Mysql fetch error while getting data from mysql data base

Question

I am fetching data from my sql it gives the error here is the my code

    <?php
    $con =     mysql_connect("productivo.db.6420177.hostedresource.com","","");
    if (!$con)
    {


    die('Could not connect: ' . mysql_error());
    }

    mysql_select_db("productivo", $con);

    $username=$_POST['UserName'];
    $password=$_POST['UserPassword'];
    echo($u);
    echo($pw);


     $query = "SELECT *  from  UserCredentials  WHERE UserName='$username' AND  UserPassword='$password'";

//$con = mysql_query($query,$con); //$cnt = mysql_num_rows($con);

  $res = mysql_query($query,$con);
  $cnt  = mysql_num_rows($res);

  echo($cnt);
  if($cnt ==0) { echo "Login Failed"; } else { echo "Yes Successful Login" ; }  ?>

here is the warning it shows

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/content/i/h/u/ihus235/html/cs/pah_brd_v1/productivo/CheckUserCredentialsAPI.php on line 21 Login Failed

you can accees using this url

http://celeritas-solutions.com/pah_brd_v1/productivo/CheckUserCredentialsAPI.html

MySQL (mysql_ functions) extension is *deprecated. I suggest to use MySQLi (mysqli*_ functions) or PDO instead. — BlitZ
– BlitZ, Commented Apr 30, 2013 at 5:40

chandresh_cool · Accepted Answer · 2013-04-30 05:40:20Z

5

The problem is you are trying to use connection variable again don't reuse the same variable again use some other variable

$res = mysql_query($query,$con);
$cnt  = mysql_num_rows($res);

answered Apr 30, 2013 at 5:40

chandresh_cool

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4 Comments

Vinoth Babu Over a year ago

Why we should now reuse it. Anyhow we are not going to use that connection after that know? please explain.

BlitZ Over a year ago

@Vinoth Babu , there might be no difference, but it's a better snippet for OP. And, yeah, connections should be manually closed, as for me.

kuldeep.kamboj Over a year ago

You are making a connection instance variable and then try to set by it with query result variable. It is very normal case for application misbehaviour. Also reusing variable is not a better idea every time.

chandresh_cool Over a year ago

edit your updated code in question. Lets double check what have you changed.

liyakat · Accepted Answer · 2013-04-30 05:43:57Z

0

you can check error with below code what actually you are doing wrong

 $check = mysql_query("SELECT *  from  UserCredentials  WHERE UserName='$username' AND       UserPassword='$password'")or die(mysql_error());

answered Apr 30, 2013 at 5:43

liyakat

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Comments

Praveen kalal · Accepted Answer · 2013-04-30 05:44:46Z

0

<?php
     $link = mysqli_connect("productivo.db.6420177.hostedresource.com","","");
     $query = "SELECT *  from  UserCredentials  WHERE UserName='$username' AND       UserPassword='$password'";

     $result = mysqli_query($link,$query);
     $cnt  = mysqli_num_rows($result);


    ?>

answered Apr 30, 2013 at 5:44

Praveen kalal

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Comments

zkanoca · Accepted Answer · 2013-04-30 05:44:50Z

0

Here you use

$con =    mysql_connect("productivo.db.6420177.hostedresource.com","","");

and then for the resultset you use

$con = mysql_query($query,$con);
$cnt  = mysql_num_rows($con);

again. Use $condom for the second.

$condom = mysql_query($query,$con);
$cnt  = mysql_num_rows($condom);

answered Apr 30, 2013 at 5:44

zkanoca

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Comments

Roy · Accepted Answer · 2013-04-30 06:07:32Z

0

Here u try with this

<?php
    $con =    mysql_connect("productivo.db.6420177.hostedresource.com","","");
   if (!$con)
    {
    die('Could not connect: ' . mysql_error());
   }
    mysql_select_db("productivo", $con);
   $username=$_POST['UserName'];
   $password=$_POST['UserPassword'];
   $result =mysql_query("SELECT *  from  UserCredentials WHERE UserName='$username' AND UserPassword='$password'",$con) or die(mysql_error());
   $cnt  = mysql_num_rows($result);
   if($cnt ==0) { echo "Login Failed"; } else { echo "Yes Successful Login" ; }  ?>

answered Apr 30, 2013 at 6:07

Roy

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Comments

Borniet · Accepted Answer · 2013-04-30 06:15:33Z

0

Try

$query = "SELECT *  from  UserCredentials  WHERE UserName='$username' AND UserPassword='$password'";
echo "query= " . $query;

to see what your resulting query looks like. Most probably there is an error in your query that you didn't spot. Best thing is to copy the echo'd result and run it straight on your database.

After that you better use:

$res = mysql_query($query,$con) or die('$query gave error: ' . mysql_error());

This will give you detailed error information from MySQL too.

And last but not least: don't use the mysql_xxx functions anymore, they're deprecated, and very unsafe. This means that your app might not work anymore after a future php-upgrade.

answered Apr 30, 2013 at 6:15

Borniet

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3 Comments

Jdeveloper Iphone Over a year ago

this is the result query= SELECT * from UserCredentials WHERE UserName='jamshaid.ali' AND UserPassword='aliraza40'$query gave error: Table 'productivo.UserCredentials' doesn't exist

Borniet Over a year ago

Well, that says it all doesn't it :-) Or your table really doesn't exist, or you miss spelled it. If all that is correct, make sure you're connected to the correct database.

Jdeveloper Iphone Over a year ago

i am copying name from the data base it is same it also shows record in admin panel

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