very simple singup script

[logged_with_bugmenot]

Hi guys another signup script, I've been reading here for a few hours now and still can't solve my problem so I decided to post here.

 

Anyways the problem is the data isn't getting to the mysql here's my code by the way the signup is more of a add user script and is only available once someone has signed in:

 

http://privatepaste.com/d915cqIufh

[aniesh82]

You did not specify the action page on the form.

 

<form method="post" name="frmadd" id="frmadd">

 

Change to : <form method="post" name="frmadd" id="frmadd" action="NameOFactionPAge.php">

 

The user name and password to connect to your mysql database is available on the page. It may be  misused. So please delete the page as soon as possible.

 

 

[logged_with_bugmenot]

Ah, I tried adding that but didn't help and my login script doesn't specify a action and continues to work.

 

I made another script which directly inserts a name and password with no form and that worked so I think it has to be something with my syntax, any other ideas?

[aniesh82]

 

Please use the following lines of code:

 

<?php

 

    if ($_POST['user'] != NULL && $_POST['pass'] != NULL )

    {

 

/*  Please add the two lines of code to connect to your MySql

  and select MySql Database */

 

$user = $_POST['user'];

$pass = $_POST['pass'];

 

 

$query=mysql_query("INSERT INTO users_logins (user_id, user_password) VALUES ('$user', PASSWORD('$pass'))")

or die('Error' . mysql_error());

 

 

mysql_close($con);

}

?>

 

<h1>Login successful</h1>

 

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

<input name="user" type="text" id="user">

<input name="pass" type="password" id="pass">

<input type="submit" name="btnadd" value="Add">

</form>

[logged_with_bugmenot]

Still no go, I thought that would be right to.

 

*looks for something to blame*

[aniesh82]

Did you add the lines for connecting to MySql and select Database. Are you sure that database and table with the same name exists ?

 

Also please check after submit the form, it goes to the same page.

 

One change needed:

 

1. Change the action page with name of your action page

 

The you assign the sql query to a variable and prints it:

    $sql ="INSERT INTO users_logins (user_id, user_password) VALUES ('$user', PASSWORD('$pass'))";   

    echo $sql; exit();

    $query=mysql_query($sql);

[aniesh82]

 

Add this lines to check whether your program makes  successful connection with database

 

$con = mysql_connect('mysql2.freehostia.com','greval5_greval5','somepassword') or die("Couldnot connect to MySql");

mysql_select_db("greval5_greval5",$con) or die("Couldnot select Database");

 

 

[logged_with_bugmenot]

I made the action the same script (main.php)

 

and also changed the query with the most current line you gave me and it did not return the error and the echo was:

 

INSERT INTO users_logins (user_id, user_password) VALUES ('john', PASSWORD('smith'))

 

also the tables do exist because I'm able to login using the users i created through phpmyadmin sql interface... very odd...

[aniesh82]

 

Please copy the line of text and run using phpmyadmin

 

INSERT INTO users_logins (user_id, user_password) VALUES ('john', PASSWORD('smith'))

[logged_with_bugmenot]

I ran INSERT INTO users_logins (user_id, user_password) VALUES ('john', PASSWORD('smith')) and it added the user:

 

Full Texts    user_id    user_password

Edit Delete cross 46c3909c48c7e704

Edit Delete sam 7888350b26333e97

Edit Delete john 1dacdbf749f8f2a9

 

somewhere it mustn't be inserting anything.

[aniesh82]

 

The records added successfully  =>  john    1dacdbf749f8f2a9

[logged_with_bugmenot]

Yeah, but I want to add it through the script this was through phpmyadmin really odd

[aniesh82]

 

I think some connection problem still existing. 

 

Could you replace the sql query with this ?

 

$sql = 'SELECT * FROM users_logins";

$re = mysql_query($sql);

echo "Th e number of Records ".mysql_num_rows($re);

 

If the connection is established, then it will output the number of records in user_logins table.

 

[logged_with_bugmenot]

Hmm, it outputs "Th e number of Records 3"

 

this is weird