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User Log in

Discuss User Log in in the PHP Coding forum on Codewalkers.
User Log in Having problems with a PHP script you are coding? This is the place to get help!

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  #1  
Old July 27th, 2002, 08:17 PM
russ russ is offline
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User Log in
Oh no, russ has another problem....sorry guys. Anyone know any decent user login tutorials? and a comments tutorial too?

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  #2  
Old July 27th, 2002, 09:07 PM
Gunblade Gunblade is offline
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RE: User Log in
Ive seena comments tutorial somewhere, ill try and find u it again.

With a login tutorial, try www.hotscripts.com then go to php>>user authentication
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  #3  
Old July 27th, 2002, 09:11 PM
EvilivE EvilivE is offline
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RE: User Log in
Single line comments, use //
Multiple line comments use /* */
php Code:
Original - php Code 
  1. <?php
  2. //This line is a comment
  3. echo "This is not a comment";
  4. /*
  5. This
  6. is
  7. a
  8. multiple
  9. line
  10. comment
  11. */
  12. ?>
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  #4  
Old July 27th, 2002, 09:34 PM
notepad notepad is offline
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RE: User Log in
lol Evi..
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  #5  
Old July 28th, 2002, 12:55 AM
russ russ is offline
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RE: User Log in
hey guys, thanks for your help. I found one at phpbeginner. but when I add this to my script

setcookie ("ck_user", "");
setcookie ("ck_password", "");
setcookie ("ck_user_id", "");

i get this error

Warning: Cannot add header information - headers already sent by (output started at /home/gmflp/public_html/main.php:3) in /home/gmflp/public_html/scripts/head.php on line 31
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  #6  
Old July 28th, 2002, 01:07 AM
EvilivE EvilivE is offline
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RE: User Log in
read this
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  #7  
Old July 28th, 2002, 01:18 AM
russ russ is offline
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RE: User Log in
Cheers man
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  #8  
Old July 28th, 2002, 01:38 AM
russ russ is offline
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RE: User Log in
okay, another error, after I submit the form, I get this error

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/gmflp/public_html/test.php on line 76

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/gmflp/public_html/test.php on line 78


that is on this line

php Code:
Original - php Code 
  1.  
  2.             //get the user_id from the result 
  3.             $num_rows = mysql_num_rows($result); 
  4.             $row = mysql_fetch_array($result); 
  5.             $user_id = $row[0]; 
  6.  
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  #9  
Old July 28th, 2002, 03:58 AM
D1NGO D1NGO is offline
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RE: User Log in
im pretty sure that means theres an error in your SQL syntax.
do this:
php Code:
Original - php Code 
  1.  
  2. $result = mysql_query($SQL) or die(mysql_error());

that should get you your problem
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  #10  
Old July 28th, 2002, 04:32 AM
EvilivE EvilivE is offline
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RE: User Log in
ditto.

line 76: $num_rows = mysql_num_rows($result);
line 78: $row = mysql_fetch_array($result);

What do these two lines have in common? tic-toc A: $result
The error msg reads "supplied argument is not a valid MySQL result resource". What is the supplied argument? A: $result

What is $result? A variable. That is most likely assigned the value mysql_query($query)

So mysql_query() did _NOT_ return a valid MySQL result resource. Why not? Probably, 9 out of 10 times, is because $query is not a properly formatted SQL statement or some indicies your trying to collect are not legit.

So even though the error is on line 76 and 78, the problem is before that ... probably line 74.

Good luck.

PS ... it is good practice to run your queries in your mysql client to writing them in code.
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  #11  
Old July 29th, 2002, 12:48 AM
russ russ is offline
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RE: User Log in
okay, I have got it narrowed down to this now

MySQL said:


You have an error in your SQL syntax near '$sql = "SELECT id "
. "FROM membersdb "
. "WHERE use' at line 1


I personally can't see anything wrong with it. I think I might be the word sql, but I've tried numbering it and stuff, and its just not working, I find this wierd as this is the code in the tutorial
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  #12  
Old July 29th, 2002, 12:48 PM
Gunblade Gunblade is offline
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RE: User Log in
Am not good wit sql but try:

'$sql = "SELECT id" ;
."FROM membersdb" ;
."WHERE USE" at line 1';
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  #13  
Old July 29th, 2002, 01:03 PM
ASk ASk is offline
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RE: User Log in
for mysql queries use:

$result=mysql_query("SELECT id FROM membersdb WHERE yourwheresyntax")
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  #14  
Old July 29th, 2002, 08:09 PM
EvilivE EvilivE is offline
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RE: User Log in
Russ,

When you execute a query using your mysql client, just send the sql statement. For example, SELECT id FROM memberdb.

It looks like you tried diving it some PHP code ... $sql = SELECT id from membersdb. Which, as you can, mysql will not understand it and send you an error msg.

HTH
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  #15  
Old July 29th, 2002, 09:54 PM
russ russ is offline
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RE: User Log in
okay guys, I got it right this time, I had to number the result var
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