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[SOLVED] can someone help?

menios

By menios
in PHP Coding Help

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menios Member

menios

Posted
Posted

I ve been trying to display items from my product table but instead of selecting them by category id it just displays everything.

I can't find what goes wrong.

so here is the code if someone could help with  :)

 


<html>


<body>
<?php
 $rpp;		//Records Per Page
 $cps;		//Current Page Starting row number
 $lps;		//Last Page Starting row number

/////////////////////////////////////////////////////////////////////////////////
//Database connection
/////////////////////////////////////////////////////////////////////////////////
$user="root";
$password="asere";
$database="smalltownbooks";
$tablename="tbl_product";
$conn=mysql_connect("localhost",$user,$password);

mysql_select_db($database,$conn) or die( "Unable to select database");
mysql_query($sql,$conn);
if (isset($_GET['1']))
{
    $cat = $_GET['1'];      //  for simplicity only. 
                              
    $sql = "SELECT * FROM tbl_product WHERE cat_id = '$cat'";
    echo $sql or die ("dead");

}
/////////////////////////////////////////////////////////////////////////////////

/////////////////////////////////////////////////////////////////////////////////
//Load for the first time First result printed user will click next to change
/////////////////////////////////////////////////////////////////////////////////	
if(empty($_GET["cps"])) 
{
	$cps = "0";
} 
else
{
	$cps = $_GET["cps"];
}
/////////////////////////////////////////////////////////////////////////////////

$a = $cps+1; 

$rpp = "15";

$lps = $cps - $rpp; //Calculating the starting row number for previous page

/////////////////////////////////////////////////////////////////////////////////
//Check for previus link If the user is viewing the first set of data then the link will
//be disabled, if on the next set then it will carry the $lps in its link and enable it
if ($cps <> 0)
{
	$prv =  "<a href='cookery.php?cps=$lps'>Previous</a>";
}
else	
{
	$prv =  "<font color='cccccc'>Previous</font>";
}
/////////////////////////////////////////////////////////////////////////////////

///////////////////////////////////////////////////////////////////////////////
$q="Select SQL_CALC_FOUND_ROWS * from tbl_product limit $cps, $rpp";
$rs=mysql_query($q) or die(mysql_error());
$nr = mysql_num_rows($rs); //Number of rows found with LIMIT in action

$q0="Select FOUND_ROWS()";
$rs0=mysql_query($q0) or die(mysql_error());
$row0=mysql_fetch_array($rs0);
$nr0 = $row0["FOUND_ROWS()"]; //Number of rows found without LIMIT in action

/////////////////////////////////////////////////////////////////////////////////
//Check  the user has reached thelast page of the items	
if (($nr0 < 50) || ($nr < 10))
{
   		$b = $nr0;
}
else
{
	$b = ($cps) + $rpp;
}

/////////////////////////////////////////////////////////////////////////////////

?>
<br>
<table border="1" cellpadding="4" cellspacing="1" width="100%" align="center">
  
  <tr><td bgcolor="#000080" align='Center'><b><font face="verdana" color="#FFFFFF">Title</font></b></td>
  <td bgcolor="#000080" align='Center'><b><font face="verdana" color="#FFFFFF">test</font></b></td>
</tr>
<?php
while ($row=mysql_fetch_array($rs))
{
	/////////////////////////////////////////////////////////////////////////////////
	//loop after fullfilling the above SQL criteria.
	/////////////////////////////////////////////////////////////////////////////////
	$cps = $cps +1; 

	$val=$row["pd_name"];
	$val2=$row ["pd_price"];
	$val3=$row ["cat_id"];
	echo "<tr><td width='10%' align='center'><font face=verdana><img</font></td><td align='center'><font fave=verdana>$val $val2 $val3 </font></td></tr>";  

}

echo "<tr><td align='right' colspan=2>$prv";


//Check whether there is a Next link if not disable

if ($cps == $nr0)
{ 	
	echo "  |  <font color='CCCCCC'>Next</font>";
}
else
{
	if ($nr0 > 5)
	{ 
		echo "  |  <a href='cookery.php?cps=$cps&lps=$lps'>Next  </a>"; 
	}
}


?>
</td>
</tr>
</table>
</body>
</html>

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https://forums.phpfreaks.com/topic/79704-solved-can-someone-help/
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revraz Member

revraz

Posted
Posted

Lets look at it from a logical view.

 

If you do

 

echo $a

 

$a = 1

 

What would $a echo out?  Nothing right, because you are not setting it until after you echo it.

 

So you are doing a mysql query before you enter the parameters.

menios Member

menios

Posted
  • Author
Posted

which parameter am i missing cause i understand what what you mean by stating the parameters first but i m not using any parameters

all i wan to do is display only the content of a certain category. so i need to set a parameter about category_id?

I m getting confused about it to tell you the truth If you could help me understant it i would re appreciate it

Thanks in advance for your time

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