i developed a simple php program with mysql and it'll store data twice to mysql database

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<!DOCTYPE html>
<html>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <title></title>

        <form action="" method="post" >
    Name:<input type="text" name="un"> <br>
    Password:<input type="password" name="un2"/> <br>
    <input type="submit" value="submit" name="submit" />
</form>
    </head>
    <body>
        <?php

        if ($_POST['submit']) {
        mysql_connect ("localhost", "root", "promod19861030") or die ('Error: ' . mysql_error());
        mysql_select_db("empdata") or die ('Data error:' . mysql_error());
        $text = mysql_real_escape_string($_POST['un']);
        $query="INSERT INTO empname (empName) VALUES ('$text')";
        mysql_query($query) or die ('Error updating database' . mysql_error());

}
        ?>
    </body>
</html>

it gives this error as well?

Notice: Undefined index: submit in C:\xampp\htdocs\PhpProject1\index.php on line 16

Line 16 is this :-

PHP

if ($_POST['submit']) {

if i input data to the two textboxes and i hit the submit button
the data will store into the database twice.

Please tell me what is the problem in the code?

Posted 4-Jan-14 22:16pm

promod madushan

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promod madushan · Answer 1 · 2014-01-04T22:20:00

I solved it myself...............

XML

<!DOCTYPE html>
<html>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <title></title>

        <form action="" method="post" >
    Name:<input type="text" name="un"> <br>
    Password:<input type="password" name="un2"/> <br>
    <input type="submit" value="submit" name="submit" />
</form>
    </head>
    <body>
        <?php

        if (isset($_POST['submit'])) {
        mysql_connect ("localhost", "root", "promod19861030") or die ('Error: ' . mysql_error());
        mysql_select_db("empdata") or die ('Data error:' . mysql_error());
        $text = mysql_real_escape_string($_POST['un']);
        $query="INSERT INTO empname (empName) VALUES ('$text')";
        mysql_query($query) or die ('Error updating database' . mysql_error());

}
        ?>
    </body>
</html>

Thanks everyone for the attention!!!!