[英]What am I doing wrong with his php mysql
我的目的是在我的網站上放置一個表格,其中包含來自datbase的一些包含ID的字段
$con = mysqli_connect('hostname','username','password','dbname');
if (!$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>";
在我運行此命令的那一刻,我正在伸出一張這樣的桌子
+---------+-------+
| Product | Price |
+---------+-------+
我的數據庫看起來像這樣
+--------------+-------+-------------+
| MenuItemName | Price | Student UID |
+--------------+-------+-------------+
| Foo | 99 | 12345 |
| Foo2 | 11 | 11111 |
| Foo2 | 11 | 12345 |
+--------------+-------+-------------+
我在做什么錯呢?
嘗試更改此行:
$sql="SELECT * FROM Orders WHERE 'Student UID' = '".$id."'";
至:
$sql="SELECT * FROM Orders WHERE `Student UID` = '".$id."'";
喊解決您的問題。
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