PHP搜索引擎

Question

我的搜索功能

 <?php 
    $i=0; 
    $column_name = 'title'; // column to search by 
    $k =$_GET['k']; 
    $terms = explode(" ",$k); 
    //connect before calling mysql_real_escape_string 
    mysql_connect("localhost","",""); 
    mysql_select_db("test"); 
    $query ="SELECT bookid,title,author 
    FROM books WHERE"; 
    foreach ($terms as $each){ 
     $i++; 
     $each = '%' . $each . '%'; // add wildcard 
     $each = mysql_real_escape_string($each); // prevent sql injection 
     if($i==1) 
      $query .= " $column_name LIKE '$each' "; 
     else 
      $query .= " OR $column_name LIKE '$each' "; 

    } 

    echo 'QUERY: ' . $query; 

    $query = mysql_query($query) OR DIE(mysql_error()); 

//Code below is for using the relationships table assuming you have a column name id that 
//references to the relationships table. Also, you should add a index on the column id. 

$results = ""; 

while($row = mysql_fetch_array($query)) { 
    $results .= '<li> 
       <a href="book-relationships.php?id='.$row['id'].'">'.$row['title'].' author: '.$row['author'].'</a> 
       </li>'; 
} 

$results = '<ul>' . $results . '</ul>'; 

echo $results; 

我需要幫助連接一個數據庫與其他搜索時，此數據庫列出所有都拿出了在搜索圖書，然後將它們與其他表進行比較，稱爲關係並打印出與搜索到的書籍有關係的書籍的標題。

當搜索運行這是在瀏覽器中

QUERY: SELECT id,title,author FROM books WHERE title LIKE '%jarrads%' Unknown column 'id' in 'field list' 

任何helpy，將不勝感激產生我100％粘

模式

書籍 - BOOKID，標題，作者，年出版，出版者..... 關係 - relationshipid，bookOne，bookTwo，relationship，relationshiplikes，relationsdislikes

bookOne和bookTwo，涉及到BOOKID

Answer 1

您有一個錯誤在您的查詢，作爲錯誤消息指出：

Unknown column 'id' in 'field list'

檢查表books爲有效字段列表。實現這一目標的一個方法是運行下面的查詢：

SHOW CREATE TABLE books;