需要一些启发,这是我在这里的第一篇文章.

我想使用PHP调用并使用我之前创建的mysql视图…是否可能？
或换句话说,
我想知道….我们可以或者我们如何调用MysqL视图,我们已经使用PHP创建了？减少长查询编码

标准通用代码如下：

$sql= " SELECT shipping.tarriF1,cITy.city_name
         From shipping JOIN city
   ON shipping.id_city = city.id_city";
$db->QueryArray($sql);   

while ($row = $db->Row()) {
echo $row->city_name. " - " . $row->tarrif1 . "<br />\n"; 
}

现在为MysqL视图：

$sql= " CREATE VIEW shipsumarry AS SELECT shipping.tarrif1,city.city_name
         From shipping JOIN city
   ON shipping.id_city = city.id_city";

纯MySQL命令：

query: SELECT * FROM shipsummary

在PHP中：

$sql = i'm badly stuck here...please help

我们如何使用PHP访问它.

谢谢你

增加1：

好的…让我重写一下这个例子：

$sql1= " CREATE VIEW shipsumarry AS SELECT shipping.tarrif1,city.city_name
         FROM shipping JOIN city
   ON shipping.id_city = city.id_city";
$sql2= "SELECT * FROM shipsummary";
$db->QueryArray($sql2);

$sql2看不到shipummary VIEW,因为它已经在另一个VAR中了

如何利用然后执行$sql1？ &安培; $sql2？

MysqL_query("SELECT * FROM shipsummary");

// Or for whatever framework you're using:
$db->QueryArray("SELECT * FROM shipsummary");

$sql1= " CREATE VIEW shipsumarry AS SELECT shipping.tarrif1,city.city_name
         FROM shipping JOIN city
   ON shipping.id_city = city.id_city";
$sql2= "SELECT * FROM shipsummary";

// Here,you need to execute $sql1 before $sql2 is useful.
$db->QueryArray($sql1);
// Now execute $sql2
$db->QueryArray($sql2);