PHP MySQL search engine problem

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May 17th, 2004, 02:48 PM

Danboy

Registered User

Join Date: May 2004

Posts: 1

Time spent in forums: < 1 sec
Reputation Power: 0

PHP MySQL search engine problem

Hi folks,

I'm new here so please bear with me on this one.....

I've been following Mitch Harper's tutorial on Developing A Site Search Engine With PHP And MySQL, which can be found here. It works too...............sort of.

The problems I'm having are these:

1) When searching with more than one keyword (ie "mysql" and "apache") the code outputs two separate results (i.e "1 article found: MySQL is a great database"; "1 article found: Apache is pronounced A-Patchy") instead of the expected "2 articles found".

2) When the search keyword is not found I'm presented with a blank page instead of telling me no results were found.

The code I'm working on is this:

<?php
$submit = $_POST["submit"];
$keywords = $_POST["keywords"];
if(isset($submit) || isset($keywords)) {
doSearch($keywords); }
else {
getKeywords(); }
function getKeywords() {
?>
<html>
<head>
<title> Enter Search Keywords </title>
</head>
<body bgcolor="#FFFFFF">
<form name="frmKW" action="searchdocs.php" method="post">
<h1>Keyword Search</h1>
Enter keywords to search on:
<input type="text" name="keywords" maxlength="100">
<br><br><input type="submit" name="submit" value="Search">
</form>
</body>
</html>

<?php
}
function doSearch($search_keywords) {
$arrWords = explode(" ", $search_keywords);
if(sizeof($arrWords) == 0 || $search_keywords == "") {
echo "You didn't enter any keywords<br>";
echo "<a href='searchdocs.php'>Go Back</a>"; }
else { // Connect to the database
$dServer = "localhost";
$dDb = "content_1";
$dUser = "root";
$dPass = "whatever";
$s = @mysql_connect($dServer, $dUser, $dPass) or die("Couldn't connect to database server");
@mysql_select_db($dDb, $s)
or die("Couldn't connect to database");
for($i = 0; $i < sizeof($arrWords); $i++) {
$query = "select articleIds from searchWords where word LIKE '{$arrWords[$i]}'";
$result = mysql_query($query) or exit('$query failed: '.mysql_error());
if (mysql_num_rows($result) > 0) {
// Get the id's of the articles
$row = mysql_fetch_array($result);
$arrIds = explode(",", $row[0]);
$arrWhere = implode(" OR articleId = ", $arrIds);
$aQuery = "select articleId, title, left(content, 100) as summary from articles where articleId = " . $arrWhere;
$aResult = mysql_query($aQuery);
$count = 0;
$articles = array();
if(mysql_num_rows($aResult) > 0) {
while($aRow = mysql_fetch_array($aResult)) {
$articles[$count] = array ("articleId" => $aRow["articleId"], "title" => $aRow["title"], "summary" => $aRow["summary"]);
$count++;
}
}
if(isset($articles)) {
$articles = array_unique($articles);
echo "<h1>" . sizeof($articles);
echo (sizeof($articles) == 1 ? " article" : " articles");
echo " found:</h1>";
foreach($articles as $a => $value)
{
?>
<a href="article.php?articleId=<?php echo $articles[$a]["articleId"]; ?>">
<b><u><?php echo $articles[$a]["title"]; ?></u></b>
</a>
<br><?php echo $articles[$a]["summary"] . "..."; ?>
<br>
<a href="article.php?articleId=<?php echo $articles[$a]; ?>">
http://www.mysite.com/article.php?articleId=<?php echo $articles[$a]["articleId"]; ?>
</a>
<br><br>
<?php
}
}
else {
echo "No results found for '$search_keywords'<br>";
echo "<a href='searchdocs.php'>Go Back</a>";
}
}
}
}
}
?>
Apologies for posting so much code, but if anyone can see what I'm doing wrong I'd really appreciate it.

By the way, I'm running PHP 4.3.3 and MySQL 3.23.49

Many thanks,

Dan

May 18th, 2004, 01:59 PM

Pheifel

Contributing User

Join Date: Mar 2004

Location: Denmark

Posts: 174

Time spent in forums: 43 m 46 sec
Reputation Power: 5

hmmm.... maybe, instead of using a foreloop with the $arrWord you could use a "wildguess" in the db with the $arrWord (after you have exploded it.) like this:

PHP Code:


		
			
 mysql_query("select * from db where word like '%$arrWord%'");

- Pheifel

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