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Posted

Am aceasta eroare la php pe localhost imi da eroarea asta desi eu am configurat panelul cu baza de date de la host dar da eroarea asta

 


 

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in D:\XAMPP\htdocs\rpg\profile.php on line 7

Codul este acesta:

< ?php
include 'includes/config.php';
include 'includes/header.php';

$id = isset($_GET['id']) ? (int)$_GET['id'] : $udata['id'];
$q = mysql_query("SELECT * FROM players WHERE id=$id"); 
while($playerdata = mysql_fetch_array($q)): // linia 7
$name = $playerdata['username'];

Uitati o poza cum arata

i60fNzi.jpg

Posted

Dupa primul apel al functiei mysql_query adauga

if($result === FALSE) 
{ 
    die(mysql_error());
}

si pune aici ce eroare primesti.

Mutat la totul si nimic.

Posted (edited)

dar nu ma ajuta cu nimic :(

EDIT: scuze pentru atatea edituri acum imi da asta

 

Unknown column 'id' in 'where clause'

Edited by Selby
Posted

Nu iti spune ce eroare ai in interogare?

Vreau sa vad eroarea, de asta te-am pus s-o pui.

Posted

Pune

$q = mysql_query("SELECT * FROM `players` WHERE `id` = ''$id'"); 

in loc de

$q = mysql_query("SELECT * FROM players WHERE id=$id"); 

 

Posted (edited)

Tot asa
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in D:\XAMPP\htdocs\rpg\profile.php on line 7

 

<?php
include 'includes/config.php';
include 'includes/header.php';

$id = isset($_GET['id']) ? (int)$_GET['id'] : $udata['id'];
$q = mysql_query("SELECT * FROM `players` WHERE `id` = '$id'");
while($playerdata = mysql_fetch_array($q))://linia 7
$name = $playerdata['username'];

 

Edited by Selby
Posted

Dupa primul apel al functiei mysql_query adauga

if($result === FALSE) 
{ 
    die(mysql_error());
}

si pune aici ce eroare primesti.

E aceeasi eroare?

De ce zice ca nu ai coloana id in tabela... Foarte ciudat.

Pune putin script-ul asta la inceput si zi-ne ce zice. (pune-l inainte de orice mysql_query din fisierul tau)

$Query = mysql_query("SHOW COLUMNS FROM `players`") or die(mysql_error());

while($Column = mysql_fetch_object($Query))
{
    $Columns[] = $Column;
}
 
foreach($Columns as $Key => $Column)
{
    echo $Column->Field."</br>";
}

die();

 

Posted (edited)

Dupa ce am  pus codul acela nu mai da  eroare dar nu se incarca nimic si m gasit asta in link

http://localhost/rpg/profile.php?id=<br /><b>Notice</b>: Undefined index: id in <b>D:\XAMPP\htdocs\rpg\includes\header.php</b> on line <b>45</b><br />

 

Iar din cate scrie acolo header linia 45

este asta

<li>
									<a href="profile.php?id=<?=$udata['id']?>">//45
										<i class="icon-user"></i>
										Profilul Meu
									</a>
								</li>							
									

 

Edited by Selby
Posted

Da stiu, dar vreau sa zice vad ce ai in tabela jucatori.

Doar atat iti arata?

Posted

da si nu inteleg de ce ce alta varianta mai am sa fac ?

Sincer, aici nu te inteleg.

Posted

Nu am cum sa testez daca nu am toate fisierele. Care este eroarea acolo vreau sa stiu.

Posted

Pai daca vrei iti dau tot panelul

iar eroarea este asta si nu inteleg de ce

Warning: mysql_result() expects at least 2 parameters, 1 given in D:\XAMPP\htdocs\rpg\profile.php on line 7

Posted

Pune ce ti-am zis

if($result === FALSE) 
{ 
    die(mysql_error());
}

sau pune

$q = mysql_query("SELECT * FROM `players` WHERE `id` = '$id'") or die(mysql_error());

in loc de

$q = mysql_query("SELECT * FROM `players` WHERE `id` = '$id'");

 

Si zi-mi ce eroare iti da.

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