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Check for data for that specific player
Posted: Tue Sep 27, 2011 1:59 pm
by vitinho444
Hey Indie resource community!
Im trying to make a simple game just to check some things and stuff..
Basicly all the info of a certain thing is on a table just for that
then each player has one/multiple rows depending the thing of course.
The structure is basicly
VARCHAR for Player Name (to indentify the player)
INT for Ammount (the thing ammount)
INT for Thing_ID (the thing id)
but if the player has 2 differents things it creates 2 rows for each one but for only that player!
Now i want to know whats the query to gather all the rows just for that player and display them...
I used:
Code: Select all
$result=mysql_fetch_array(mysql_query("SELECT * from negocios where player='$player'"));
$num=mysql_numrows($result);
$i=0;
while ($i < $num) {
$variable = mysql_result($result,$i,"variable");
$i++;
but it didnt work
This error appears: Warning: mysql_numrows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\game\all.bus.php on line 23
line 23 is: $num=mysql_numrows($result);
Thanks for the help
Re: Check for data for that specific player
Posted: Tue Sep 27, 2011 2:55 pm
by Chris
You are trying to count an array with mysql_numrows. What you should be doing is counting what the query is retrieving:
Code: Select all
$query = mysql_query("SELECT * FROM `negocios` WHERE `player`='$player'");
$result = mysql_fetch_array($query);
$num = mysql_numrows($result);
This will now count the amount of rows found where the field `player` = $player
Another solution to this would simply to do what you are doing and then count the amount of values $result contains:
Code: Select all
$result=mysql_fetch_array(mysql_query("SELECT * FROM `negocios` WHERE `player`='$player'"));
$num=count($result);
The only problem with this is calling mysql_fetch_array only retrieves on row at a time. So calling it once will only retrieve one row, leaving you array with only one value. Meaning $num will always equal 1.
Re: Check for data for that specific player
Posted: Tue Sep 27, 2011 3:01 pm
by vitinho444
same error
my code:
Code: Select all
$query = mysql_query("SELECT * FROM `negocios` WHERE `player`='$player'");
$result = mysql_fetch_array($query);
$num = mysql_numrows($result);
Warning: mysql_numrows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\game\all.bus.php on line 24
Re: Check for data for that specific player
Posted: Tue Sep 27, 2011 3:05 pm
by Chris
Oops my bad:
Code: Select all
$num = mysql_numrows($query); // not $result
Re: Check for data for that specific player
Posted: Wed Sep 28, 2011 4:51 pm
by vitinho444
Well.. now i got the that error on another line:
Code: Select all
Warning: mysql_result() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\game\all.bus.php on line 38
Warning: mysql_result() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\game\all.bus.php on line 39
Warning: mysql_result() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\game\all.bus.php on line 40
Warning: mysql_result() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\game\all.bus.php on line 41
Warning: mysql_result() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\game\all.bus.php on line 42
Warning: mysql_result() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\game\all.bus.php on line 43
Warning: mysql_result() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\game\all.bus.php on line 44
Here are that lines:
Code: Select all
$negocio = mysql_result($result,$i,"negocio");
$lucro = mysql_result($result,$i,"lucro");
$gastos = mysql_result($result,$i,"gastos");
$numero = mysql_result($result,$i,"numero");
$nivel = mysql_result($result,$i,"nivel");
$alugado = mysql_result($result,$i,"alugado");
$comprado = mysql_result($result,$i,"comprado");
My code from the query:
Code: Select all
$info=mysql_fetch_array(mysql_query("SELECT * from players where username='$player'"));
$query = mysql_query("SELECT * FROM `negocios` WHERE `player`='$player'");
$result = mysql_fetch_array($query);
$num = mysql_numrows($query); // not $result
$i=0;
while ($i < $num) {
$negocio = mysql_result($result,$i,"negocio");
$lucro = mysql_result($result,$i,"lucro");
$gastos = mysql_result($result,$i,"gastos");
$numero = mysql_result($result,$i,"numero");
$nivel = mysql_result($result,$i,"nivel");
$alugado = mysql_result($result,$i,"alugado");
$comprado = mysql_result($result,$i,"comprado");
$i++;
}
if($alugado)
{
echo "<b><br>Negocio: " . $negocio . " | Lucro: " . $lucro . " | Gastos: " . $gastos . "Nivel: " . $nivel . "Renda: " . $alugado . "<br></b>";
}
else
{
echo "<b><br>Negocio: " . $negocio . " | Lucro: " . $lucro . " | Gastos: " . $gastos . "Nivel: " . $nivel . "(Loja Comprada!)<br></b>";
}
echo "<br><h3><br><a href='index.php'>Voltar</a><br>";
Re: Check for data for that specific player
Posted: Wed Sep 28, 2011 6:42 pm
by Chris
I think you need to use $query instead of $result.
Re: Check for data for that specific player
Posted: Thu Sep 29, 2011 6:28 pm
by vitinho444
same error...
isnt there another way?
Re: Check for data for that specific player
Posted: Thu Sep 29, 2011 7:17 pm
by Chris
No. I'll try explain again.
What's happening is you are sending you database a query:
Code: Select all
$query = mysql_query("SELECT * FROM `negocios` WHERE `player`='$player'");
This is then stored as a resource in $query. This resource can be used to read information from.
What you are then doing is retrieving information from $query:
Code: Select all
$reuslt = mysql_fethc_array($query);
This is now storing data as an array in PHP. This is not a resource. $query is still the resource.
So if you are going to try and pass the first parameter in mysql_result() as an array when it should be a resource, it's going to throw an error:
Code: Select all
$negocio = mysql_result($result,$i,"negocio");
What you should be doing is using $query instead of $result:
Code: Select all
$negocio = mysql_result($query,$i,"negocio");
$lucro = mysql_result($query,$i,"lucro");
$gastos = mysql_result($query,$i,"gastos");
$numero = mysql_result($query,$i,"numero");
$nivel = mysql_result($query,$i,"nivel");
$alugado = mysql_result($query,$i,"alugado");
$comprado = mysql_result($query,$i,"comprado");
Re: Check for data for that specific player
Posted: Thu Sep 29, 2011 7:41 pm
by vitinho444
Ok first.. Thank you for your time
second:
I got the code in this way now:
Code: Select all
$query = mysql_query("SELECT * FROM `negocios` WHERE `player`='$player'");
$result = mysql_fetch_array($query);
$num = mysql_numrows($query);
echo "<h3>" . $player . " aqui podes gerir os teus negócios:</h3>";
$i=0;
while ($i < $num) {
$negocio = mysql_result($query,$i,"negocio");
$lucro = mysql_result($query,$i,"lucro");
$gastos = mysql_result($query,$i,"gastos");
$numero = mysql_result($query,$i,"numero");
$nivel = mysql_result($query,$i,"nivel");
$alugado = mysql_result($query,$i,"alugado");
$comprado = mysql_result($query,$i,"comprado");
$i++;
}
Now the good notice is: There are no errors.
The bad new is: Nothing appears (it means nothing is fetched)
And yes i got at least 1 record inside the table for this player :/
It just shows the Title of the page and the Back link...
Thanks for the help but i dont see the problem here
Re: Check for data for that specific player
Posted: Thu Sep 29, 2011 9:14 pm
by Ark
Try echoing the mysql_results the inside the while statement.
Code: Select all
<?php
$query = mysql_query("SELECT * FROM `negocios` WHERE `player`='$player'");
$result = mysql_fetch_array($query);
$num = mysql_numrows($query);
echo "<h3>" . $player . " aqui podes gerir os teus negócios:</h3>";
$i=0;
while ($i < $num) {
$negocio = mysql_result($query,$i,"negocio");
$lucro = mysql_result($query,$i,"lucro");
$gastos = mysql_result($query,$i,"gastos");
$numero = mysql_result($query,$i,"numero");
$nivel = mysql_result($query,$i,"nivel");
$alugado = mysql_result($query,$i,"alugado");
$comprado = mysql_result($query,$i,"comprado");
echo $negocio."<br />";
echo $lucro."<br />";
echo $gastos."<br />";
echo $numero."<br />";
echo $nivel."<br />";
echo $alugado."<br />";
echo $comprado."<br />";
$i++;
}
?>