如何内连接3表

时间:2015-05-29 07:23:03

标签: mysql database inner-join

如何加入三个表,我有3个表是buku,sedia和perpus,buku有属性idbuku,namapengarang,namabuku,penerbit,genre和tahunterbit.perpus有属性idperpus,namaperpus和alamat.and sedia有属性stok,idperpus和idbuku 我想问的是我试图用语法

内连接3表
select * from buku inner join sedia where buku.idbuku =sedia.idbuku;

它有效,但是当我尝试

select * from buku inner join sedia where buku.idbuku =sedia.idbuku inner join perpus where sedia.idperpus = perpus.idperpus;

代码是错误的,它说警告:mysql_fetch_array()期望参数1是资源,第64行给出布尔值。

任何人都可以告诉我该怎么办?

它的我的PHP代码仍然无法正常工作

<?php
                include "koneksi.php";

                        $cari=$_POST["cari"];
                        $rbut=$_POST["rbut"];

                    if (isset($cari)||isset($rbut)){
                        $sql = "select * from buku where $rbut like '%$cari%'inner join sedia on buku.IDbuku = sedia.IDbuku inner join perpus on sedia.IDperpus = perpus.IDperpus;";
                        $result = mysql_query($sql);
                    }
                    elseif(!isset($cari)||!isset($rbut)){
                        $cari=$_POST["cari"];
                        $rbut=$_POST["rbut"];
                        $sql = "select * from buku";
                        $result = mysql_query($sql);
                    }
                    echo "
                        <table border='1'><tr><th>ID</th><th>Nama Pengarang</th><th>Nama Buku</th><th>Penerbit</th><th>Genre</th><th>Tahun Terbit</th></tr>";

                    while($row=mysql_fetch_array($result)){
                        echo "
                        <tr>
                            <td>$row[IDbuku]</td>
                            <td>$row[NamaPengarang]</td>
                            <td>$row[NamaBuku]</td>
                            <td>$row[Penerbit]</td>
                            <td>$row[Genre]</td>
                            <td>$row[TahunTerbit]</td>
                        </tr>";
                    }
                    echo "
                        </table>";
                ?>

1 个答案:

答案 0 :(得分:0)

您的where条件必须是on

select * 
from buku 
inner join sedia on buku.idbuku =sedia.idbuku 
inner join perpus on sedia.idperpus = perpus.idperpus;

有关联接语法的详细信息,请参阅documentation

**更新** 您的新查询错误:

$sql = "select * from buku inner join sedia on buku.IDbuku = sedia.IDbuku inner join perpus on sedia.IDperpus = perpus.IDperpus where $rbut like '%$cari%'";

如果您阅读文档,您将看到where语句必须在joun语句之后。