我正在用 mysql 在 php 中创建一个表单。当用户输入新数据时，该数据会自动保存到数据库中，并在我需要时显示浏览器。此表单用于在浏览器中显示 mysql 数据。
一切运作良好。但我的问题是当用户提交数据时，该数据显示最后一个。但我需要先在浏览器中显示该详细信息。
<?PHP
$connection=Mysql_connect('xxxresource.com','xxx','xxxx');
if(!$connection)
{
echo 'connection is invalid';
}
else
{
Mysql_select_db('tions',$connection);
}
//check if the starting row variable was passed in the URL or not
if (!isset($_GET['startrow']) or !is_numeric($_GET['startrow'])) {
//we give the value of the starting row to 0 because nothing was found in URL
$startrow = 0;
//otherwise we take the value from the URL
} else {
$startrow = (int)$_GET['startrow'];
}
//this part goes after the checking of the $_GET var
$fetch = mysql_query("SELECT * FROM customer_details LIMIT $startrow, 10")or
die(mysql_error());
$num=Mysql_num_rows($fetch);
if($num>0)
{
echo "<table margin=auto width=999px border=1>";
echo "<tr><td><b>ID</b></td><td><b>Name</b></td><td><b>Telephone</b></td><td> <b>E-mail</b></td><td><b>Couttry Applying for</b></td><td><b>Visa-Category</b> </td><td><b>Other Category</b></td><td><b>Passport No</b></td><td> <b>Remarks</b></td></tr>";
for($i=0;$i<$num;$i++)
{
$row=mysql_fetch_row($fetch);
echo "<tr>";
echo"<td>$row[0]</td>";
echo"<td>$row[1]</td>";
echo"<td>$row[2]</td>";
echo"<td>$row[3]</td>";
echo"<td>$row[4]</td>";
echo"<td>$row[5]</td>";
echo"<td>$row[6]</td>";
echo"<td>$row[7]</td>";
echo"<td>$row[8]</td>";
echo"</tr>";
}//for
echo"</table>";
}
//now this is the link..
echo '<a href="'.$_SERVER['PHP_SELF'].'?startrow='.($startrow+10).'"><img align=left height=50px width=50px src="next.png"/></a>';
$prev = $startrow - 10;
//only print a "Previous" link if a "Next" was clicked
if ($prev >= 0)
echo '<a href="'.$_SERVER['PHP_SELF'].'?startrow='.$prev.'"><img align=right height=50px width=50px src="previous.png"/></a>';
?>
我的问题是当用户将数据提交到 mysql 数据库时，然后首先在浏览器中显示该详细信息。
谢谢。