Dont find in the db ?

Anon

Hello!

What might be wrong here please, regardless what I write in the two forms it says it is wrong, even if there is a(there can be one at the most) corsponding "email" and "bnumber" in the table 🙁

Thanks in advance.

Klas

On the first page

<FORM METHOD="Post" ACTION="pwdcheck.php?">
E-mail:<br>
<INPUT TYPE="text" NAME="mail" value="33"><p>
Lösenord:<br>
<INPUT TYPE="text" NAME="bnummer" value="5845"><p>
<INPUT TYPE="submit" value="Press here">
</FORM>

And on the next(pwdcheck.php):

<?PHP $db = @mysql_connect("localhost:3307","root","root");
mysql_select_db("bokning",$db);

$result = mysql_query("SELECT * FROM deltagare WHERE 'email'='$mail'",$db);

if ($myrow = mysql_fetch_array($result))
{
printf("$myrow[number]");
}
else
{
printf("Wrong e-mail and/or number");
}
?>

sarahk

("SELECT * FROM deltagare WHERE 'email'='$mail'",$db);

should be

("SELECT * FROM deltagare WHERE email ='$mail'",$db);

try that for starters.

Another good idea is to have

$sqlstr = "SELECT * FROM deltagare WHERE email='$mail'";
echo $sqlstr;
$result = mysql_query( $sqlstr, $db);

that way you can copy the sqlstr from the browser and paste directly into mysql to see what the problem is. Often seeing it printed out can be enough.

Anon

Hi!

Thanks for the advice, espially the 2:nd one, very bright 🙂 It's great that I can try the thing directly in SQL!!! Thanks, I will do this way from now 🙂

Anyway, I corrected the SQL statement 🙂 and I got:

SELECT * FROM deltagare WHERE email='eeee'
Warning: Supplied argument is not a valid MySQL result resource...line 15.

So, I pasted SELECT * FROM deltagare WHERE email='eeee' into mySQL-front which I use for mySQL, and there it found the correct row, how strange, after further brainstorming, I think this is the incorrect line(line 15). This is "true" as long anythis has went into $result, right? So I dont understand why this line is an error 🙁

if ($myrow = mysql_fetch_array($result))

Klas

Anon

Yes, when I took this(se beneith) away, the error dissapeard. How strange. How do I please write to check is there is any result from the SQL question, and if there is print something out, and else print "wrong id and/or password" please?

Huge thanks in advance again.

Klas

if ($myrow = mysql_fetch_array($result))
{
printf("$myrow[kvittensnummer]");
}
else
{
printf("Fel e-mail och/eller bokningsnummer");
}

Anon

Hello!

It is working now, thanks for the help. I send the code here if anyone is intrested.

Thanks Sarah 🙂

Klas

<?PHP $db = @mysql_connect("localhost","root","root");
mysql_select_db("bokning",$db);

$sqlstr = "SELECT kvittensnummer FROM deltagare WHERE email='$mail'";
echo $sqlstr;
$result = mysql_query( $sqlstr, $db);
echo mysql_error();
?>

<?
if ($myrow = mysql_fetch_array($result))
{
printf("$myrow[kvittensnummer]");
}
else
{
printf("Wrong e-mail and/or number");
}