Question

我一直在使用stackoverflow.com一段时间来回答问题，但这是我第一次发帖。提前谢谢！

我正在尝试获取与此查询相关联的实际数字：

$purchasenumber = mysql_query("SELECT `purchase_order_number` FROM `purchaseorder` ORDER BY `purchase_order_number` DESC LIMIT 1");
echo $purchasenumber;

我收到“资源ID＃”而不是实际数字。

然后，我想使用检索到的数字从数据库中提取更多数据。我实际上是在创建一个可打印的页面，其中包含数据库中两个表的信息。

$result = mysql_query("SELECT * FROM purchaseorder, customers WHERE purchase_order_number = $purchasenumber AND purchaseorder.customer_ID = customers.customer_ID");
while ($row = mysql_fetch_assoc($result)) {}

有人有任何想法将资源ID转换为实际数据吗？我可能会以错误的方式解决这个问题，如果我是，请告诉我！

感谢。

Answer 1

您需要使用mysql_fetch_array（）从mysql_query获取数据：

    $result = mysql_query("SELECT `purchase_order_number` FROM `purchaseorder` ORDER BY `purchase_order_number` DESC LIMIT 1");
    $data = mysql_fetch_array($result);
    $purchasenumber = $data['purchase_order_number'];

More Info可以在PHP.net上找到

Answer 2

谢谢，kimbarcelona。你的回复有效。

$result = mysql_query("SELECT `purchase_order_number` FROM `purchaseorder` ORDER BY `purchase_order_number` DESC LIMIT 1");
$data = mysql_fetch_array($result);
$purchasenumber = $data['purchase_order_number'];

然后我只是为下一个结果使用了另一个变量

$result2 = mysql_query("SELECT * FROM purchaseorder, customers WHERE purchase_order_number = $purchasenumber AND purchaseorder.customer_ID = customers.customer_ID");
while ($row = mysql_fetch_assoc($result2)) {}

再次感谢您的回复！

需要来自DB的数据而不是资源ID

2 个答案: