使用PHP和MySQL搜索結果

Question

我最近發布了另一個與我正在研究的搜索引擎相關的線程，基於Kevin Yank的“構建自己的數據庫驅動的網站等”。 搜索引擎正在運行，但是我想修復一些錯誤並且不知道如何。 我將在這個帖子中只關注其中一個，這樣就不會太混亂了。

在數據庫中，有一個用於笑話的表（稱為“笑話”）和另一個用於主題的表（稱為“主題”）。 這兩個表與另一個名為“joketheme”的表相關聯。 每個笑話應該能夠有超過1個主題，我希望結果列出所有主題而不重復輸入。 到目前為止，我還沒有實現這一目標。 事實上，我不確定如何在MySQL中為1個笑話指定2個主題。

    table 1: joke
    id~~~joketext~~~other data
    1~~~joke1~~~
    2~~~joke2~~~
    3~~~joke3~~~

    table 2: theme
    id~~~name
    1~~Knock knock
    2~~Lawyers

    table 3: joketheme
    jokeid~~~themeid
    1~~~~~~1
    1~~~~~~2
    2~~~~~~2
    3~~~~~~1

您是否知道我需要在代碼（或MySQL）中更改以在每個結果中列出多個主題（如果該條目具有多個主題？）？

這是搜索結果頁面中的代碼。 預先感謝您的任何幫助！：

    <?php

    $dbcnx = @mysql_connect('localhost', 'root', 'password'); 

if (!$dbcnx) {
    exit('<p>Unable to connect to the ' . 'database server at this time.</p>');
}

if (!@mysql_select_db('ijdb')) { 
    exit('<p>Unable to locate the joke ' . 'database at this time.</p>');
}
    $authors = @mysql_query('SELECT id, name FROM author'); 
if (!$authors) {
    exit('<p>Unable to obtain author list from the database.</p>');
}

$cats = @mysql_query('SELECT id, name FROM category'); 
if (!$cats) {
    exit( '<p>Unable to obtain category list from the database.</p>');
} 

$themes = @mysql_query('SELECT id, name FROM theme'); 
if (!$themes) {
    exit( '<p>Unable to obtain category list from the database.</p>');
}

$geofoci = @mysql_query('SELECT id, name FROM geofocus'); 
if (!$geofoci) {
    exit( '<p>Unable to obtain category list from the database.</p>');
}

?>

搜索表單：

    <form class="searchField" name="input" action="fundfetch_search.php" method="post">
    <ul>
<li>
    <label>Search by keyword:</label>
    <input type="text" name="searchtext" class="styleSearchbox" placeholder="By keyword" value="<?php echo $_POST['searchtext']; ?>">
</li>
<li>
        <label>OR by the following: </label>
        <label><select name="aid" size="1" class="styleDropdown">
        <option selected value="">Any Author</option> 
        <?php
            while ($author = mysql_fetch_array($authors)) { 
                $aid = $author['id']; 
                $aname = htmlspecialchars($author['name']); 
                echo "<option value='$aid'>$aname</option>\n";
        } 
        ?> 
        </select></label>           
    </li>
    <li>
        <label><select name="cid" size="1" class="styleDropdown">
            <option selected value="">Any Category</option> 
        <?php
        while ($cat = mysql_fetch_array($cats)) { 
            $cid = $cat['id']; 
            $cname = htmlspecialchars($cat['name']); 
            echo "<option value='$cid'>$cname</option>\n";
        } 
        ?>
        </select></label>
    </li>
    <li>
        <label><select name="tid" size="1" class="styleDropdown">
            <option selected value="">Any Theme</option> 
        <?php
        while ($theme = mysql_fetch_array($themes)) { 
            $tid = $theme['id']; 
            $tname = htmlspecialchars($theme['name']); 
            echo "<option value='$tid'>$tname</option>\n";
        } 
        ?>
        </select></label>
    </li>
    <li>
        <label><select name="gfid" size="1" class="styleDropdown">
            <option selected value="">Any Region</option>
        <?php
        while ($geofocus = mysql_fetch_array($geofoci)) { 
            $gfid = $geofocus['id']; 
            $gfname = htmlspecialchars($geofocus['name']); 
            echo "<option value='$gfid'>$gfname</option>\n";
        } 
        ?>
        </select></label>
    </li>
    <li style="visibility:hidden"><a href="../FUNDER.COM website/searchfilteroption">Closing</a></li>
 <li><input type="submit" value="Search" class="searchButton"></li>
</ul>
    </form>

查詢：

    <?php

$dbcnx = @mysql_connect('localhost', 'root', 'password'); 

if (!$dbcnx) {
    exit('<p>Unable to connect to the ' . 'database server at this time.</p>');
}

if (!@mysql_select_db('ijdb')) { 
    exit('<p>Unable to locate the joke ' . 'database at this time.</p>');
    }



    $select = 'SELECT DISTINCT joke.id, joke.joketext, joke.jokedate, 
            author.id AS author_id, author.name AS author_name, 
            jokecategory.jokeid AS cat_jokeid, jokecategory.categoryid AS joke_catid, category.id AS cat_id, category.name as cat_name, 
            joketheme.jokeid AS theme_jokeid, joketheme.themeid AS joke_themeid, theme.id AS theme_id, theme.name AS theme_name,
            jokegeofocus.jokeid AS geofocus_jokeid, jokegeofocus.geofocusid AS joke_geofocusid, geofocus.id AS geofocus_id, geofocus.name AS geofocus_name';
$from   = ' FROM joke, author, jokecategory, category, joketheme, theme, jokegeofocus, geofocus'; 
$where = ' WHERE joke.authorid = author.id AND joke.id = jokecategory.jokeid AND jokecategory.categoryid = category.id AND joke.id = joketheme.jokeid AND joketheme.themeid = theme.id AND joke.id = jokegeofocus.jokeid AND jokegeofocus.geofocusid = geofocus.id';
$in = ' ORDER BY jokedate DESC';

$aid = $_POST['aid']; 
if ($aid != '') { // An author is selected
    $where .= " AND authorid='$aid'";
}

$cid = $_POST['cid']; 
if ($cid != '') { // A category is selected
    $from .= ''; // usually written as ' ,tablename'
    $where .= " AND joke.id=jokecategory.jokeid AND categoryid='$cid'";
}

$tid = $_POST['tid'];
if ($tid != '') { // A theme is selected
    $from .= '';
    $where .= " AND joke.id=joketheme.jokeid AND themeid='$tid'";
}

$gfid = $_POST['gfid'];
if ($gfid != '') { // A region is selected
    $from .= '';
    $where .= " AND joke.id=jokegeofocus.jokeid AND geofocusid='$gfid'";
}

$searchtext = $_POST['searchtext'];
if ($searchtext != '') { // Some search text was specified
    $where .= " AND keywords LIKE '%$searchtext%'";
    }
    ?>  

結果

    <?php   

    $jokes = @mysql_query($select . $from . $where . $in); 
    if (!$jokes) {
        echo '</table>'; exit('<p>Error retrieving jokes from database!<br />'.
        'Error: ' . mysql_error() . '</p>');
    }

    $numrows = mysql_num_rows($jokes);
    if ($numrows>0){
    while ($joke = mysql_fetch_array($jokes)) { 
        $id = $joke['id'];
        $joketext = htmlspecialchars($joke['joketext']);
        $jokedate = htmlspecialchars($joke['jokedate']);
        $aname = htmlspecialchars($joke['author_name']);
        $category = htmlspecialchars($joke['cat_name']);
        $theme = htmlspecialchars($joke['theme_name']);
        $geofocus = htmlspecialchars($joke['geofocus_name']);
        $position = 200;
        $post = substr($joketext, 0, $position);
        echo "<li id=\"jump\">
                <article class=\"entry\">
                    <header>
                        <h3 class=\"entry-title\"><a href=''>$aname</a></h3>
                    </header>
                    <div class=\"entry-content\">
                        <p>$post...</p>
                    </div>
                    <div class =\"entry-attributes\">
                        <p>> Category: $category</p>
                        <p> > Theme(s): $theme</p>
                        <p> > Region(s) of focus: $geofocus</p>
                        </div>
                    <footer class=\"entry-info\">
                        <abbr class=\"published\">$jokedate</abbr>
                    </footer>
                </article>
            </li>";
    }
}
    else
        echo "Sorry, no results were found. Please change your search parameters and try again!";
     ?>

Answer 1

聽起來你需要對SQL查詢以及它們的工作方式進行一些研究。 如果你想要一個單獨的笑話的主題列表，你可以做一個這樣的簡單查詢：

  SELECT DISTINCT(theme.name) FROM theme
     INNER JOIN joketheme ON joketheme.theme_id = theme.id
     INNER JOIN joke ON joke.id = joketheme.joke_id
     WHERE joke.id = {insert joke id}

我們在這里做的是一個連接。 我們將三個表連接在一起，根據所有表中常見的ID匹配結果。 這些被稱為密鑰。 基本上，我們試圖從主題表中選擇與特定笑話主題相匹配的所有主題。

查詢的DISTINCT部分確保我們只獲得唯一的結果。

我建議您查看一些SQL教程，以了解使用查詢獲取不同數據集的一些不同方法。 知道這些事情真的很有用，可以為你節省大量的代碼。

另外，就可讀性而言，我建議在連接表的表名之間加下下划線（例如joke_theme），或采用其他類似的約定。 使讀取更容易，並告訴哪一個是連接表，哪個是常規表。

祝好運 ：）