Question

很抱歉，如果有一种简单的Mysql方法可以做到这一点，但我发现很难说出来获得任何有意义的搜索结果。

我正在搜索我的网站，搜索工作正常，直到我尝试通过基于下拉列表中所选类型的单词进行搜索。

此查询最能代表我正在尝试做的事情：

SELECT *
FROM Books
JOIN bookauthor ON books.BookID = bookauthor.BookID 
JOIN authors ON bookauthor.AuthorID = authors.AuthorID 
JOIN bookgenre ON books.BookID = bookgenre.BookID 
JOIN genre ON bookgenre.GenreID = genre.GenreID 
WHERE genre.Genre = 'Fantasy' 
WHERE books.BookName LIKE '%$variable%' OR authors.Forename LIKE '%$variable%' OR authors.Surname LIKE '%$variable%' 
GROUP BY books.BookName ORDER BY authors.AuthorID

我想选择所有具有所选类型的书籍，然后从他们那里做其他检查，因为现在让OR覆盖其他检查并最终吐出每个记录。

我猜我可能必须有一个查询选择该类型的所有书籍，然后运行另一个查询以从之前的查询中进行选择。只有我以前从未这样做过，我需要将其纳入我的搜索功能......

这是我当前的搜索代码，除了在尝试搜索选择了“全部”以外的类型的内容时，它的工作正常，如果您只选择一种类型并且不输入任何内容，它将显示该类型的书籍，只是如果你也输入了一个搜索词，那就不行了。

if(isset($_POST['search']))
            {
                if($_POST['genre']!="All")
                    {
                    $where3 = "WHERE genre.GenreID = '".$_POST['genre']."'";        
                    }
                if($_POST['field'] == "All")
                    {
                        if(str_word_count($_POST['find'])>=2)
                        {
                        $findEx = explode(' ', $_POST['find']);
                        $where2 = "OR authors.Forename LIKE  '%".($findEx[0])."%' AND authors.Surname LIKE  '%".$findEx[1]."%'";
                        }
                    $where = "
                    WHERE books.BookName LIKE  '%".($_POST['find'])."%'
                    OR  authors.Forename LIKE  '%".($_POST['find'])."%' OR authors.Surname LIKE  '%".($_POST['find'])."%'
                    $where2
                    ";
                    }
                else if($_POST['field'] == "Books")
                    {
                    $where = "WHERE books.BookName LIKE  '%".($_POST['find'])."%'";
                    }
                else if($_POST['field'] == "Authors")
                    {
                    if(str_word_count($_POST['find'])>=2)
                        {
                        $findEx = explode(' ', $_POST['find']);
                        $where = "WHERE authors.Forename LIKE  '%".($findEx[0])."%' AND authors.Surname LIKE  '%".$_POST[1]."%'";
                        }
                    else
                        {
                        $where = "WHERE authors.Forename LIKE  '%".($_POST['find'])."%' OR authors.Surname LIKE  '%".($_POST['find'])."%'";
                        }
                    }                       
                    $sql = "SELECT * FROM Books
                        JOIN bookauthor ON books.BookID        = bookauthor.BookID
                        JOIN authors    ON bookauthor.AuthorID = authors.AuthorID
                        JOIN bookgenre  ON books.BookID        = bookgenre.BookID
                        JOIN genre      ON bookgenre.GenreID   = genre.GenreID
                        ".$where3."
                        ".$where."
                        GROUP BY books.BookName                                
                        ORDER BY authors.AuthorID"; 
                        echo $sql;
            }

我对这一切仍然相当新，所以很抱歉，如果我的搜索代码看起来像一场噩梦，它会以最少量的代码完成我所需要的，但欢迎任何提示。

所有帮助表示赞赏-Tom

Answer 1

也许

SELECT *
FROM Books
JOIN bookauthor ON books.BookID = bookauthor.BookID 
JOIN authors ON bookauthor.AuthorID = authors.AuthorID 
JOIN bookgenre ON books.BookID = bookgenre.BookID 
JOIN genre ON bookgenre.GenreID = genre.GenreID 
WHERE genre.Genre = 'Fantasy' 
AND books.BookName LIKE '%$variable%' (OR authors.Forename LIKE '%$variable%' OR authors.Surname LIKE '%$variable%') 
GROUP BY books.BookName ORDER BY authors.AuthorID

MySQL WHERE结果基于以前的WHERE子句

1 个答案: