my php script only displays page 1 of results

[kingberrill]

Hi all, I have a php script and sql database so that people can search for products on my website but when the results appear it will not go past the first page of results.

here is the form

<form action="results.php#results" method="get" id="q">
<input type="text" name="q" class="searchbox"  value="Search" onfocus="if(this.value==this.defaultValue)this.value='';" onblur="if(this.value=='')this.value=this.defaultValue;" id="q2" />
<input type="image" name="button" id="button" src="images/search_button.jpg" class="searchbox_submit" value="Go" />
</form> 

here is the script from results.php

<?php

//retreive script to connect to mysql database
require('database.php');

//get the info SB
require('search_database.php');

?>

here is the main script that searches the database

<?php
 // Get the search variable from URL

  $var = @$_GET['q'] ;
  $trimmed = trim($var); //trim whitespace from the stored variable

// rows to return
$limit=10; 

// check for an empty string and display a message.
if ($trimmed == "")
  {
  echo "<p>Please enter a search...</p>";
  exit;
  }

// check for a search parameter
if (!isset($var))
  {
  echo "<p>We dont seem to have a search parameter!</p>";
  exit;
  }

//connect to your database ** EDIT REQUIRED HERE **
mysql_connect("localhost","dectecc1_dectec","password"); //(host, username, password)

//specify database ** EDIT REQUIRED HERE **
mysql_select_db("dectecc1_products") or die("Unable to select database"); //select which database we're using

// Build SQL Query  

$query = "select * from products_tb where Description like \"%$trimmed%\" OR Code like \"%$trimmed%\" order by description"; 
// EDIT HERE and specify your table and field names for the SQL query

 $numresults=mysql_query($query);
 $numrows=mysql_num_rows($numresults);

// If we have no results, offer a google search as an alternative

if ($numrows == 0)
  {
  echo "<h1>Results</h1>";
  echo "<p>Sorry, your search: &quot;" . $trimmed . "&quot; returned zero results</p>";

// google
 echo "<p><a href=\"http://www.google.com/search?q=" 
  . $trimmed . "\" target=\"_blank\" title=\"Look up 
  " . $trimmed . " on Google\">Click here</a> to try the 
  search on google</p>";
  }

// next determine if s has been passed to script, if not use 0
  if (empty($s)) {
  $s=0;
  }

// get results
  $query .= " limit $s,$limit";
  $result = mysql_query($query) or die("Couldn't execute query");

// display what the person searched for
echo "<p>You searched for: &quot;" . $var . "&quot;</p>";

echo "<br />";

// begin to show results set
echo "Results"; 
echo "<br />";
echo "<br />";
$count = 1 + $s ;

// now you can display the results returned
  while ($row= mysql_fetch_array($result)) {
  $title = $row["Description"]; 

  echo "$count.)&nbsp;$title" ;
  $count++ ;
  }

$currPage = (($s/$limit) + 1);

//break before paging
  echo "<br />";

  // next we need to do the links to other results
  if ($s>=1) { // bypass PREV link if s is 0
  $prevs=($s-$limit);
  print "&nbsp;<a href=\"$PHP_SELF?s=$prevs&q=$var\">&lt;&lt; 
  Prev 10</a>&nbsp&nbsp;";
  }

// calculate number of pages needing links
  $pages=intval($numrows/$limit);

// $pages now contains int of pages needed unless there is a remainder from division

  if ($numrows%$limit) {
  // has remainder so add one page
  $pages++;
  }

// check to see if last page
  if (!((($s+$limit)/$limit)==$pages) && $pages!=1) {

  // not last page so give NEXT link
  $news=$s+$limit;

  echo "&nbsp;<a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 &gt;&gt;</a>";
  }

$a = $s + ($limit) ;
  if ($a > $numrows) { $a = $numrows ; }
  $b = $s + 1 ;
  echo "<p>Showing results $b to $a of $numrows</p>";

?>

feel free to test the results live at www.dec-tec.co.uk (search for "brushes")

any help with this would be greatly appreciated 🙂

[Sops21]

Just looking over the code, you never get the value of $s. Because of that $s is always going to be empty and will be assigned the value of 0 because your statement:

 if (empty($s)) { 
  $s=0; 
  } 

You need to use

$s = $_GET['s']; 

Before you if statement so you can get the value.

[kingberrill]

Sops21;10968145 wrote:

Just looking over the code, you never get the value of $s. Because of that $s is always going to be empty and will be assigned the value of 0 because your statement:

 if (empty($s)) { 
  $s=0; 
  } 

You need to use

$s = $_GET['s']; 

Before you if statement so you can get the value.

wow! thanks alot

I have another problem as well, I need the search_database to look in to another field (info) on my database could you help me with that as well?

here is the code

// Build SQL Query  

$query = "select * from products_tb where Description like \"%$trimmed%\" OR Code like \"%$trimmed%\" order by description"; 
// EDIT HERE and specify your table and field names for the SQL query

[Alf_Bennett]

Knowing that the previous answer to your problem with $s will work fine. Here is how I would assign $s

$s = (is_int($_GET['s']))? $_GET['s'] : 0;

this is really just a coding preference, but this also validates the $_GET input