嗨,有人可以帮我把这两个代码放在一起。
我是PHP的新手,我花了4个小时才得到一个代码才能工作,但现在我需要将相同的两个代码组合起来,只需调用另一个表。但无论我做什么,它都不会使输出只输出第一个输出。我得到的错误的第二个代码是:
“错误:无法执行SELECT * FROM treasure。”
我认为这是因为我在第一个代码中关闭了连接,所以我尝试删除了关闭,但这没有帮助
这是我的代码:
<?php
if($instance=="1")
{
$sql = "SELECT * FROM treasure";
if($result = mysqli_query($link, $sql))
{
if(mysqli_num_rows($result) > 0)
{
echo "<table id='Cheader'>";
echo "<tr>";
echo "<th>Day</th>";
echo "<th>Green Fee</th>";
echo "<th>Cart</th>";
echo "<th>Caddy</th>";
echo "<th>Visitor Green Fee</th>";
echo "<th>Note</th>";
echo "</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['day'] . "</td>";
echo "<td>" . $row['cf'] . "</td>";
echo "<td>" . $row['cart'] . "</td>";
echo "<td>" . $row['caddy'] . "</td>";
echo "<td>" . $row['viscf'] . "</td>";
echo "<td>" . $row['note'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_free_result($result);
} else
{
echo "No records matching your query were found.";
}
} else
{
echo "ERROR: Could not able to execute $sql. " .
mysqli_error($link);
}
mysqli_close($link);
}
if($instance=="2")
{
$sql = "SELECT * FROM hill";
if($result = mysqli_query($link, $sql))
{
if(mysqli_num_rows($result) > 0)
{
echo "<table id='Cheader'>";
echo "<tr>";
echo "<th>Day</th>";
echo "<th>Green Fee</th>";
echo "<th>Cart</th>";
echo "<th>Caddy</th>";
echo "<th>Visitor Green Fee</th>";
echo "<th>Note</th>";
echo "</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['day'] . "</td>";
echo "<td>" . $row['cf'] . "</td>";
echo "<td>" . $row['cart'] . "</td>";
echo "<td>" . $row['caddy'] . "</td>";
echo "<td>" . $row['viscf'] . "</td>";
echo "<td>" . $row['note'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_free_result($result);
} else
{
echo "No records matching your query were found.";
}
} else
{
echo "ERROR: Could not able to execute $sql. " .
mysqli_error($link);
}
mysqli_close($link);
}
?>