SQL搜索引擎

Question

我知道的HTML和其他很少。

我想爲我的網站放在一起一個簡單的搜索引擎。我有3個表，我想加入，然後有數據可供搜索。

MESSAGES (ID_TOPIC, SUBJECT) 
TAGS (ID_TAG, TAG) 
TAGS_LOG (ID_TAG, ID_TOPIC) 

我試圖創建具有三個表的陣列（消息，標籤，和TAGS_LOG），所以我可以創建與MESSAGES.SUBJECT搜索標籤和輸出該數據作爲超鏈接，其中超鏈接的搜索引擎是www.website.com/ID_TOPIC

我看了所有，我已經接近，但沒有錢。

到目前爲止，我有;

CREATE TABLE new_table 
    AS (SELECT tags.id_tag, messages.subject, tags.tag, tags_log.id_topic 
     FROM messages, tags, tags_log); 

它用於將信息組合成單個表進行搜索。接下來我創建了search.php;

<h2>Search</h2> 
<form name="search" method="post" action="<?=$PHP_SELF?>"> 
Seach for: <input type="text" name="find" /> in 
<Select NAME="field"> 
<Option VALUE="tag">tag</option> 


</Select> 
<input type="hidden" name="searching" value="yes" /> 
<input type="submit" name="search" value="Search" /> 
</form> 

<? 
$field = $_POST['field'] ; 
$find = $_POST['find'] ; 
$searching = $_POST['searching'] ; 

//This is only displayed if they have submitted the form 
if ($searching =="yes") 
{ 
echo "<h2>Results</h2><p>"; 

//If they did not enter a search term we give them an error 
if ($find == "") 
{ 
echo "<p>You forgot to enter a search term"; 
exit; 
} 

// Otherwise we connect to our Database 
mysql_connect("host", "user", "pass") or die(mysql_error()); 
mysql_select_db("database") or die(mysql_error()); 

// We preform a bit of filtering 
$find = strtoupper($find); 
$find = strip_tags($find); 
$find = trim ($find); 

//Now we search for our search term, in the field the user specified 
$data = mysql_query("SELECT * FROM new_table WHERE upper($field) LIKE'%$find%'"); 

//And we display the results 
while($result = mysql_fetch_array($data)) 
{ 
echo " "; 
echo "Topic ID: {$result['id_topic']}"; 
echo "<br>"; 
Print "<a href=/index.php?topic={$result['id_topic']} target=new>{$result['subject']} </a>"; 
echo "<br>"; 
echo "<br>"; 
} 

//This counts the number or results - and if there wasn't any it gives them a little message explaining that 
$anymatches=mysql_num_rows($data); 
if ($anymatches == 0) 
{ 
echo "Sorry, but we can not find an entry to match your query<br><br>"; 
} 

//And we remind them what they searched for 
echo "<b>Searched For:</b> " .$find; 
} 
?> 

但是我在主題方面遇到問題，我從Subject和RE：Subject獲得結果。我不知道如何清除「RE：Subject」，所以沒有重複的條目。

Answer 1

嘗試：SELECT TAGS.TAG, MESSAGES.SUBJECT, MESSAGES.ID_TOPIC FROM MESSAGES INNER JOIN TAGS_LOG ON MESSAGES.ID_TOPIC = TAGS_LOG.ID_TOPIC INNER JOIN TAGS ON TAGS_LOG.ID_TAG = TAGS.ID_TAG

但由於在評論中提到的，這是一個SQL 101的問題。然而，這並不像建議的那麼微不足道，因爲在那裏有一個額外的內部連接，這必然會使任何新手都不確定自己。