Update Field Text

ORLYman · March 3, 2009

I need help modifying/adding/fixing my script so that after the NPC id field is filled in and the user hits enter, the Display Id field updates with the number from the query result.

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>index</title>
<style type="text/css">
<!--
body {
background-color: #000;
}
body,td,th {
color: #FFF;
}
-->
</style></head>

<body>
<h1 align="center">Display Id Lookup Tool</h1>
<p align="center">Created By: Vindicated</p>
<form id="form1" name="form1" method="post" action="">
  <p align="center"> </p>
  <p align="center">
    <label>NPC 
      
      Id:
<input type="text" name="npcid" id="npcid" />
      Display Id:
    </label> 
    <input name="displayid" type="text" id="displayid" value="" size="45" />
  </p>
  <p align="center">
    <input type="submit" name="submit" id="submit" value="Submit" />
  </p>
</form>
<?php
$servername = "localhost:3306";
$username = "root";
$password = "gojags";
$worlddb = "twink_world";

mysql_connect($servername, $username, $password);

mysql_select_db($worlddb, $connection);

$result = mysql_query("SELECT male_displayid FROM creature_names");

while($row = mysql_fetch_array($result))
{
echo $_GET[$result];
}

echo '<displayid>'.$row[male_displayid].'</displayid>';

mysql_close($connection)
?>
<p>*Created by Vindicated for MMOwned.com. Unacceptable for public use without permission.*</p>
</body>

</html>

ORLYman · March 4, 2009

Bump!

samshel · March 4, 2009

r u looking for something liek this ? u have to change the field name to the one in ur table for npcid, also i have moved php code above HTML code.

<?php
$servername = "localhost:3306";
$username = "root";
$password = "gojags";
$worlddb = "twink_world";

mysql_connect($servername, $username, $password);
mysql_select_db($worlddb, $connection);

if($_POST['npcid']) {

$result = mysql_query("SELECT male_displayid FROM creature_names WHERE npcid = '".$_POST['npcid']."'"); // change WHERE npcid to field name in ur DB.
$row = mysql_fetch_array($result))
$DisplayId = $row['male_displayid'];
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>index</title>
<style type="text/css">
<!--
body {
background-color: #000;
}
body,td,th {
color: #FFF;
}
-->
</style></head>

<body>
<h1 align="center">Display Id Lookup Tool</h1>
<p align="center">Created By: Vindicated</p>
<form id="form1" name="form1" method="post" action="">
  <p align="center"> </p>
  <p align="center">
    <label>NPC 
      
      Id:
<input type="text" name="npcid" id="npcid" />
      Display Id:
    </label> 
    <input name="displayid" type="text" id="displayid" value="<?php if (isset($DisplayId)) echo $DisplayId; ?>" size="45" />
  </p>
  <p align="center">
    <input type="submit" name="submit" id="submit" value="Submit" />
  </p>
</form>

<p>*Created by Vindicated for MMOwned.com. Unacceptable for public use without permission.*</p>
</body>

</html>

ORLYman · March 4, 2009

Parse error: parse error in C:\wamp\www\Vindicated's Display Id Lookup Tool\index.php on line 13

samshel · March 4, 2009

sorry forgot semicolon on line 13

$row = mysql_fetch_array($result));

ORLYman · March 5, 2009

One more minor problem. here is the script in its entirety

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>index</title>
<style type="text/css">
<!--
body {
   background-color: #000;
}
body,td,th {
   color: #FFF;
}
-->
</style></head>

<body>
<h1 align="center">Display Id Lookup Tool</h1>
<p align="center">Created By: Vindicated</p>
<form id="form1" name="form1" method="post" action="">
  <p align="center"> </p>
  <p align="center">
    <label>NPC
     
      Id:
<input type="text" name="npcid" id="npcid" />
      Display Id:
    </label>
    <input name="displayid" type="text" id="displayid" value="<?php if (isset($DisplayId)) echo $DisplayId; ?>" size="45" />
  </p>
  <p align="center">
    <input type="submit" name="submit" id="submit" value="Submit" />
  </p>
</form>

<p>*Created by Vindicated for MMOwned.com. Unacceptable for public use without permission.*</p>
</body>

</html>
<?php
$servername = "localhost:3306";
$username = "root";
$password = "gojags";
$worlddb = "twink_world";

$connection = mysql_connect($servername, $username, $password);
mysql_select_db($worlddb, $connection);

if($_POST['npcid']) {

   $result = mysql_query("SELECT male_displayid FROM creature_names WHERE npcid = '".$_POST['npcid']."'"); // change WHERE npcid to field name in ur DB.
   $row = mysql_fetch_array($result);
   $DisplayId = $row['male_displayid'];
}
?>

And the error I get is:

Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'root'@'localhost' (using password: YES) in C:\wamp\www\Vindicated's Display Id Lookup Tool\index.php on line 46

Warning: mysql_select_db(): supplied argument is not a valid MySQL-Link resource in C:\wamp\www\Vindicated's Display Id Lookup Tool\index.php on line 47

Notice: Undefined index: npcid in C:\wamp\www\Vindicated's Display Id Lookup Tool\index.php on line 49

before using it

and

*Created by Vindicated for MMOwned.com. Unacceptable for public use without permission.*

Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'root'@'localhost' (using password: YES) in C:\wamp\www\Vindicated's Display Id Lookup Tool\index.php on line 46

Warning: mysql_select_db(): supplied argument is not a valid MySQL-Link resource in C:\wamp\www\Vindicated's Display Id Lookup Tool\index.php on line 47

Warning: mysql_query() [function.mysql-query]: Access denied for user 'ODBC'@'localhost' (using password: NO) in C:\wamp\www\Vindicated's Display Id Lookup Tool\index.php on line 51

Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in C:\wamp\www\Vindicated's Display Id Lookup Tool\index.php on line 51

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\wamp\www\Vindicated's Display Id Lookup Tool\index.php on line 52

After. Also how could I add a message to display if the ID does not exsist?

samshel · March 5, 2009

the error speaks for itself, it is not able to connect to mysql as user/password are not correct.

ORLYman · March 6, 2009

But it is :S

revraz · March 6, 2009

Access Denied means just that. The ID/PW doesn't have connect rights.

Sign In

Update Field Text

Recommended Posts

ORLYman

Link to comment

Share on other sites

ORLYman

Link to comment

Share on other sites

samshel

Link to comment

Share on other sites

ORLYman

Link to comment

Share on other sites

samshel

Link to comment

Share on other sites

ORLYman

Link to comment

Share on other sites

samshel

Link to comment

Share on other sites

ORLYman

Link to comment

Share on other sites

revraz

Link to comment

Share on other sites

Archived

Browse

Activity

Important Information