mysql_num_rows(): supplied argument is not a valid MySQL result resource

[pes96]

PHP Error Message

 

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource /home/a4199910/public_html/login2.php on line 28

http://teste-vieirasousa.tk/login2.php

 

 

THE CODE: (the values of database, password, username, server that i use here are an example)

<?php

   $link = mysql_connect("localhost", "username", "password");
   mysql_select_db("database_name", $link);

function login($username, $password) {
   $username = addslashes($username);
   $password = md5($password);
   $query = mysql_query("SELECT * FROM user_accounts WHERE username='$username' AND password='$password'");
   if(mysql_num_rows($query) == 1) {
      $info = mysql_fetch_array($query);
      $userid = $info[userid];
      $sessionid = md5($userid . time());
      $time = time();
      @setcookie ('test_account', $sessionid, $time+3600, '/', '');
      mysql_query("DELETE FROM user_sessions WHERE userid='$userid'");
      mysql_query("INSERT INTO user_sessions (sessionid,userid,timestamp) VALUES('$sessionid','$userid','$time')");
      return $userid;
   } else {
      return 0;
   }
}

function status() {
   $sessionid = $_COOKIE[test_account];
   $oldtime = time() - 3600;
   $query = mysql_query("SELECT * FROM user_sessions WHERE sessionid='$sessionid' AND timestamp>$oldtime");
       if(mysql_num_rows($query) == 1) {
      $info = mysql_fetch_array($query);
      return $info[userid];
   }
   return 0;
}

function logout() {
   $sessionid = $_COOKIE[test_account];
   @setcookie ("test_account",'', time()-99999, '/', '');
   mysql_query("DELETE FROM user_sessions WHERE sessionid='$sessionid'");
}

if($_POST[username] !='' || $_POST[password] != '') {
   $login_status = login($_POST[username], $_POST[password]);
} else if($_GET[logout]) {
   logout();
}
$userid = status();


if($userid > 0) { echo "Welcome to our site, user #$userid (<a href='?logout'>Click here to logout</a>)"; } else {

if($login_status != '' && $login_status == 0) { echo "Invalid username/password combo.<br>"; }
?>

<form action="sample.php" method="POST">
<input type=text name=username>
<input type=password name=password>
<input type=submit value="Log In">
</form>

<?php } ?>

 

 

LINE 28:

if(mysql_num_rows($query) == 1) {

 

Anyone can help me to resolve this error?

Thank you

[xflawless]

Add LIMIT 1 to your query.

 

SELECT * FROM user_accounts WHERE username='$username' AND password='$password' LIMIT 1

[jazzman1]

Your query is wrong I think.

Copy/paste this:

$query = mysql_query("SELECT * FROM user_accounts WHERE username='".$username."' AND password='".$password."'");

 

[pes96]

And i have to put this on line 28?

 

Thanks

[jazzman1]

You have to learn escaping single quotes in "php varibles".

Take a look at the example:

$str1 = 'string';
$str2 = '$str1';
printf('%s and %s', $str1, $str2);

 

 

[PFMaBiSmAd]

The query that is failing, based on the line number being listed is the following - $query = mysql_query("SELECT * FROM user_sessions WHERE sessionid='$sessionid' AND timestamp>$oldtime");

 

You need to have error checking and error reporting logic in your code to get your code to tell you when and why something is failing. Try the following for debugging purposes -

 

$query = mysql_query("SELECT * FROM user_sessions WHERE sessionid='$sessionid' AND timestamp>$oldtime") or die(mysql_error());

[pes96]

I do this, but appears the same error.

[pes96]

You have to learn escaping single quotes in "php varibles".

Take a look at the example:

$str1 = 'string';
$str2 = '$str1';
printf('%s and %s', $str1, $str2);

 

So you can explain me where is the error in my code?

 

Thanks for all

[PFMaBiSmAd]

I do this, but appears the same error.

 

What is your current code? And is there any chance you are using a cheap/free web host where changes you make to your files don't take affect immediately due to disk caching?

[pes96]

I do this, but appears the same error.

 

What is your current code? And is there any chance you are using a cheap/free web host where changes you make to your files don't take affect immediately due to disk caching?

 

Now i have done and appears the message: "No database selected"

[PFMaBiSmAd]

Your mysql_select_db() statement is failing, either because the database name you are putting into it doesn't exist or if it does, your database user has not been assigned permission to access it.

 

Again, all code needs error checking and error reporting logic to get it to tell you when and why it is failing. Use the following in place of your mysql_select_db statement -

 

mysql_select_db("database_name", $link) or die("Select db failed: " . mysql_error());

[pes96]

Your mysql_select_db() statement is failing, either because the database name you are putting into it doesn't exist or if it does, your database user has not been assigned permission to access it.

 

Again, all code needs error checking and error reporting logic to get it to tell you when and why it is failing. Use the following in place of your mysql_select_db statement -

 

mysql_select_db("database_name", $link) or die("Select db failed: " . mysql_error());

 

Now appears: "Select db failed: Access denied for user 'a4199910_login'@'10.1.1.43' to database 'Login'"

[jazzman1]

So you can explain me where is the error in my code?

Thanks for all

 

My apologies, your sql  syntax is correct  :-\

First, I thought that the line 28 is just before the query I gave you above.

Secondly, you can make a simple debugging test to check, what actually values you will send to database.

 

<?php
$username = 'jazzman';
$password = 'password';
$id = 1;
$sql1 = "SELECT * FROM user_accounts WHERE username='".$username."' AND password='".$password."' AND user_id='".$id."'";
$sql2 = 'SELECT * FROM user_accounts WHERE username ='.$username.' AND password='.$password.' AND user_id='.$id; // integer id
$sql3 = "SELECT * FROM user_accounts WHERE username='$username' AND password='$password' AND user_id='$id'"; 
$sql4 = 'SELECT * FROM user_accounts WHERE username=$username AND password=$password'; // invalid query

printf('%s<br />%s<br />%s <br />%s', $sql1, $sql2, $sql3, $sql4);

[PFMaBiSmAd]

Now appears: "Select db failed: Access denied for user 'a4199910_login'@'10.1.1.43' to database 'Login'"

 

After you created the database user and the database, did you assign that user permissions to access that database?

[pes96]

So you can explain me where is the error in my code?

Thanks for all

 

My apologies, your sql  syntax is correct  :-\

First, I thought that the line 28 is just before the query I gave you above.

Secondly, you can make a simple debugging test to check, what actually values you will send to database.

 

<?php
$username = 'jazzman';
$password = 'password';
$id = 1;
$sql1 = "SELECT * FROM user_accounts WHERE username='".$username."' AND password='".$password."' AND user_id='".$id."'";
$sql2 = 'SELECT * FROM user_accounts WHERE username ='.$username.' AND password='.$password.' AND user_id='.$id; // integer id
$sql3 = "SELECT * FROM user_accounts WHERE username='$username' AND password='$password' AND user_id='$id'"; 
$sql4 = 'SELECT * FROM user_accounts WHERE username=$username AND password=$password'; // invalid query

printf('%s<br />%s<br />%s <br />%s', $sql1, $sql2, $sql3, $sql4);

 

So i have to put this code where?

[jazzman1]

So i have to put this code where?

I think, you need to start learning php from beginning - step by step .

[pes96]

So i have to put this code where?

I think, you need to start learning php from beginning - step by step .

 

But you can only help me correcting this code, (saying what i have to do) after i will learn php, but now i need this code working fine.

 

Thanks for all

[xyph]

So i have to put this code where?

I think, you need to start learning php from beginning - step by step .

 

But you can only help me correcting this code, (saying what i have to do) after i will learn php, but now i need this code working fine.

 

Thanks for all

 

We are here to guide, not do for you.

[pes96]

So someone can guide me to what i have to do?

 

[jazzman1]

Go back to post #13.

[pes96]

So you can explain me where is the error in my code?

Thanks for all

 

My apologies, your sql  syntax is correct  :-\

First, I thought that the line 28 is just before the query I gave you above.

Secondly, you can make a simple debugging test to check, what actually values you will send to database.

 

<?php
$username = 'jazzman';
$password = 'password';
$id = 1;
$sql1 = "SELECT * FROM user_accounts WHERE username='".$username."' AND password='".$password."' AND user_id='".$id."'";
$sql2 = 'SELECT * FROM user_accounts WHERE username ='.$username.' AND password='.$password.' AND user_id='.$id; // integer id
$sql3 = "SELECT * FROM user_accounts WHERE username='$username' AND password='$password' AND user_id='$id'"; 
$sql4 = 'SELECT * FROM user_accounts WHERE username=$username AND password=$password'; // invalid query

printf('%s<br />%s<br />%s <br />%s', $sql1, $sql2, $sql3, $sql4);

 

So and now?

 

Thanks for your patience

[fenway]

If you can't even prove that you can connect the database, there's really nothing we can do.