Question

有我的桌子：

|category|
|--------|
|id_cat  |
|name_cat|

填写表类别：

cat1 | BOX A
cat2 | BOX B

第二张表：

|display   |
|----------|
|id_display|
|name_1    |
|name_2    |

表格显示填写：

dis1 | cat1 | cat2

然后我用这个QUERY加入两个表

$sql = "SELECT * FROM display d JOIN category c ON d.name_1=c.id_cat JOIN category c ON d.name_2=c.id_cat ORDER BY $id_display DESC";
$result = mysql_query($sql) or exit("Failed : ".$sql);
while($data = mysql_fetch_assoc($result)){
    $id_display = $data['id_display'];
    $name1 = $data['name_cat'];
    $name2 = $data['name_cat'];
    echo "
        <div>
            $name1<br/>
            $name2<br/>
        </div>
    ";
}

结果是：

BOX A
BOX A

如何显示这样的结果？

BOX A
BOX B

Answer 1

$sql = "select d.name_1 as display1, d.name_2 as display2, c.name_cat as cat, c2.name_cat as cat2 FROM display d 
INNER JOIN category c ON(d.name_1 = c.id_cat) 
INNER JOIN category c2 ON(d.name_2 = c2.id_cat)
ORDER BY $id_display DESC"
$result = mysql_query($sql) or exit("Failed : ".$sql);
while($data = mysql_fetch_assoc($result)){
    $id_display = $data['id_display'];
    $name1 = $data['cat'];
    $name2 = $data['cat2'];
    echo "
        <div>
            $name1<br/>
            $name2<br/>
        </div>
    ";
}

查看联接中列和表的命名别名。

如何在JOIN表上显示数据作为不同的结果？

1 个答案: