Not fetching result into Array in PHP

Question

I have my data in MySQL database is

'1', 'Demo', 'Demo1', 'Test'

On line 'echo $row[2]' it gives error "Notice: Undefined offset: 2"

Any solutions..?? PHP Code is as below...

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

    $uname=$_POST['txtUserName'];
    $passwd=$_POST['txtPassword']; 

    $query="SELECT usernm,passwd FROM tbuserdemo WHERE usernm='$uname'";
    $result=  mysql_query($query);
    $rows=  mysql_num_rows($result); 

    if($rows==0)
    {
        echo "User Name is Wrong";
    }
    else
    {
        $row=mysql_fetch_row($result);
        $encpasswd=  encrypt(1, $passwd);        


        if ($row[1]==$encpasswd)
        {            
            echo "Welcome $uname";
            echo $row[2];
        }
        else
        {
            echo "Password is Wrong";
        }
    }


    ?>
</body>
</html>

Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial. — Arnold Daniels
– Arnold Daniels, Commented Feb 12, 2013 at 21:33
Your code is vunarable to SQL injection. Please take care of that before using this code in a live environment. — Arnold Daniels
– Arnold Daniels, Commented Feb 12, 2013 at 21:37

sbeliv01 · Accepted Answer · 2013-02-12 21:31:04Z

3

You're only fetching 2 fields from the database, SELECT usernm,passwd so you should only have $row[0] and $row[1] with values. Array indexes are zero-based.

answered Feb 12, 2013 at 21:31

sbeliv01

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Comments

UnholyRanger · Accepted Answer · 2013-02-12 21:31:05Z

0

Welcome to programming. Many languages do this, Array indexes start at 0 NOT 1. So you need to do

if ($row[0]==$encpasswd)
        {            
            echo "Welcome $uname";
            echo $row[1];
        }

answered Feb 12, 2013 at 21:31

UnholyRanger

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Comments

Marc B · Accepted Answer · 2013-02-12 21:31:19Z

0

It'd be $row[0] and $row[1]. Remember, PHP's arrays are zero-based. To get a $row[2] you'd have to be selecting THREE fields in your query.

answered Feb 12, 2013 at 21:31

Marc B

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Comments

Nimrod007 · Accepted Answer · 2013-02-12 21:31:44Z

0

change to this :

   if ($row[0]==$encpasswd)
    {            
        echo "Welcome $uname";
        echo $row[1];
    }
    else
    {
        echo "Password is Wrong";
    }
}


?>

first index in the array is in location 0 (not 1)

answered Feb 12, 2013 at 21:31

Nimrod007

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Comments

Sangar82 · Accepted Answer · 2013-02-12 21:32:06Z

0

The arrays begin with 0.

Try this:

    if ($row[0]==$encpasswd)
    {            
        echo "Welcome $uname";
        echo $row[1];
    }
    else
    {
        echo "Password is Wrong";
    }

answered Feb 12, 2013 at 21:32

Sangar82

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Comments

illuzive · Accepted Answer · 2013-02-12 21:37:38Z

0

You might want to read up a bit on SQL Injection attacks. At least if you intend to publish that code outside a development/testing environment.

answered Feb 12, 2013 at 21:37

illuzive

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