A small question concerning modifying db infomation through php.

[Hitoshi]

Ok, here's my problem :
I want to make a RPG type thing for my website. I am a total rookie. I have been learning from a book, I have read pretty much the whole of it but I still don't understand how to do this. I made a table of users, and people can sign up and log in whatever. Now I also added a new collumn "HP" I need to know how I would trigger the HP of a certain user to decrease. I'm sure this is very simple and its just a case of a short paragraph which modifiys the infomation but I can't work it out. I have done very similar things with javascript but this is different because you have to connect to the db to change the stats. please help? If possible can you post the lines of code here?

Thanks.

[drag0n]

Are you trying to change the number for "HP"?
if that is it, then you should use this code:

$sql_up = "UPDATE tablename SET hp = 1 WHERE id = id ";
$result_up = mysql_query( $sql_save ) or die( mysql_error() );

That updates the table "tablename" and sets hp to 1 where the id is "id".

[Hitoshi]

😃 that helps alot! thank you!
one last thing..
I keep getting this error : Query empty
What does this mean?

The whole php now looks like this :

<?php

$page_title = 'Reduce HP';

include ('templates/header.inc');

require_once ('mysql_connect.php');
$sql_up = "UPDATE users SET HP = 1 WHERE id = 1 ";
$result_up = mysql_query( $sql_save ) or die( mysql_error() );

mysql_close();
?>

Whats wrong here?
This should make user_id "1", in table "users" have HP reduced to "1" (It's currenlty 50) How do I reduce is by 1 instead of making it drop to 1? and how do I eliminate my error message? 'mysql_connect.php' is definatly correct as it work on other pages. please help.

[ahundiak]

$sql_up = "UPDATE users SET HP = 1 WHERE id = 1 ";
$result_up = mysql_query( $sql_save ) or die( mysql_error() );
Change $sql_save to $sql_up

[drag0n]

oops my bad:
change:
$result_up = mysql_query( $sql_save ) or die( mysql_error() );
to
$result_up = mysql_query( $sql_up ) or die( mysql_error() );

[Hitoshi]

yeah cool!!! im making this into a proper script now. thanks. how do I increase or decrease the HP by a certain number? Is it a different from SET?

Something like Hitoshi loses 3 HP, or Rob gains 5 HP.
I can do the pop ups in javascipt, just need to work out how to change the values up and down rather than setting them.\

Also, how would u call upon these values in another page? clicking on a link to attack or to restore HP.

I promise not to bother u if it works lol.

EDIT : wait there's more lol.
how do i generate an IF, so that when ur HP drops to 0 or below u get some kind of signal.

[drag0n]

yeah cool!!! im making this into a proper script now. thanks. how do I increase or decrease the HP by a certain number? Is it a different from SET?

No its not different from set, but what you would do is make hp a variable before and then decrease/increase the variable and then in the set, just do SET hp = ".$hp." therefore hp would be dynamic
the rest of the stuff your talking about, unfortuneatley I am very busy with clients, but you should look up:
http://us4.php.net/manual/en/function.mysql-fetch-array.php
and use that to do all those things and you should read a book about PHP youll learn alot more and you wont have to deopend on us for answers 😉

[Hitoshi]

Thanks! lol I have a book but it was for the wrong sort of thing. Is there a php book on .. how to make a forum (I dont want to make one, but there could be a lot of useful content) or how to make multi user games in php? I will have a look 😉

[Hitoshi]

How do I make HP into the variable $hp that comes from the database? I could set
$hp_current = 50
$heal = $hp + 20
$hp = $heal
or sumthing similar. But that wouldnt be coming from the database. I need the 50 bit to be a reference to the database rather than a number

EDIT : appologize for the double posting :S

[drag0n]

Yes I know what your doing
You have to do a select statement and then do mysql_query on that statement then do mysql_fetch_array which puts all the data into an array for ex:

$id = 3 //or you can do dynamic variables.. $_GET or $_POSt etc...
$sql = 'SELECT * FROM game WHERE id = $id';
$res = mysql_query($sql);
$data = mysql_fetch_array($res);
//Now you have all the data in $data and its an array (assoc)
//Now do the calculations on $hp
$hp = $data[hp]
$heal = $hp + 20 
$hp = $heal
$sql_up = "UPDATE game SET HP = $hp WHERE id = $id "; 
$result_up = mysql_query( $sql_up ) or die( mysql_error() ); 

I hope that helps, code is not tested and probably very sloppy, so you mgiht get errors but other then that I think the code should get the job done
Tell me if it works 😉

[Hitoshi]

Parse error: parse error in /home/hitoshi/public_html/heal_user_hp.php on line 13

I tryed this too :
<?php

$page_title = 'HP';

include ('templates/header.inc');

require_once ('mysql_connect.php');

$id = 1
$sql = 'SELECT users FROM HP WHERE id = $id';
$res = mysql_query($sql);
$data = mysql_fetch_array($res);

$hp = $data[HP]
$heal = $hp + 20
$hp2 = $heal
$sql_up = "UPDATE users SET HP = $hp2 WHERE id = $id ";
$result_up = mysql_query( $sql_up ) or die( mysql_error() );

mysql_close();
?>
<script>
alert( 'RECOVERED 20HP!' );
</script>

I thought that prehaps the $hp being used twice had messed it up but... it still doesnt work lol and i get the same error

[tsinka]

Hi,

some ; are missing

<?php 
$page_title = 'HP'; 
include ('templates/header.inc'); 
require_once ('mysql_connect.php'); 

$id = 1;
$sql = 'SELECT users FROM HP WHERE id = $id'; 
$res = mysql_query($sql); 
$data = mysql_fetch_array($res); 

$hp = $data[HP]; 
$heal = $hp + 20; 
$hp2 = $heal; 
$sql_up = "UPDATE users SET HP = $hp2 WHERE id = $id"; 
$result_up = mysql_query( $sql_up ) or die( mysql_error() ); 

mysql_close(); 
?> 
<script language="JavaScript"> 
<!--
alert( 'RECOVERED 20HP!' ); 
//-->
</script> 

Thomas

[JdL]

You can also do this in the SQL statement without requesting the number for HP first:

To increase:

"UPDATE users SET HP = HP+$heal WHERE id = $id"

To decrease do this:

"UPDATE users SET HP = HP-$heal WHERE id = $id"

Saves a few lines of code (and a request to the database)

$query = <one of the update statements>
$result . . . (see previous posts)

[Hitoshi]

ok!
this is my code :
<?php

$page_title = 'HP';

include ('templates/header.inc');

require_once ('mysql_connect.php');

$id = 1;
$hp = $data[HP];

$heal = 20;

$query = "UPDATE users SET HP = HP+$heal WHERE user_id = $id";

$result_up = mysql_query( $query ) or die( mysql_error() );

mysql_close();
echo 'HP increased by 20!';

?>

I don't get errors anymore... but it doesnt update my HP lol. I tried the method before, and it updated my HP but had lots of error messages that I was unable to solve. I would prefer to use this method if somebody could get it to work plz? ultimatly.. I want this script to also charge the user a certain amount on money. I will call this G (gold.) I am sure this only involves doubling the code.. but if u have time plz show me how. thanks.

Hitoshi

[Hitoshi]

ok.. ive been messing around with the code for ages and this is the best i can get :

<?php

$page_title = 'HP';

include ('templates/header.inc');

require_once ('mysql_connect.php');

$id = 1;
$sql = 'SELECT users FROM HP WHERE user_id = $id';

$res = mysql_query($sql);

$data = mysql_fetch_array($res);
$hp = $data[HP];

$heal = 20;

$query = "UPDATE users SET HP = HP+$heal WHERE user_id = $id";

$result_up = mysql_query( $query ) or die( mysql_error() );

mysql_close();
echo 'HP increased by 20!';

?>

it updates as it should but I also get the following message :
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/hitoshi/public_html/hp.php on line 8

[tsinka]

Change the following lines:

$res = mysql_query($sql); 
$data = mysql_fetch_array($res); 

to

$res = mysql_query($sql) or die("Query Error: ".mysql_error()); 
$data = mysql_fetch_array($res) or die("Fetch Error: ".mysql_error()); 

to get an error description.

Thomas

[Hitoshi]

Query Error: Table 'hitoshi_web.HP' doesn't exist

the database name is hitoshi_web
the table is called users
HP is a column

:S😕

[tsinka]

Ok,

what happens if you change

$sql = 'SELECT users FROM HP WHERE user_id = $id';

to

$sql = 'SELECT HP FROM users WHERE user_id = $id';

Thomas

[Hitoshi]

ahh!!! now i get another one!!
Query Error: Unknown column '$id' in 'where clause'
will it ever end!?:o

[tsinka]

Sorry,

didn't see that 🙁

$sql = 'SELECT HP FROM users WHERE user_id = '.$id;

or

$sql = "SELECT HP FROM users WHERE user_id = $id";

or

$sql = "SELECT HP FROM users WHERE user_id = ".$id;