seyidesh's Profile User Rating: -----

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  1. In Topic: .HTACCESS Error

    Posted 15 May 2013

    Thank you guys, i have been a able to revolved this problem myself.

    I only remove the # sign before the LoadModule rewrite_module modules/mod_rewrite.so in httpd.conf file of my wamp server and everything work perfect.
  2. In Topic: WYSIWYG Error

    Posted 14 May 2013

    View Postno2pencil, on 14 May 2013 - 10:47 AM, said:

    A quick & dirty way would be to use the php function strip_tags().


    strip_tags() is removing all the html tag from the post making it look as if i have not apply wysiwyg editor at all.

    What i want is to be able to display the WYSIWYG effect as i have applied it. But what am getting is the raw code of the effect from WYSIWYG looking like: I don <font color="#660066"><font size="6">Port</font> </font>oooo instead of giving me something like this: I don Port oooo
  3. In Topic: Login page Query Error

    Posted 9 May 2013

    View Postandrewsw, on 09 May 2013 - 06:25 AM, said:

    I note also that you are referring to $row:

    if($row['active'] !=NULL){
    

    but it won't exist because you've commented out the preceding lines.

    I can't run the code without a database. I suppose you could post the CREATE TABLE statement that created your table, but it shouldn't be necessary to do this: some sensible debugging should discover the problem.

    Open the database table in PhpMyAdmin and copy the password from there: does it match
    52caba604eb8817b9b62d66a19941a50087919dd
    (Perhaps the field length is too short and this value is truncated in the table :)/>)


    Thanks, you have given me a progressive cue. i think the error is coming from this line of codes:

    	if (preg_match('%\A(?=[-_a-zA-Z0-9])\S{4,}\z%',  stripslashes(trim($_POST['userpass'])))){
    			$p = escape_data($_POST['userpass']);
    			}else {
    		$p = FALSE;
    		echo 'Please enter a valid Password';
    	}
    
    and

    user_pass='".sha1($p) ."'
    


    The password generated when print_r($sql) is 52caba604eb8817b9b62d66a19941a50087919dd and the password save in PhpMyAdmin is 66ed1c42a2a8250d3dea00330e6483, The length is longer and it is not the same.

    Can this codes: user_pass='".sha1($p) ."' be the source of the error? or why is the value generated not the same thing with the one the database?
  4. In Topic: Login page Query Error

    Posted 9 May 2013

    View Postandrewsw, on 09 May 2013 - 05:27 AM, said:

    View Postseyidesh, on 09 May 2013 - 12:17 PM, said:

    Please also note that when i print_r($result) the result returned is: Resource id #6 which i did not know the meaning and where it is coming from.

    Yes, $result is a resource and 6 is a number used internally by PHP, so printing $result isn't (usually) useful. You might print out the number of rows, or individual fields.

    I would also correct the alignment of brackets, to make it easier to read and spot errors in logic.


    I try to print out the number of $row, non was able to print. Can you please do me a favour?

    Kindly copy the code and i posted in post 1 and run it on your system to see if it work. I have the form and the action codes all posted.

    Maybe my system need some config, i don't know? but i was able to create my registration form and it works perfectly. am surprise this is not working?

    Thanks.
  5. In Topic: Login page Query Error

    Posted 9 May 2013

    Please also note that when i print_r($result) the result returned is: Resource id #6 which i did not know the meaning and where it is coming from.

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