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Posted by Jerry Stuckle on 09/27/05 04:16
wgerry wrote:
> Thanks for all your kind help guys. I have tried what you have
> suggested but I am still having trouble. I have made the following
> changes but it still comes up with the warning mysql_num_rows is not a
> valid SQL resource and my query does not return a result and it comes
> up with couldnt execute query,
> <html>
> <head>
> <title>search</title>
>
> </head>
>
> <body>
>
> <form name="form" action="search09.php" method="get">
> <input type="text" name="q" />
> <input type="submit" name="Submit" value="Search" />
> </form>
> </body>
> </html>
>
> <?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("host ","usename",password"); //(host, username,
> password)
>
> //specify database ** EDIT REQUIRED HERE **
> mysql_select_db("online") or die("Unable to select database"); //select
> which database we're using
>
>
> // Build SQL Query
> $query = "SELECT * FROM arrow WHERE surname LIKE %'$trimmed %.'" ;
>
> $results= mysql_query($query);
> $numrows= mysql_num_rows($results);
>
> // 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";
> $results = mysql_query($query) or die ("Couldnt execute query");
> }
> // now you can display the results returned
> while ($row = mysql_fetch_array($results)){
> $title =$row["surname"];
>
> 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>";
> ?>
>
> any help give me with the above greatfully recieved and thanks again
> for the help already given.
>
> Kind Regards
> Gerry
>
Gerry,
You should ALWAYS check the results of any call to MySQL. If the result
is false, echo mysql_error() to help find the problem.
--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
==================
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