change img src on select box selection

Posted by user1871596 on Stack Overflow See other posts from Stack Overflow or by user1871596
Published on 2012-12-03T05:02:27Z Indexed on 2012/12/03 5:03 UTC
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i have a form that auto populates input fields using jquery and ajax i can not get the img url to change in the img src when i select the option from dropdown

my dropdown is dynamicaly populated here is is my function

    <script type="text/javascript">
$(document).ready(function(){
    $("#id").change(function(){
        $.ajax({
            url     : 'get_driver_data2.php',
            type    : 'POST',
            dataType: 'json',
            data    : $('#ContactTrucks').serialize(),
            success: function( data ) {
                   for(var id in data) {        
                          $(id).val( data[id] );
                   }
            }
        });
    });
});
</script>

here is get_driver_data2.php

    <?php include ('dbc.php');

    $id_selected = $_POST['id']; // Selected  Id
    $query  = "SELECT * from admin_dispatch_records where id = '$id_selected' AND     driver LIKE '%$username%'";
    $result = mysqli_query($dbcon, $query);
    while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
    {
    $eta = $row['eta'];
    $time = $row['dispatch_time'];
    $date = $row['dispatch_date'];
    $name = $row['name'];
    $phone = $row['phone'];
    $vehicleyear = $row['vehicleyear'];
    $color = $row['color'];
    $make = $row['make'];
    $model = $row['model'];
    $vin = $row['vin'];
    $plate = $row['plate'];
    $mileage = $row['mileage'];
    $pickup = $row['pickup'];
    $dropoff = $row['dropoff'];
    $price = $row['price'];
    $invoice = $row['invoice'];
    $cash = $row['cash'];
    $credit = $row['credit'];
    $check = $row['check'];
    $po = $row['po'];
    $billed = $row['billed'];
    $need_to_bill = $row['need_to_bill'];
    $getphoto = $row['image_path'];
    }

    $arr = array( 'input#eta' => $eta, 'input#dispatch_time' => $time, 'input#dispatch_date' => $date, 'input#name' => $name, 'input#phone' => $phone, 'input#vehicleyear' => $vehicleyear, 'input#color' => $color, 'input#make' => $make, 'input#model' => $model, 'input#vin' => $vin, 'input#plate' => $plate, 'input#mileage' => $mileage, 'textarea#pickup' => $pickup, 'textarea#dropoff' => $dropoff, 'input#price' => $price, 'input#invoice' => $invoice, 'input#cash' => $cash, 'input#credit' => $credit, 'input#check' => $check, 'input#po' => $po, 'input#billed' => $billed, 'input#need_to_bill' => $need_to_bill, 'image#image_path' => $getphoto);
   echo json_encode( $arr );
   ?>

a bit of the html

        <td>

        <img  id="image_path" src="????" />
        </td>
        </tr>
        </table>

        <p><strong>
        <input type="submit" value="Complete Dispatch">
        </strong></p>

how do i if at all possible populate the src with the database vaule ajax recieved when i change the select box all other data is populated and the string is returned correctlly i have tested that by placing an input box and calling input#image_path => $getphoto. is there syntax for an img tag like the input....textarea....etc. I have tried including the get....php inline and assigning the src to $getphoto no luck there I was looking at trying to make a hidden input field with the ajax passed data and then taking that data and making it a var but can not figure that out either.

thanks

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