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php - 如果表中没有数据,如何在php中显示消息

转载 作者:塔克拉玛干 更新时间:2023-11-03 06:07:55 26 4
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<分区>

我有一个报表,里面有几个提示,根据提示的选择,有时候报表里面可能没有记录。在这种情况下,我想在封面上显示一条消息 NO RECORDS FOUND。

谁能告诉我怎么做?

<?php      
error_reporting(0);
include("connection.php");
session_start();
if(!($_SESSION['email']))
{
echo "please login first to access this page";
header("refresh:3;url=index.php");
exit();
}
?>
<!DOCTYPE html>
<html>
<head>
<style>
body
{
margin: 0;
padding: 0;
}
table
{
border-collapse: collapse;
width: 100%;
font-family: sans-serif;
font-size: 15px;
}
th,td
{
padding: 5px;
text-align: left;
}
tr:nth-child(even)
{
background: #edf0f5;
}
th
{
background: #00A800;
color: white;
padding: 5px;
font-family: sans-serif;
}
a
{
text-decoration: none;
color: #00A800;
}
a:hover
{
text-decoration: underline;
}
</style>
<title>View all the data</title>
</head>
<body>
<table>
<tr>
<th>Roll NO</th>
<th>Student Name</th>
<th>Father Name</th>
<th>Class</th>
<th>Class Section</th>
<th>Phone number</th>
<th>Email Address</th>
<th>Address</th>
<th>Edit</th>
<th>View Fees</th>
<th>Attendance</th>
</tr>
<?php
$select="select *from student where class='1' AND section='B' ";
$run=mysqli_query($con,$select);
$i=0;
while($row=mysqli_fetch_array($run))
{
$id=$row['id'];
$s_name=$row['s_name'];
$f_name=$row['f_name'];
$class=$row['class'];
$section=$row['section'];
$phone=$row['phone'];
$email=$row['email'];
$address=$row['address'];
$i++;
?>
<tr>
<td><?php echo $i;?></td>
<td><?php echo $s_name;?></td>
<td><?php echo $f_name;?></td>
<td><?php echo $class;?></td>
<td><?php echo $section;?></td>
<td><?php echo $phone;?></td>
<td><?php echo $email;?></td>
<td><?php echo $address;?></td>
<td><a href="edit.php?edit=<?php echo $id;?>" target="_blank">Edit</a></td>
<td><a href="view_fees.php?view_fees=<?php echo $id;?>" target="_blank">Fees</a></td>
<td><a href="attendance.php" target="_blank">Attendance</a></td>
</tr>
<?php
}
?>
</table>
</body>
</html>

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