Possible Duplicate:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
mysql_fetch_array():提供的参数不是有效的MySQL结果资源
`
mysql_select_db("test", $con);
$result = mysql_query("SELECT * FROM order");
echo "<table border='1' bgcolor='#99CCCC' >"
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>". $row['orderid']."</td>";
echo "<td>". $row['odate'] . "</td>";
echo "<td>". $row['pdtdetail']."</td>";
echo "<td>". $row['unitprice']."</td> ";
echo "<td>".$row['quantity']."</td> ";
echo "<td>".$row['netprice']."</td> ";
echo "<td>".$row['status']."</td>";
echo "</tr>";
}
echo "</table>";
mysql_close($con);
?>`
最佳答案
SELECT * FROM order <-- this is a wrongful SQL
订单是reserved word in mysql。
如果您有一个名为
order
的表,您应该尽快将其重命名
或在查询过程中反选表名:
SELECT * FROM `order`
当您编写SQL代码时,这是您的责任
以确保查询正常工作...
关于php - mysql_fetch_array():提供的参数在第48行的C:\wamp\www\mahesh\login\orderhistory.php中不是有效的MySQL结果资源,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/8441806/