Jquery/Ajax return data but doesn't alert

Question

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.

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.

You misunderstood the option dataType which is (according to documentation) the type of data that you're expecting back from the server. So your PHP should echo/print JSON instead of expecting JSON from the javascript client — devnull69
– devnull69, Commented Nov 5, 2015 at 11:47
Or maybe I misunderstood you ... then the answer to your question 2. would be NO — devnull69
– devnull69, Commented Nov 5, 2015 at 11:48

memo · Accepted Answer · 2015-11-05 14:17:52Z

3

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

edited Nov 5, 2015 at 14:17

answered Nov 5, 2015 at 11:47

memo

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7 Comments

devnull69 Over a year ago

on top of that it should be $_POST['command'] rather than $_SESSION['command'] as command is part of the POST-Request-Body

memo Over a year ago

Ah yes, you're correct. I didn't even bother to look at code above since OP said that the response is being returned.

meph Over a year ago

Thanks that worked. Now if 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.

memo Over a year ago

What does actually $db->query('/**/'') return? If it's an array json_encode() will work.

meph Over a year ago

Thanks memo for all you help. I appreciate it :)

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