query not working

Discussion in 'PHP' started by kharearch, Dec 11, 2007.

0
  1. #1
    I have written a query to find matching record from the table. At the end of query I want to find out whether any row is selected in response of query. I have written following query.

    <body>
    <?php
    $con = mysql_connect("localhost","root");
    if (!$con)
    {
    die('Could not connect: ' . mysql_error());
    }mysql_select_db("shopping_cart", $con);
    $s="SELECT * FROM user where name ='" . $_POST[name] . "' and password = '" .$_POST[passwd]."'";
    $result = mysql_query($s);
    $row = mysql_fetch_array($result);
    if(mysql_num_rows($row>0))
    {
    echo "welcome";
    }
    else
    {
    echo "not a valid user";
    }

    ?>
    </body>

    But this query is giving error. I am not clear, what is error in following query. Please help me.


    Following error is coming.

    Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in E:\xampplite\htdocs\practice\shop_cart\query.php on line 21

    Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in E:\xampplite\htdocs\practice\shop_cart\query.php on line 22
    not a valid user
     
    kharearch, Dec 11, 2007 IP
  2. Brewster

    Brewster Active Member

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    #2
    Try this:

    <?php
$con = mysql_connect("localhost","root");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}mysql_select_db("shopping_cart", $con);
$s="SELECT * FROM user where name ='" . $_POST[name] . "' and password = '" .$_POST[passwd]."'";
$result = mysql_query($s);

if(mysql_num_rows($result))
{
echo "welcome";
}
else
{
echo "not a valid user";
}

?>
    PHP:
    Brew
     
    Brewster, Dec 11, 2007 IP
  3. nico_swd

    nico_swd Prominent Member

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    #3
    nico_swd, Dec 11, 2007 IP
  4. sm9ai

    sm9ai Active Member

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    #4
    Echo out the value of $s to see what it contains.

    Then paste it directly into sql on phpmyadmin (or equivalent)
     
    sm9ai, Dec 11, 2007 IP
  5. Gawk

    Gawk Peon

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    #5
    You have no result to call mysql_fetch_array() for, your query returns 0 results - Brewster accounts for this in his code example.

    Once you fix this make sure you cleanse the $_POST's or you will have a huge security hole!
     
    Gawk, Dec 11, 2007 IP