how to pass dropdown dynamic value in php/mysql

Question

i have a dropdown list the value of each item generate dynamic

          <?php
  include('../db_inc.php');
$sql="select * from genre";
$result = mysql_query($sql) or die(mysql_error());
while($row = mysql_fetch_object($result)){
    $option =$row->genre_name;
    $value =$row->genre_id;
    echo '<option value='.$value.'>'.$option.'</option>';

    }
?>

i want pass selected item value from dropdown to server and save that to database, i can do that with ajax,but i want to know php can do that without ajax function?

PHP cannot do it alone, unless you reload the whole page. That's what AJAX is for. PHP has no connection to the client browser, really. — Michael Berkowski
– Michael Berkowski, Commented Nov 1, 2013 at 19:58
In other words, if you just submit the form as normal (assuming this is in a proper <form> tag) and PHP will get the value in $_POST. — Michael Berkowski
– Michael Berkowski, Commented Nov 1, 2013 at 19:59
if i add dropdown in <form> how can access the value of selected dropdown item on server? — Hamed mayahian
– Hamed mayahian, Commented Nov 1, 2013 at 20:32

unconnected · Accepted Answer · 2013-11-01 20:00:01Z

1

PHP runs on server side. Dropdown list on client side. You need somehow send the value from client to server (post, get, ajax, websocket, etc). The answer is NO.

answered Nov 1, 2013 at 20:00

unconnected

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2 Comments

Hamed mayahian Over a year ago

i mean how can perform that with post?

unconnected Over a year ago

Use mehdi's sample. Add to select property name. <select name="blabla" ...> At server side $selected=$_POST["blabla"]

mehdi · Accepted Answer · 2013-11-01 20:02:48Z

0

you can send it with a <form> tag and submit form with on dropdown onchange attribute. then get it with php and save that to database.

<form name='form1' method='post' action=''>
    <?php
      include('../db_inc.php');
    $sql="select * from genre";
    $result = mysql_query($sql) or die(mysql_error());
    echo '<select onchange="form1.submit();">';
    while($row = mysql_fetch_object($result)){
        $option =$row->genre_name;
        $value =$row->genre_id;
        echo '<option value='.$value.'>'.$option.'</option>';

        }
    echo '</select>';
    ?>
</form>

answered Nov 1, 2013 at 20:02

mehdi

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1 Comment

Hamed mayahian Over a year ago

ok, if i use onchange() method,how can i access the value on server?

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