Preserved Topic: mysql_num_rows problem (Page 1 of 1)

Hi peoples,

I created the following login form and script...

here is my code - it is only a short form and login script.

<form action="logoncheck.php" method="post" name="Logon" id="Logon" >

<table width="0%" border="0" align="center" cellpadding="0" cellspacing="0">

  <tr>

      <td bgcolor="#003366"><font color="#FFFFFF">Logon:</font></td>

    <td><input name="logon" type="text" id="logon"></td>

  </tr>

  <tr>

      <td bgcolor="#003366"><font color="#FFFFFF">Password:  </font></td>

    <td><input name="password" type="password" id="password"></td>

  </tr>

  <tr bgcolor="#000000">

      <td align="center"><font color="#FFFFFF">

        <input type="submit" name="Submit" value="Submit">

        </font></td>

      <td align="center"><font color="#FFFFFF">

<input type="reset" name="Submit2" value="Reset">

        </font></td>

  </tr>

</table>

</form>

<?

$link = mysql_connect("localhost", "gee", "ftm");

mysql_select_db("auth", $link);



$_POST['name'] = mysql_query("SELECT name FROM auth WHERE name = '$logon' AND pass = PASSWORD('$password')", $link);

$_POST['num_rows'] = mysql_num_rows($name);



if ($num_rows <= 0)

    {

        echo "<p align= center>You either entered your user name or password wrong.</p>";

        exit;

    }



include ("includes/private.php");

echo "<p align= center>You are now logged in</p>";

echo "<br>";

?>

I created a table called auth - with 2 fields, name and pass. I dunno why it is not working - maybe because it has something to do with the password()function in mysql?

I keep getting 'you either entered youe username or password wrong' even when I enter them correctly.

Any help would be greatly appreciated!

Update - I changed the following line to :
if ($num_rows = 0)

now I always get logged in even if I eneter the wrong password... I also get the following error message.

Warning: Failed opening 'includes/private.php' for inclusion (include_path='.;c:\php4\pear') in C:\FoxServ\www\logoncheck.php on line 15

I have never used pear before so now I am completely lost

[This message has been edited by Gee (edited 12-05-2002).]

Gee: I'm not sure why you are using this:

quote:

---

PASSWORD('$password')

---

___________________
Emps

FAQs: Emperor

thnX emporer - I fixed it.

I was playing around with $_POST and my variables - somehow I forgot to
get rid of PASSWORD($password); sorry!

One other thing - I use php with globals turned off. If I include " extract($_POST); "
at the beginning of my scripts then I dont have to write all the extra syntax is true.

But does this defeat the object of having globals turned off or is it just as safe as
using $_POST everywhere?

You only need to use $_POST["thing"] when accessing a variable that was sent to your script from the previous page (i.e. user input). You don't need to use it for anything else.

So $_POST['num_rows'] = mysql_num_rows($name); really isn't necessary, and it's probaby what's messing you up. Setting $_POST["thing"] sets an element of the POST superglobal array. That means it's part of the variable $_POST, not a variable of its own.

For this kind of thing, just use normal variables.

$query = "SELECT blah blah blah";
$result = mysql_query($query);
$num_rows = mysql_num_rows($result);

if ($num_rows < $value) doThing();