PHP and HTML form issue

Question

Mycode

    $show=mysql_query("SELECT * FROM room_group_options");
    while ($array = mysql_fetch_array($show))
    {
        $op_id = $array['op_id'];
        $group = $array['group'];
        echo "<form action = 'options.php' method = 'post'>";
        echo "$op_id <input type='hidden' name='op_id' value='$op_id'>";
        echo "<input type = 'text' name = 'group' value = '$group'>";

        echo "<input type='button' name='edit' value='edit'>";
        echo "<input type='button' name='del' value='delete'>";
        echo "</form><br>";

    }
    echo "<form action = 'options.php' method = 'post'>
          <input type = 'text' name = 'group'><br>
          <input type = 'submit' name = 'addgroup'></form>";

I am a PHP newbie, I want those "edit" and "delete" buttons to send a POST request to a options.php page, but with this code when I hit a button nothing happens. How can I fix this error?

The second form option is ok for now

Welcome to Stack Overflow! Please, don't use `mysql_*` functions for new code. They are no longer maintained and the community has begun the [deprecation process](http://goo.gl/KJveJ). See the [**red box**](http://goo.gl/GPmFd)? Instead you should learn about [prepared statements](http://goo.gl/vn8zQ) and use either [PDO](http://php.net/pdo) or [MySQLi](http://php.net/mysqli). If you can't decide, [this article](http://goo.gl/3gqF9) will help to choose. If you care to learn, [here is good PDO tutorial](http://goo.gl/vFWnC). — Madara's Ghost, Jun 29 '12 at 06:43
is your `options.php` in the same directory with this html form? — ianace, Jun 29 '12 at 06:58

score 1 · Answer 1 · edited May 23 '17 at 10:09

1

Open and close the <form> tag outside the while loop
Use <input type='submit'>. Refer this link to see how can you use two submit buttons in one form
And finally as Truth mentioned above, use PDO / MySQLi

edited May 23 '17 at 10:09

Community

1
1

answered Jun 29 '12 at 06:47

skos

4,102
8
36
59

OK. are you using `` OR ` – skos Jun 29 '12 at 06:59
it's work, I forgot to use submit instead of button. Thank you. – Jay Sithiporn Jun 29 '12 at 07:00

score 0 · Answer 2 · answered Jun 29 '12 at 06:45

You must add some action to the button, i.e. <input type='button' name='edit' value='edit' onClick='edit()'>. When you click it, a javascript function edit() wil be called. It should have some AJAX request to your php page, POST or GET. You can use jQuery for that.

score 0 · Answer 3 · answered Jun 29 '12 at 06:57

0

You either need to add a javascript function to the button's onclick that fires a form submission, or change the button type from "button" to "submit" in order for it to post the page.

answered Jun 29 '12 at 06:57

Ekster

353
1
4

score 0 · Accepted Answer · edited Jun 29 '12 at 12:50

0

Change type='button' to type='submit'.

You should close input tags simply by adding a whitespace and a slash to the end. Like this:

<input type='text' name='username' />

I don't know if you want to have multiple forms or just 1. If you want 1 form, than put <form></form> tags outside the while loop.

edited Jun 29 '12 at 12:50

j0k

22,600
28
79
90

answered Jun 29 '12 at 07:07

Pixel

104
1
10

score -2 · Answer 5 · answered Jun 29 '12 at 06:45

echo "<form action = 'options.php' method = 'post'>";
$show=mysql_query("SELECT * FROM room_group_options");
while ($array = mysql_fetch_array($show))
{
    $op_id = $array['op_id'];
    $group = $array['group'];

    echo "$op_id <input type='hidden' name='op_id' value='$op_id'>";
    echo "<input type = 'text' name = 'group' value = '$group'>";

    echo "<input type='button' name='edit' value='edit'>";
    echo "<input type='button' name='del' value='delete'>";
}
echo "</form><br>";

echo "<form action = 'options.php' method = 'post'>
<input type = 'text' name = 'group'><br>
<input type = 'submit' name = 'addgroup'></form>";

Try something like this.

it still won't work - there is no submit button for first form. — maialithar, Jun 29 '12 at 06:55

PHP and HTML form issue

5 Answers5