CODEWITHSUNDEEP
phpmysqlsql
Jdeveloper Iphone
Jdeveloper Iphone

Reputation: 217

Mysql fetch error while getting data from mysql data base

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

Upvotes: 0

Views: 1088

Answers (6)

Borniet
Borniet

Reputation: 3546

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.

Upvotes: 0

Roy
Roy

Reputation: 42

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" ; }  ?>

Upvotes: 0

zkanoca
zkanoca

Reputation: 9928

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);

Upvotes: 0

Praveen kalal
Praveen kalal

Reputation: 2129

<?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);


    ?>

Upvotes: 0

liyakat
liyakat

Reputation: 11853

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());

Upvotes: 0

chandresh_cool
chandresh_cool

Reputation: 11830

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);

Upvotes: 5

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