我是什么，他PHP的MySQL做错了

Question

什么我的本意是把一个表在我的网站方含由含有一个ID

$con = mysqli_connect('hostname','username','password','dbname'); 
if (!$con) { 
    die('Could not connect: ' . mysqli_error($con)); 
} 



$sql="SELECT * FROM Orders WHERE `Student UID` = '".$id."'"; 
$result = mysqli_query($con,$sql); 
echo 
echo "<table><tr><th>Product</th><th>Cost</th></tr>"; 
while($row = mysqli_fetch_array($result)) { 
    echo "<tr>"; 
    echo "<td>" . $row['MenuItemName'] . "</td>"; 
    echo "<td>" . $row['Price'] . "</td>"; 


    echo "</tr>"; 
} 
echo "</table>"; 

此刻datbase某些字段，当我运行此，我上午突出部分得到一个表像这样

+---------+-------+ 
| Product | Price | 
+---------+-------+ 

我的数据库看起来像这样

+--------------+-------+-------------+ 
| MenuItemName | Price | Student UID | 
+--------------+-------+-------------+ 
| Foo   | 99 |  12345 | 
| Foo2   | 11 |  11111 | 
| Foo2   | 11 |  12345 | 
+--------------+-------+-------------+ 

我在做什么毛病日是什么？

Answer 1

尝试修改此行：

$sql="SELECT * FROM Orders WHERE 'Student UID' = '".$id."'"; 

到：

$sql="SELECT * FROM Orders WHERE `Student UID` = '".$id."'"; 

怒吼解决您的问题。