Listing items with different links

jopereira · March 2, 2013

I've got 3 listed items, like this:

Item one

Item two

Item three

each one of this items have a link, but all the 3 items are different rows from the same table, but all the links open the same

how can I distinct each one of the links?

I got this code:

<?PHP
//echo "<link href='default.css' rel='stylesheet' type='text/css' media='all' />";
$conn = mysql_connect("localhost", "root", "") or die("Impossivel conectar");

if($conn)
	{
		mysql_select_db("eventos", $conn);	
		} 
		$sql = "SELECT id_eventos, nome FROM eventos";		
		$rs = mysql_query($sql, $conn);		
		$num = mysql_num_rows($rs);
		if($num > 0)
		{
			//echo "<table border=0 cellpadding=0 cellspacing=0>\n";

			while($dados = mysql_fetch_array($rs))
			{
				$ref 	= $dados["id_eventos"];
				$nome 	= $dados["nome"];
			/*	$img = $dados["imagens"];
				echo"		<tr>";
				echo"		<td>$ref</td>";
				echo"		<td>$desc</td>";
				echo"		<td>$pvp</td>";
				echo"		<td>$func</td>";
				echo"		<td>$garant</td>";*/
			//	echo "<td>";
			
				//echo "<li><a href=\"mostra_evento.php?id=$ref\"></li>";
				echo"<li><a href=\"listar_imagem.php?id=$ref\">$nome</a></li>";
				//echo"</td>";		
				//echo"	</tr>\n";
			}
	   	}
   // echo "</table>\n";
?>

<?PHP
$conn = mysql_connect("localhost", "root", "") or die("Impossivel conectar");

if($conn)
	{
		mysql_select_db("eventos", $conn);	
		}  
		$sql = "SELECT * from eventos where id_eventos=1;		
		$rs = mysql_query($sql, $conn);		
		$num = mysql_num_rows($rs);
		if($num > 0)
		{
			echo "<table border=0 cellpadding=0 cellspacing=0>\n";
			echo "<table border=0 cellpadding=3 cellspacing=0>
			<tr>
			<td>Nº de Evento</td>
			<td>Nome</td>
			<td>Descrição</td>
			<td>Requisitos</td>
			<td>Data</td>
			<td>Vencedor</td>
			</tr>";
			while($dados = mysql_fetch_array($rs))
			{
				$ref 	= $dados["id_eventos"];
				$nome 	= $dados["nome"];
				$desc    = $dados["descricao"];
				$req   = $dados["requisitos"];
				$data = $dados["data"];
				$vence   = $dados["vencedor"];
				//$img = $dados["imagens"];
				echo"		<tr>";
				echo"		<td>$ref</td>";
				echo"		<td>$nome</td>";
				echo"		<td>$desc</td>";
				echo"		<td>$req</td>";
				echo"		<td>$data</td>";
				echo"		<td>$vence</td>";
				echo "<td>";
				echo"<a href=\"listar_imagem.php?id=$id\"></a>";
				
				echo"</td>";		
				echo"	</tr>\n";
			}
	   	}
		echo "</table>";
    echo "</table>\n";
?>

can you help?

jopereira · March 2, 2013

anyone?!

Jessica · March 2, 2013

Your SQL statement is only selecting id #1 because you explicitly say so.

jopereira · March 2, 2013

Your SQL statement is only selecting id #1 because you explicitly say so.

i know.. i just put it there cuz i dont know what to put

i think it's some parameter that passes the id, which relates it to the row of the table i want

AyKay47 · March 2, 2013

Here is a hint, $_GET the id from the query string you passed.

You also need to implement proper debugging steps, use Jessica's signature for that!

Edited March 2, 2013 by AyKay47

jopereira · March 2, 2013

i'm complity a newbie..

i barely can do a log in system, mutch less debugg. i read what she said, and I understand a good hint to discover the error is to echo where they happear, but.. reading is different than doing :/

btw can you give me more detail info about that $get?

Edited March 2, 2013 by jopereira

Jessica · March 2, 2013

Look in the manual under $_GET

jopereira · March 2, 2013

Look in the manual under $_GET

i read into some info about $_GET but I found nothing i could use

my next guess it's in the script responsible to list the items I got this

echo"<li><a href=\"list_event.php?id=$ref\">$name</a></li>";

and on the script responsible to load its info is:

$sql = "SELECT * from eventos where id_event=$ref";
...

echo"<a href=\"list_event.php?id=$ref\"></a>";

i think this $ref parameter could help, but how it is now it gives the error:

Notice: Undefined variable: ref in C:\xampp\htdocs\xampp\eventos\list_event.php on line 8

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\xampp\eventos\list_event.php on line 10

Edited March 2, 2013 by jopereira

AyKay47 · March 2, 2013

Sigh, $_GET['id'] on the receiving page will hold the value you are looking for. Just make sure to cast it to an int data type before using it in the SQL statement.

Barand · March 2, 2013

Assuming ref is numeric

$ref = intval($_GET['ref']);
 
$sql = "SELECT * from eventos where id_eventos=$ref";
 
...

jopereira · March 2, 2013

Assuming ref is numeric

$ref = intval($_GET['ref']);
 
$sql = "SELECT * from eventos where id_eventos=$ref";
 
...

it is a INT value, auto incremet, and I did as you said, but it keeps saying:

Notice: Undefined index: ref in C:\xampp\htdocs\xampp\eventos\listar_imagem.php on line 8

Edited March 2, 2013 by jopereira

Jessica · March 2, 2013

Your url is sending id, not ref.

Try to think about it logically instead of simply claiming you don't understand it. If you read the documentation on $_GET it's pretty clear.

http://www.php.net/manual/en/reserved.variables.get.php

jopereira · March 2, 2013

Your url is sending id, not ref.

Try to think about it logically instead of simply claiming you don't understand it. If you read the documentation on $_GET it's pretty clear.

http://www.php.net/manual/en/reserved.variables.get.php

nope.. I already saw that and corrected

id_eventos is the value that distincts the line of my table (the primary key) and the variable that holds its value is $ref

look (pay attention to the <-----------------------------):

<?PHP
//echo "<link href='default.css' rel='stylesheet' type='text/css' media='all' />";
$conn = mysql_connect("localhost", "root", "") or die("Impossivel conectar");

if($conn)
	{
		mysql_select_db("eventos", $conn);	
		} 
		$sql = "SELECT id_eventos, nome FROM eventos";		
		$rs = mysql_query($sql, $conn);		
		$num = mysql_num_rows($rs);
		if($num > 0)
		{
			//echo "<table border=0 cellpadding=0 cellspacing=0>\n";

			while($dados = mysql_fetch_array($rs))
			{
				$ref 	= $dados["id_eventos"]; <----------------------------------
				$nome 	= $dados["nome"];
			/*	$img = $dados["imagens"];
				echo"		<tr>";
				echo"		<td>$ref</td>";
				echo"		<td>$desc</td>";
				echo"		<td>$pvp</td>";
				echo"		<td>$func</td>";
				echo"		<td>$garant</td>";*/
			//	echo "<td>";
			
				//echo "<li><a href=\"mostra_evento.php?id=$ref\"></li>";
			        echo"<li><a href=\"listar_imagem.php?id_eventos=$ref\">$nome</a></li>";<--------------------------
				//echo"</td>";		
				//echo"	</tr>\n";
			}
	   	}
   // echo "</table>\n";
?>

<?PHP
$conn = mysql_connect("localhost", "root", "") or die("Impossivel conectar");

if($conn)
    {
        mysql_select_db("eventos", $conn);    
        }
        $ref = intval($_GET['id_eventos']);<----------------------------------
        $sql = "SELECT * from eventos where id_eventos=$ref";        <----------------------------------
        $rs = mysql_query($sql, $conn);        
        $num = mysql_num_rows($rs);
        if($num > 0)
        {
            echo "<table border=0 cellpadding=0 cellspacing=0>\n";
            echo "<table border=0 cellpadding=3 cellspacing=0>
            <tr>
            <td>Nº de Evento</td>
            <td>Nome</td>
            <td>Descrição</td>
            <td>Requisitos</td>
            <td>Data</td>
            <td>Vencedor</td>
            </tr>";
            while($dados = mysql_fetch_array($rs))
            {
                $ref     = $dados["id_eventos"];
                $nome     = $dados["nome"];
                $desc = $dados["descricao"];
                $req = $dados["requisitos"];
                $data = $dados["data"];
                $vence = $dados["vencedor"];
                //$img = $dados["imagens"];
                echo"        <tr>";
                echo"        <td>$id_eventos</td>";
                echo"        <td>$nome</td>";
                echo"        <td>$descricao</td>";
                echo"        <td>$requisitos</td>";
                echo"        <td>$data</td>";
                echo"        <td>$vencedor</td>";
                echo "<td>";
                
                echo"</td>";        
                echo"    </tr>\n";
            }
         }
echo "</table>\n";
?>

Edited March 2, 2013 by jopereira

Jessica · March 2, 2013

Your URL says id.

Look at the example in the documentation. And think about it.

If you go to a URL ending in id=5 and you want to get 5, what would you need to "GET"?

Barand · March 2, 2013

As Jessica pointed out you querystring uses "id"

href=\"list_event.php?id=$ref\"

So you would need

$ref = intval($_GET['id']);
 
$sql = "SELECT * from eventos where id_eventos=$ref";

jopereira · March 2, 2013

Your URL says id.

Look at the example in the documentation. And think about it.

If you go to a URL ending in id=5 and you want to get 5, what would you need to "GET"?

do'h

damn I missed the logical thing.. I thought I had already changed that id, but I actually didn't //.-

I realy appreciate the way all you people help here, it's such a shame that some teachers dont do that way, resulting in my unknowledge and "personal hate" for php, when it is so easy when you get thing done

sorry for beeing dumb :/

my issue is solved here

thanks!

Edited March 2, 2013 by jopereira

AyKay47 · March 3, 2013

Sigh, $_GET['id'] on the receiving page will hold the value you are looking for. Just make sure to cast it to an int data type before using it in the SQL statement.

If only you would have read it when I posted it.

Edited March 3, 2013 by AyKay47

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