Script help??

[MobiTec]

EDITED
Hi, i'm wantin to get a WAP enabled login script so that ppl have to login to enter my wap site.. i think i've managed to find 1 here.. http://www.1phpstreet.com/vb/scripts/ShowCodeAsText.asp?txtCodeId=326&lngWId=8

i have uploaded it all and i found it didnt work.. i think it cos there is no .SQL file, but there is a create_db.php file that i think is suppose to create it for me, that didnt work..

so i created this .SQL table out of the create_db.php file

CREATE TABLE login (
user varchar(21) NOT NULL default '3',
pass varchar(21) NOT NULL default '3',
name varchar(35) NOT NULL default '3',
email varchar(42) NOT NULL default '3'
) TYPE=MyISAM;

and uploaded it to my MySql database and it created it..

the thing is.. now i get the error >>

Warning: Supplied argument is not a valid MySQL result resource in /home/httpd/vhosts/Mobitec.me.uk/httpdocs/members/login.php on line 7 .... WHY!!! 😕 😕

<? 
    require("config.php");
    mysql_connect($server,$user,$pass); 
    mysql_select_db($dbname); 
    $select_data=("SELECT name FROM logs WHERE pass='$password' AND user='$username' "); 
    $ver=mysql_query($select_data); 
    $string=mysql_fetch_array($ver);     <<<<<<<<<<
    if ($string==0){ 
    echo"<center>You are not in database<br> 
    <a href=register.html>Register</a></center> 
    "; 
    } 
    else 
    echo"<center><font color=blue size=+2>Welcome $string[0]</font></center>"; 
    ?>

can some 1 help me fix this problem.. pls😕

Thank you xx 🙂 🙂

[tasistro]

Try changing

$select_data=("SELECT name FROM logs WHERE pass='$password'AND user='$username' ");

To exactly this

$select_data="SELECT name FROM logs WHERE pass='$password' AND user='$username' ";

There appeared to be a missing space between ' and AND

[MobiTec]

thanx.. but it did not work.. 🙁

the prob seams to be line 7 but i cant figure it out.. lol

can any 1 else help???

[feldon23]

You have a loose else, too.

should be:

if (condition==true) {
do stuff
} else {
do other stuff
}

[MobiTec]

oops.. missed that.. lol 😉

thats still not the problem tho...

this is all new to me so, not quite got to know everything but
its sumthin to do with line 7

and i dont know wat.. 🙁 😕

[agoe]

I think you use wrong tablename. You create a table, but use another in the query.

Check out what mysql_error says. It probably states that the table "logs" doesn't exist.

[Frederick]

$sql = mysql_query("SELECT * FROM login WHERE pass='$password' AND user='$username' ")
or die(mysql_error());

if (mysql_numrows($sql)) {
$var = mysql_fetch_array($sql);
    if ($var["pass"] == $password && $var["user"] == $username) {
        echo"<center><font color=blue size=+2>Welcome ".$var["uname"]."</font></center>";
    } else {
        echo "Incorrect Login";
    }
} else {
    echo"<center>You are not in database<br><a href=register.html>Register</a></center> ";
}

[dustbuster]

change

SELECT name FROM logs WHERE pass='$password' AND user='$username'

to

SELECT name FROM login WHERE pass='$password' AND user='$username'

Unless you provided us with an innacurate table definition you are accessing the wrong table

For troubleshooting

$ver=mysql_query($select_data)  or die ("Invalid Query: ".mysql_error());

[MobiTec]

Thank you ALL very much

you have all been a great help!!!!!

(Shakes all yr hands)

😃 😃 😃 😃 Much appreciated 😃 😃 😃 😃