Chapter 13 login failure

I am using:
phpMyAdmin 2.6.2-pl1
MySQL 4.1.18-log running on localhost

Chapter 13.
I can register successfully.
I click on the link in the email generated by the registration and receive a message that my account is activated and I can log in.
I enter the correct email and password, but I get the error message that either the password/username is not right or else my account is not activated. No match is found by the query, and 0 rows are affected...but when I run the relevant query from phpMyAdmin, it works!

I try logging in using 3 different browsers. All fail.

I am using all scripts that accompany the book verbatim except for making the needed changes for the database connection info and the activating link definition.

Where should I be looking for the solution?

Have you considered the change in the PASSWORD() function in MySQL with version 4.1? See this post for details: [www.dmcinsights.com]


____________

Paul Swanson
Portland, OR
USA
____________

PHP Version 4.4.0
MySQL Version 4.1.22
Apache Version 1.3.37
Server OS: Linux Version 2.4.20-8
Workstation OS: Windows XP Pro SP2

Thanks for tip.
I looked at the thread you referred to, and I changed the password field to 41 chars

But the login still does not work.

After you changed the password field, did you re-register? You need to change the value held in the password column for it to run correctly. Try re-registering so that the new (encoded) password value is stored instead of the old value and try logging in again.

VGMusic

Yes, I did reregister after changing the password field, but the login still fails.

PHP 5.1.4
Apache 1.3.34 (UNIX)
MySQL 4.1.18
phpMyAdmin 2.6.2-pl1

The example scripts from earlier chapters of the book which used a password worked fine! And as I said, when I run the relevant query from phpMyAdmin, it works. I just can't get it to work from any browser.

I don't think the PASSWORD() field is the issue as I don't use that function in this book. What is the query the PHP script is trying to run?

Best Wishes,
Larry

Writer/Web Developer/Instructor
Forum Moderator

The query is from line 34 of script 13.8 login.php:

$query = "SELECT user_id, first_name FROM users WHERE (email='$e' AND pass=SHA('$p')) AND active IS NULL";

Right, but what is the query that the PHP script is actually trying to run WITH THE VARIABLE VALUES? There's nothing wrong with that query unless...
- the variable values aren't getting in there properly
- you don't have a connection to the database
- your table differs from the one in the book

If you print the value of $query, you can confirm the first possible cause. If you use the mysql_error() function, you can confirm the second and third.

Best Wishes,
Larry

Writer/Web Developer/Instructor
Forum Moderator

//EXCERPT FROM SCRIPT 13.8
...
if ($e && $p) { // If everything's OK.

// Query the database.
$query = "SELECT user_id, first_name FROM users WHERE (email='$e' AND pass=SHA('$p')) AND active IS NULL";
$result = mysql_query ($query) or trigger_error("Query: $query\n<br />MySQL Error: " . mysql_error());

//THIS IS CODE I ADDED TO TRY TO TEST AS YOU SUGGESTED
$errors[] = mysql_error().'<br /><br />Query: '.$query;
if (!empty($errors))
echo '<h1 id="mainhead">Error!</h1>
<p class="error">The errors are:<br />';
foreach ($errors as $msg){
echo " - $msg<br />\n";
}
echo 'Number of rows affected: '.mysql_num_rows($result);
//END OF TEST CODE

if (@mysql_num_rows($result) == 1) { // A match was made.
...

//THIS IS RESULT
Error!

The errors are:
-

Query: SELECT user_id, first_name FROM users WHERE (email='info@iplusnow.com' AND pass=SHA('sayid')) AND active IS NULL
Number of rows affected:0

QUESTION
Looks to me like the values are inserted correctly and no errors are produced. The trigger_error() function never produced any result. Have I not tested for all 3 of the possiblities you listed?

Yep, that all looks good. Okay, so you've confirmed that the query is valid and that it is executing in MySQL. There are two possible problems:

- your table differs from the one in the book (you haven't confirmed this). Specifically you might have problems if your active column is defined as NOT NULL.

- You are referring to the number of "affected" rows but a SELECT query NEVER has any "affected" rows. I think this is just a terminology issue, as you are using mysql_num_rows(), not mysql_affected_rows(), correct? However, it would be a good idea to remove the @ from in front of mysql_num_rows() just in case.

Finally, just to confirm, if you run the query "SELECT user_id, first_name FROM users WHERE (email='info@iplusnow.com' AND pass=SHA('sayid')) AND active IS NULL" using phpMyAdmin you do get exactly one row returned, right? And both phpMyAdmin and your PHP scripts are using the same database?

Best Wishes,
Larry

Writer/Web Developer/Instructor
Forum Moderator

My table was different. I had the pass field set to UNIQUE instead of the email field, and I did not have the pass field indexed. I had overlooked that. Works like a charm now!

I don't know how you have time to answer everybody's questions, but I sure do appreciate the help. Mille Grazie! (A thousand thanks.)

Matteo

You're very welcome. I did suspect there was a table definition problem and what you described would certainly be the cause.

Best Wishes,
Larry

Writer/Web Developer/Instructor
Forum Moderator