Mod-Jay Posted September 18, 2010 Share Posted September 18, 2010 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>  "; } // 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.. Link to comment https://forums.phpfreaks.com/topic/213728-error-in-search/ Share on other sites More sharing options...
Pikachu2000 Posted September 18, 2010 Share Posted September 18, 2010 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); Link to comment https://forums.phpfreaks.com/topic/213728-error-in-search/#findComment-1112468 Share on other sites More sharing options...
Mod-Jay Posted September 18, 2010 Author Share Posted September 18, 2010 LOLL Thank You Pikachu2000 Ur a Beast Link to comment https://forums.phpfreaks.com/topic/213728-error-in-search/#findComment-1112471 Share on other sites More sharing options...
Pikachu2000 Posted September 18, 2010 Share Posted September 18, 2010 LOLL Thank You Pikachu2000 Ur a Beast A pocket-sized battle monster, actually . . . Link to comment https://forums.phpfreaks.com/topic/213728-error-in-search/#findComment-1112472 Share on other sites More sharing options...
Mod-Jay Posted September 18, 2010 Author Share Posted September 18, 2010 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? Link to comment https://forums.phpfreaks.com/topic/213728-error-in-search/#findComment-1112474 Share on other sites More sharing options...
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