Error in Search

[Mod-Jay]

Error:

Warning: mysql_query() expects parameter 1 to be string, resource given in C:\Program Files\xampp\htdocs\NEW\search.php on line 35

Warning: mysql_num_rows() expects parameter 1 to be resource, null given in C:\Program Files\xampp\htdocs\NEW\search.php on line 36
Results

Sorry, your search: "mod" returned zero results

Click here to try the search on google
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #3 limit 0,10' at line 1

 

My PHP code

<?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","root",""); //(host, username, password)

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

// Build SQL Query  
$query = mysql_query("SELECT * FROM main WHERE username LIKE '%".$trimmed."%'  
  ORDER BY username") or die(mysql_error()) ; // 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 "<h4>Results</h4>";
  echo "<p>Sorry, your search: "" . $trimmed . "" 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(mysql_error()) ;

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

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

// now you can display the results returned
  while ($row= mysql_fetch_array($result)) {
  $title = $row["1st_field"];

  echo "$count.) $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-$limit);
  print " <a href=\"$PHP_SELF?s=$prevs&q=$var\"><< 
  Prev 10</a>&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 " <a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 >></a>";
  }

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

 

Lines:

35 == $numresults=mysql_query($query);
36 == $numrows=mysql_num_rows($numresults);
59 ==   $result = mysql_query($query)  or die(mysql_error()) ; --
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #3 limit 0,10' at line 1

 

Help Please..

[Pikachu2000]

You're trying to run a second query with the resource from the first one as the string. That doesn't work so well.

 

$query = "SELECT * FROM main WHERE username LIKE '%$trimmed%' ORDER BY username"; // EDIT HERE and specify your table and field names for the SQL query
$numresults=mysql_query($query) or die(mysql_error());
$numrows=mysql_num_rows($numresults);

[Mod-Jay]

LOLL Thank You Pikachu2000 Ur a Beast

[Pikachu2000]

LOLL Thank You Pikachu2000 Ur a Beast

 

A pocket-sized battle monster, actually . . .

[Mod-Jay]

LOLL Thank You Pikachu2000 Ur a Beast

 

A pocket-sized battle monster, actually . . .

 

Totally Haha, Hey Take A look at my other post if you will, in the html section! Also Do you have a msn?