Assign MySQL database value to a PHP variable

Question

I am trying to get the password of a specified username, assign it to a PHP variable, and then print it. Here's the code for it -

$user = $_POST["uname"];
$pass = mysqli_query($con, "SELECT pass FROM users WHERE username = '$user'");                                              
$result = mysqli_fetch_array($pass);

$specific = $result['pass'];

echo $specific;

The problem is that nothing is being printed at all! Even no error. What do I do?

You aren't checking for errors though.. php.net/manual/en/mysqli.error.php You also should use prepared statements, this is open to SQL injections. Also hopefully the passwords are hashed.. — chris85
– chris85, Commented Jan 29, 2016 at 16:45
$result =mysqli_fetch_array($pass,MYSQLI_ASSOC); use this. this is to fetch the db value into array format. — ameenulla0007
– ameenulla0007, Commented Jan 29, 2016 at 16:46
I am just testing the code, this isn't gonna be used anywhere. Just checking the basics. — Shiven Sinha
– Shiven Sinha, Commented Jan 29, 2016 at 16:54

user5742934 · Accepted Answer · 2016-01-29 16:47:34Z

1

$pass = mysqli_query($con, "SELECT pass FROM users WHERE username = '$user'") or die(mysqli_error($con));

This will Show you the Correct Error in your query

answered Jan 29, 2016 at 16:47

user5742934

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Blake A. Nichols · Accepted Answer · 2016-01-29 17:01:17Z

1

Escape your values and check for errors:

$user = mysqli_escape_string( $con, $_POST["uname"] );
$pass = mysqli_query( $con, "SELECT pass FROM users WHERE username = '$user'");

# Error checking
if( $pass === false ) {
    echo 'Error: ', mysqli_error( $con );
}

# Check for no user with that password
if( mysqli_num_rows( $pass ) == 0 ) {
    echo 'No user with that username.';
}

# Use as associate arary              
$result = mysqli_fetch_assoc($pass);

$specific = $result['pass'];

echo $specific;

edit: Added check for no results.

answered Jan 29, 2016 at 17:01

Blake A. Nichols

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