我有两张桌子,
agent(id, name, password, ..., shop_id);
shop(id, name, address, ...);现在我想获取代理商的店名,我该怎么做?我是否应该使用两个序列sql查询,
<?php
$qstr = <<<SQL
SELECT * FROM agent WHERE id=$id;
SQL;
$result = $conn->query($qstr);
$row = $result->fetch_assoc();
foreach ($row as $key => $value)
$$key = $value;
$qstr = <<<SQL
SELECT name FROM shop WHERE id=$shop_id;
SQL;
$result = $conn->query($qstr);
$row = $result->fetch_assoc();
$shop_name = $row['name'];
?>或者我应该为此select创建一个视图,
CREATE VIEW f_agent AS SELECT agent.id AS id ,agent.name AS name, shop.id AS shop_id, shop.name AS shop_name FROM agent, shop WHERE agent.shop_id=shop.id;或者我应该只使用sql,
SELECT agent.id AS id ,agent.name AS name, shop.id AS shop_id, shop.name AS shop_name FROM agent, shop WHERE agent.shop_id=shop.id and agent.id=$id;做这件事最好的方法是什么?
谢谢。
发布于 2011-01-17 11:26:29
如果这只是您将在前端的页面中使用的特定查询,那么简单的SQL应该工作得最好。没有必要为您希望从MySQL获得的每个可能的结果集创建视图。
如果我可以提出一个建议,请考虑SQL92 ANSI语法
SELECT agent.id AS id ,agent.name AS name, shop.id AS shop_id, shop.name AS shop_name
FROM agent
INNER JOIN shop ON agent.shop_id=shop.id
WHERE agent.id=$id;在PHP 环境下,您应该多次从使用循环查找结果集的列来访问MySQL。
发布于 2011-01-17 11:27:33
我不会对这样一个简单的查询使用视图,而且使用两个单独的结果似乎没有意义,所以我建议使用最后一个查询
发布于 2011-01-17 11:58:31
我刚开始运行这样的查询,但我发现这个帖子非常有用。像这样的东西也能工作吗?
$query = "SELECT shop.id, shop.name, agent.name FROM shop LEFT JOIN agent ON shop.id = agent.id";
$res = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($res)){
echo $row['shop']['name']. " - ". $row['agent']['name'];
echo "<br />";
}https://stackoverflow.com/questions/4709755
复制相似问题