Unable to echo value from html <option> from while loop

Question

i want to echo selected parent value. but i am getting error- Notice: Undefined index: How can i echo selected parent value then? Whats wrong i am doing?

$q = mysql_query("SELECT * FROM menu");
    echo '<form action="" method="post">
      Menu name:<input type="text" name="mname"><br>
      <select>';
    while ($row = mysql_fetch_array($q)) {
        $menu_name = $row['menu_name'];

    echo '<option value="'.$menu_name.'">'.$menu_name.'</option>';
    }
    echo '</select><br>
      <input type="submit" name="submit" value="Add Menu">
    </form>';



    if (isset($_POST['submit'])) {
        echo $mname = $_POST['mname'];
        echo $parent = $_POST[$menu_name];
    }

Note: mysql_query() and mysql_fetch_array() are deprecated, please use mysqli_query() and mysqli_fetch_array() in the future — aley
– aley, Commented Jun 9, 2016 at 14:34
$_POST[$menu_name]? $menu_name is only ever going to contain the LAST item you fetched from the db. And since your <select> has no name, it'll never get submitted with the rest of the form anyways. — Marc B
– Marc B, Commented Jun 9, 2016 at 14:40
WARNING: If you're just learning PHP, please, do not use the mysql_query interface. It’s so awful and dangerous that it was removed in PHP 7. A replacement like PDO is not hard to learn and a guide like PHP The Right Way explains best practices. — tadman
– tadman, Commented Jun 9, 2016 at 15:28

shilpi Singh · Accepted Answer · 2016-06-09 14:36:22Z

1

add name to the select box and get the value of select box by name. Updated code:-

$q = mysql_query("SELECT * FROM menu");
echo '<form action="" method="post">
  Menu name:<input type="text" name="mname"><br>
  <select name="menu_name">';
while ($row = mysql_fetch_array($q)) {
    $menu_name = $row['menu_name'];

echo '<option value="'.$menu_name.'">'.$menu_name.'</option>';
}
echo '</select><br>
  <input type="submit" name="submit" value="Add Menu">
</form>';



if (isset($_POST['submit'])) {
    echo $mname = $_POST['mname'];
    echo $parent = $_POST['menu_name'];
}

answered Jun 9, 2016 at 14:36

shilpi Singh

4041 gold badge5 silver badges15 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

David · Accepted Answer · 2016-06-09 14:37:04Z

0

$_POST[$menu_name] probably doesn't exist, because only two elements in your form have name attributes. The text input and the submit input.

option elements aren't posted as part of the form, but rather the selected option's value for the select element. But your select element has no name, therefore no key to use in the key/value pair, so it isn't posted.

Give the element a name:

<select name="someName">

Then in the POST, you would be able to fetch the selected value just as you do for any other form element:

$_POST['someName']

answered Jun 9, 2016 at 14:37

David

221k42 gold badges245 silver badges337 bronze badges

Comments

Ravinder Reddy · Accepted Answer · 2016-06-09 14:37:49Z

0

You need to add name attribute to select tag.

echo '<form action="" method="post">
  Menu name:<input type="text" name="mname"><br>
  <select name="any_name">';
$q = mysql_query("SELECT * FROM menu");
while ($row = mysql_fetch_array($q)) {
    $menu_name = $row['menu_name'];
    echo '<option value="'.$menu_name.'">'.$menu_name.'</option>';
}
echo '</select><br>
  <input type="submit" name="submit" value="Add Menu">
</form>';



if (isset($_POST['submit'])) {
    echo $mname = $_POST['mname'];
    echo $select_option_name = $_POST['any_name'];
}

Note: mysql_* functions are depricated, use mysqli_* functions

answered Jun 9, 2016 at 14:37

Ravinder Reddy

3,8791 gold badge16 silver badges24 bronze badges

Collectives™ on Stack Overflow

Unable to echo value from html <option> from while loop

3 Answers 3

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related