使用PHP和MySQL搜索結果

Question

我最近發佈了另一個與我正在從事的基於Kevin Yank的「構建您自己的DB驅動的網站等」的搜索引擎相關的主題。搜索引擎正在工作，但是我想修復一些錯誤並且不知道如何去做。我將專注於這個線程中的其中一個，以免它太混亂。

在數據庫中，有一個笑話表（稱爲「笑話」）和另一個表格（稱爲「主題」）。這兩個表與另一個名爲「joketheme」的表相關。每個笑話應該能夠有超過1個主題，並且我希望結果列出所有主題而不重複條目。到目前爲止，我一直無法做到這一點。事實上，我不知道如何在MySQL中爲一個笑話指定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 

你知道我需要在代碼（或在MySQL）列出每個結果不止一個主題（如果該條目有不止一個主題？）來改變？

這是我在搜索結果頁面的代碼。預先感謝您的幫助！：

<?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

聽起來像是你需要做一些研究SQL查詢以及它們如何工作。如果你想主題的列表爲個人的笑話，你可以做一個簡單的查詢是這樣的：

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} 

我們在這裏所做的被稱爲連接。我們一起加入這三個表格，根據所有表格中常見的ID匹配結果。這些被稱爲密鑰。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。

我建議你查看一些SQL教程，以瞭解使用查詢獲取不同數據集的一些不同方法。瞭解這些東西非常有用，並可以爲您節省大量的代碼。就可讀性而言，我建議在連接表的表名之間加上下劃線（例如joke_theme），或採用其他類似的約定。使讀取和分辨哪一個是連接表更容易，哪個是常規表。

祝你好運:)