Dear I have following codes

<?php
// Connection variables
$host="localhost";
$username="root";
$password="";
$db_name="asia"; 

// Connect to database
$con=mysqli_connect("$host", "$username", "$password");

// Connect result
if(!$con){
die('Error Connecting to Database: ' . mysqli_error());
//die '<script type="text/javascript">alert("Error Connecting to Database"). mysqli_error($con);</script>'; 
}
else
{
echo '<script type="text/javascript">alert("Connected"). mysqli_error($con);</script>'; 
}

// Database Selection
$sel =mysqli_select_db($con,"$db_name");

// Database Selection result
if(!$sel){
die('Error selecting Database: ' . mysqli_error());
}
else
{
//echo "Database Selected" . mysqli_error($con);
echo '<script type="text/javascript">alert("Database Selected"). mysqli_error($con);</script>'; 
}
$result = mysql_query("select sno,packing,weight from ghee");  
 if(!$result){
  echo 'select query error or data not found';
  }
  else
  {
  echo 'data found';

echo "<table border='1' style='border-collapse: collapse'>";  
echo "<th>SNo</th><th>Name</th><th>Weight</th>";  
while ($row = mysql_fetch_array($result))
{  
echo"<tr><td>".$row['sno']."</td><td>".$row['packing']."</td><td>".$row['weight']."</td></tr><br>";
echo "</table><br>";
}
}
mysqli_close($con);  

?>

on this line number 35
echo 'select query error or data not found';

it says:select query error or data not found

connection with mysql is ok
and
Database is selected properly

but why I get above error message

Here is screen shot of database

[IMG]http://i41.tinypic.com/2w5lx8y.jpg[/IMG]

Please help me to get rid of error message

You are mixing MySQLi with MySQL, that doesn't work, choose one of them.

Hi,

You can rewrite it to something like this..

<?php
    // Connection variables
    $host="localhost";
    $username="root";
    $password="";
    $db_name="asia"; 

    // Connect to database
    $con= new mysqli( $host, $username, $password); // you can add the fourth parameter here as you wish.

    // Connect result

    if($con->connect_errno){
    die('Error Connecting to Database: ' . mysqli_error());
    //die '<script type="text/javascript">alert("Error Connecting to Database"). mysqli_error($con);</script>';
    ## must use this
    // exit;
    }

    else
    {
    echo '<script type="text/javascript">alert("Connected"). mysqli_error($con);</script>';
    }
    // Database Selection
    //$sel = mysqli_select_db($con,"$db_name");
    // Database Selection result
    if(!$con->select_db($db_name)){
    die('Error selecting Database: ' . mysqli_error());
    }
    else
    {
    //echo "Database Selected" . mysqli_error($con);
    echo '<script type="text/javascript">alert("Database Selected"). mysqli_error($con);</script>';
    }
    $result = $con->query("select sno,packing,weight from ghee");  
    if(!$result){
    echo 'select query error or data not found';
    }
    else
    {
    echo 'data found';
    echo "<table border='1' style='border-collapse: collapse'>";
    echo "<th>SNo</th><th>Name</th><th>Weight</th>";
    while ($row = $result->fetch_array(MYSQLI_ASSOC))
    {
    echo"<tr><td>".$row['sno']."</td><td>".$row['packing']."</td><td>".$row['weight']."</td></tr><br>";
echo "</table><br>";
    }
    }
    mysqli_close($con);
    ?>

You can also shorten your codes by adding a fourth parameter on the mysqli object $db_name. Instead of selecting the database table midway on your codes. You can always redefine a new database table as you wish after the mysqli object.

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