);
});

answered Mar 24, 2013 at 0:19

Musa

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1 Comment

StenW Over a year ago

Totally right. Thank you! I still think there is something wrong in the script since it does not load properly.

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