This may be a dumb comment, but those variable don't make any sense, you'll
have to replace them with stuff that relates to the query and the database.
-Micah
On Mon October 27 2003 8:34 pm, Shannon Doyle wrote:
> Thanks Micah,
>
>
> while ($d = mysql_fetch_assoc($return)) {
> ?><option value="<?=$d['value']?>"><?=$d['name']?></option> ----- I get
> a parse error on this line.
>
> while ($d = mysql_fetch_assoc($return)) {
> ?><option value="<?=$d['value']?>"<?
> if ($other_table_value == $d['value'])
> echo " selected";
> ?>><?=$d['name']?></option>
> <?
> }
>
> Ok cool, I am assuming that this will be OK once I work out the parse
> error on the above option.
>
> > Third:
> > If you're using MySQL, check this out:
> > http://www.mysql.com/doc/en/Fulltext_Search.html
> >
> > If you're using something else, it's more complex I think. Someone may
> >
> > have an
> > elegant solution, but I would do something like:
>
> // Get total number of keywords:
> $numofkeywords = count(str_replace ( " ", " ", $keywords));
>
> // Split up your search words:
> $search = explode(" ", $keywords);
> $total_matched = 0;
> // loop through the array of search terms and get number of returns.
> foreach ($search as $searchword) {
> $total_matched += count(str_replace($searchword, $searchword,
> $keywords));
> }
> // echo out the result in percent. (to one decimal place even!)
> echo "Percent Matched: ".round(($total_matched/$numofkeywords)*100),
> 1)." %";
>
>
> This only partly resolves my problem, I am not looking for a word count,
> I am looking to return those records that have one or more of the
> keywords in them, and display a percentage result next to each record
> that matches.
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