[SOLVED] $sqlcheckn = mysql_query help

[NSW42]

Heya all, what im trying to do and without success is to get my nickname change script to detect the nickname is available or taken, it updates the nick fine but i need it to say nick in use try again, ive tried playing around with it but im still very new to all this, my coding posted below and any help is appreciated, thanks...

 

<? 

include("config.php");

 

include("user_check.php");

 

print("<left><a href='$PHP_SELF?action=pass'><a href='$PHP_SELF?action=profile'><font color=red></a></center><BR><BR>");

$sql = mysql_query("SELECT * FROM users WHERE username = '$_COOKIE[username]'");

$row = mysql_fetch_array($sql);

 

 

if($action==update)

{

$update4 = mysql_query("UPDATE users SET nickname = '$_POST[nickname]' WHERE username = '$_COOKIE[username]'");

print("Your information is now updated.");

}

 

 

 

{

print("

<table>

<tr><td align=center><font color=ffffcc><b>Nickname Information</b></td></tr>

 

 

<form action=$PHP_SELF?action=update method=post>

<tr>

  <td><font color=red>Current NickName</td>

  <td><input type=text name=nickname value='$row[nickname]' size=25 maxlength=60></td>

 

</tr>

<tr>

 

 

</td>

</tr>

<tr>

 

</tr>

<tr>

<tr><td align=center><input type=submit value='Click To Update'></td></tr>

</form>

</table>

 

 

this is the $sqlcheckn that works fine for new signups that ive been trying to add into the above script with no luck..

 

 

$sqlcheckn = mysql_query("SELECT * FROM users WHERE nickname = '$_POST[nickname]'");

$checkn = mysql_num_rows($sqlcheckn);

 

if((!$nickname) || ($checkn > 0))

 

if($checkn > 0)

{

print("Nickname already in use!<BR>");

unset($nickname);

}

 

 

 

[btherl]

Instead of

 

$update4 = mysql_query("UPDATE users SET nickname = '$_POST[nickname]' WHERE username = '$_COOKIE[username]'");

 

use

 

$sql = "UPDATE users SET nickname = '$_POST[nickname]' WHERE username = '$_COOKIE[username]'";
$update4 = mysql_query($sql) or die("Mysql error in $sql\nError: " . mysql_error());

 

Make that change to each mysql_query() call.  Otherwise you will not be notified of errors.

[NSW42]

it updates the nickname without a hiccup now, but if someone puts in the same nick it adds it to the db, thats where i need the sqlcheck so it can say no nickname taken or if its not, then just update it, sorry if  I may have gotten your reply wrong as im still very new to this and trying very hard to learn it as i go along...

[NSW42]

ok i kinda worked out what you were saying and added in this below, it now detects if nickname is taken, but it wont halt and inserts the nickname, half way there i guess, anymore ideas so i can finsih it off would be good thanks...

 

 

<? 

include("config.php");

 

include("user_check.php");

 

print("<left><a href='$PHP_SELF?action=pass'><a href='$PHP_SELF?action=profile'><font color=red></a></center><BR><BR>");

$sql = mysql_query("SELECT * FROM users WHERE username = '$_COOKIE[username]'");

$row = mysql_fetch_array($sql) or die("Mysql error in $sql\nError: " . mysql_error());

$sqlcheckn = mysql_query("SELECT * FROM users WHERE nickname = '$_POST[nickname]'");

$checkn = mysql_num_rows($sqlcheckn);

 

if((!$nickname) || ($checkn > 0))

 

if($checkn > 0)

{

print("Nickname already in use!<BR>");

unset($nickname);

}

if($action==update)

{

$sql = "UPDATE users SET nickname = '$_POST[nickname]' WHERE username = '$_COOKIE[username]'";

$update4 = mysql_query($sql) or die("Mysql error in $sql\nError: " . mysql_error());

print("Your information is now updated.");

}

 

 

 

{

print("

<table>

<tr><td align=center><font color=ffffcc><b>Nickname Information</b></td></tr>

 

[Waldir]

<?
include("config.php");

include("user_check.php");

print("<left><a href='$PHP_SELF?action=pass'><a href='$PHP_SELF?action=profile'><font color=red>[/url]</center><BR><BR>");
$sql = mysql_query("SELECT * FROM users WHERE username = '$_COOKIE[username]'");
$row = mysql_fetch_array($sql) or die("Mysql error in $sql\nError: " . mysql_error());
$sqlcheckn = mysql_query("SELECT * FROM users WHERE nickname = '$_POST[nickname]'");

if($action==update) {
if(mysql_num_rows($sqlcheckn) > 0)
{
print("Nickname already in use!<BR>");
unset($nickname);
} else {
$sql = "UPDATE users SET nickname = '$_POST[nickname]' WHERE username = '$_COOKIE[username]'";
$update4 = mysql_query($sql) or die("Mysql error in $sql\nError: " . mysql_error());
print("Your information is now updated.");
        }
}
?>

[NSW42]

i tried that and got errors everywhere  :-\ heres the complets script if u could look at it and see whats going on, sorry to be a pain and thanks agian..

 

<H3>Edit Your Nickname</H3>

 

<td width='240'>

 

<? 

include("config.php");

 

include("user_check.php");

 

print("<left><a href='$PHP_SELF?action=pass'><a href='$PHP_SELF?action=profile'><font color=red></a></center><BR><BR>");

$sql = mysql_query("SELECT * FROM users WHERE username = '$_COOKIE[username]'");

$row = mysql_fetch_array($sql) or die("Mysql error in $sql\nError: " . mysql_error());

$sqlcheckn = mysql_query("SELECT * FROM users WHERE nickname = '$_POST[nickname]'");

$checkn = mysql_num_rows($sqlcheckn);

 

if((!$nickname) || ($checkn > 0))

 

if($checkn > 0)

{

print("Nickname already in use!<BR>");

unset($nickname);

}

if($action==update)

{

$sql = "UPDATE users SET nickname = '$_POST[nickname]' WHERE username = '$_COOKIE[username]'";

$update4 = mysql_query($sql) or die("Mysql error in $sql\nError: " . mysql_error());

print("Your information is now updated.");

}

 

 

 

{

print("

<table>

<tr><td align=center><font color=ffffcc><b>Nickname Information</b></td></tr>

 

 

<form action=$PHP_SELF?action=update method=post>

<tr>

  <td><font color=red>Current NickName</td>

  <td><input type=text name=nickname value='$row[nickname]' size=25 maxlength=60></td>

 

</tr>

<tr>

 

 

</td>

</tr>

<tr>

 

</tr>

<tr>

<tr><td align=center><input type=submit value='Click To Update'></td></tr>

</form>

</table>

 

");

}

 

 

{

 

include("$sitepath/html/footer.php");

exit;

 

}

include("$sitepath/html/footer.php");

?>

[NSW42]

ok thanks very much for all your help, I have managed to fix it, once again thank you.