Simple php and sql login form not working

Question

I am trying to make a simple php and sql login form, but it is not working

Can anyone help me to fix my code?

<form  method="post" action="form.php">
    Username <input type="text" name="username"><br>
    Password <input type="password" name="password">
    <br>
    <input type="submit" name="submit" value="submit">
</form>


<?php
if (isset($_POST['submit'])) {

    $username = $_POST['username'];
    $password = $_POST['password'];
    $con = mysql_connect("localhost", "root", "");
    mysql_select_db($con, "formcolumn");
    $sql = mysql_query("select * from data1_table where username='$username' and password='$password' ");
    $row = mysql_fetch_array($sql);
    $uname = $row['username'];
    $pass = $row['password'];
    if ($username == $uname && $password == $pass) {
        header("Location: main.php");
    } else {
        echo "invalid username and password ";
    }
}
?>

Your script is vulnerable to SQL injections. You should learn how to prevent them. — Gumbo
– Gumbo, Commented Dec 27, 2013 at 13:03
"it is not not working" is not an acceptable error description. — user330315
– user330315, Commented Dec 27, 2013 at 13:03
where do you get error and what error do you get please explain — semirturgay
– semirturgay, Commented Dec 27, 2013 at 13:03
Once you've fixed the problem please consider that there's no point in checking username and password equal the values that you used to select the row. — Popnoodles
– Popnoodles, Commented Dec 27, 2013 at 13:03

Popnoodles · Accepted Answer · 2013-12-27 13:13:29Z

1

Replace mysql_select_db($con, "formcolumn"); with mysql_select_db("formcolumn",$con); where formcolumn is your Database name

edited Dec 27, 2013 at 13:13

Popnoodles

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answered Dec 27, 2013 at 13:09

Mahavir Nahata

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1 Comment

Mohit gupta Over a year ago

thanks bro still getting 1 warning Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\php\zz\m\form.php on line 19

Orville Patterson · Accepted Answer · 2013-12-27 13:20:05Z

0

Interchange the positions of your parameters in your mysql_select_db function bass the name of your database first then pass the connection. You must remember this is a predefined function and it is defined to accept parameters in a certain order therefore it expects that the first parameter passed to the function is going to be the database name.

You should have

mysql_select_db("formcolumn",$con);

also why not try something along the lines of this:

$sql = "select * from data1_table where username='$username' and password='$password'";
$result = mysql_query($sql,$con);
$row= mysql_fetch_assoc($result);

edited Dec 27, 2013 at 13:20

answered Dec 27, 2013 at 13:14

Orville Patterson

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