CODEWITHSUNDEEP
phpmysqldatabase
SuicideSheep
SuicideSheep

Reputation: 5550

Unable to retrieve data from table?

Please consider the following code. I believe it's very straight forward however I couldn't see what's wrong with it.

<?php
        $dbhost = 'localhost';
        $dbuser = 'root';
        $dbpass = '';
        $conn = mysql_connect($dbhost, $dbuser, $dbpass);
        if(! $conn )
        {
          die('Could not connect: ' . mysql_error());
        }
        $sql = 'SELECT cif, br, 
                       fullname, id, id_type,issuance_country,class
                FROM ciftable';

        mysql_select_db('ciftable');
        $retval = mysql_query( $sql, $conn );
        if(! $retval )
        {
          die('Could not get data: ' . mysql_error());
        }
        while($row = mysql_fetch_array($retval, MYSQL_ASSOC))
        {
            echo "CIF :{$row['cif']}  <br> ".
                 "BR : {$row['br']} <br> ".
                 "Full Name : {$row['fullname']} <br> ".
                 "ID : {$row['id']} <br> ".
                 "ID Type: {$row['id_type']} <br> ".
                 "Issuance Country : {$row['issuance_country']} <br> ".
                 "Class : {$row['class']} <br> ".
                 "--------------------------------<br>";
        } 
        echo "Fetched data successfully\n";
        mysql_close($conn);
    ?>

This is the fresh installation of xampp so will be using default username and password. Kindly ignore all the possible SQL injection or best practices.

The above code throwing exception:

Could not get data: No database selected

Kindly advise.

Upvotes: 0

Views: 162

Answers (1)

Vladimir Ramik
Vladimir Ramik

Reputation: 1930

Missing 2nd parameter:

$db1 = mysql_connect($dbhost, $dbuser, $dbpass);
$rv = mysql_select_db($dbname1, $db1);

Upvotes: 2

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