CODEWITHSUNDEEP
phpjqueryajax
meph
meph

Reputation: 209

Jquery/Ajax return data but doesn't alert

I have these code:

show.php:

<div id="show" style="background-color: white; color: #2b2b2b "></div>
<script>

    $(document).ready(function(){
        $('#btnsearch').click(function(){
            var key = {
                'command': 'search',
                'data': $("#inputsearch").val()
            };
            $.ajax({
                type: 'POST',
                url: 'query.php',
                data: key,
                dataType: 'json',
                    success: function(msg){
                        $('#show').html(msg);
            }
            })
        });
    });

</script>

and query.php:

$command = $_SESSION['command'];
if($command == 'search'){
    $db = Db::getInstance();
    $str = $_POST['data'];
    $records = $db->query("SELECT * FROM m_cinfo LEFT OUTER JOIN m_jinfo ON m_cinfo.cinfo_id=m_jinfo.cinfo_id where fullName LIKE '%$str%'");

    //echo json_encode($records, JSON_UNESCAPED_UNICODE);
    echo 'تست';
}

Parrams and Responce are correct and input text + search send to query.php and تست returns but nothing shows in my div. It doesn't even alert anything in the success area.

  1. if i uncomment echo json_encode($records, JSON_UNESCAPED_UNICODE);, is the way correct to retrieve json data?

Like i want to use the data from my db. what should i do? i get json like this: 

[{"cinfo_id":"1","fullName":"علی علوی","phone":"09151234576","mail":"[email protected]","description":"در نمایشگاه آشنا شدیم","jinfo_id":"1","jobTitle":"شرکت","jobName":"بازرگانی","city":"مشهد"}]

and echo that but when i say 

$('#show').html(msg.fullName);

for example, nothing would show.

Thanks in advance.

Upvotes: 0

Views: 1336

Answers (1)

memo
memo

Reputation: 273

That's because ajax expects json response and the query.php script does not return json. If you add error callback to your ajax you will see that it's triggered.

$.ajax({
            type: 'POST',
            url: 'query.php',
            data: key,
            dataType: 'json',
            success: function(msg){
                $('#show').html(msg);
            },
            error: function (err) {
                console.log(err);
            }

To fix this in your query.php you should echo json_encoded string like this

echo json_encode(array('message' => 'تست'));

And then in your success callback you can access the message like this

success: function(response){
            $('#show').html(response.message);
}

Update: okay if your $db->query method returns json as string then you can do the following

$records = $db->query("SELECT * FROM m_cinfo LEFT OUTER JOIN m_jinfo ON m_cinfo.cinfo_id=m_jinfo.cinfo_id where fullName LIKE '%$str%'");

$decoded = json_decode($records, true);

echo json_encode($decoded);

Note that your query is vulnerable to SQL injection . You should escape the $str variable or use prepared statements. For example if somebody enters this string in the search field it could drop your 'm_cinfo' table

somestring'; DROP TABLE m_cinfo; #
                                 ^ this will comment the rest of your query

Upvotes: 3

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