我试图创建一个表单,以允许用户将表单中的数据更新为数据库中的现有金额。这是我迄今为止所看到的价值的两倍。我想我需要从数据库中提取值,然后从表单中添加数据。如何将HTML表单中的值添加到数据库中的值
<?php
$username = "username";
$password = "password";
$hostname = "localhost";
//connection to the database
$dbhandle = mysql_connect($hostname, $username, $password)
or die("Unable to connect to MySQL");
echo "<font face=tahoma color=#ff000><b>Connected to MySQL</b></font><br><br>";
//select a database to work with
$selected = mysql_select_db("pdogclan_points",$dbhandle)
or die("Did this change");
// Formulate Query
$_POST["filter"];
$memid = mysql_real_escape_string($_POST["Member_ID"]);
$query = sprintf("SELECT Member_ID, Bank, Reward_1, Reward_2, Reward_3 FROM Points_Rewards WHERE Member_ID = '$memid'") or die("Could Not Formulate the Query");
//execute the SQL query and return records
$result = mysql_query($query);
// Check result
// This shows the actual query sent to MySQL, and the error. Useful for debugging.
if (!$result) {
$message = 'Invalid query: ' . mysql_error() . "\n";
$message .= 'Whole query: ' . $query;
die($message);
}
//fetch tha data from the database
while ($row = mysql_fetch_array($result))
echo "<table width=750 cellspacing=2 cellpadding=2 border=2>
<tr>
<td bgcolor=#000000 width=150><font face=tahoma color=white>ID: {$row['Member_ID']}</font></td>".
"<td width=150><font face=tahoma>Bank: {$row['Bank']}</td>".
"<td width=150><font face=tahoma>Reward 1: {$row['Reward_1']}</td>".
"<td width=150><font face=tahoma>Reward 2: {$row['Reward_2']}</td> ".
"<td width=150><font face=tahoma>Reward 3: {$row['Reward_3']}</td>
</tr>
</table><br></font>";//display the results
// Formulate Update Query
$_POST["submit"];
$memid = mysql_real_escape_string($_POST["Member_ID"]);
$query = sprintf("SELECT Member_ID, Bank, Reward_1, Reward_2, Reward_3 FROM Points_Rewards WHERE Member_ID = '$memid'") or die("Could Not Formulate the Query");
while ($row = mysql_fetch_array($result))
{
$bankdb = $row['Bank'];
$reward1db = $row['Reward_1'];
$reward2db = $row['Reward_2'];
$reward3db = $row['Reward_3'];
}
echo $bank;
echo $reward1;
echo $reward2;
echo $reward3;
$memid = mysql_real_escape_string($_POST["Member_ID"]);
$bank = $_POST['bank'];
$reward1 = $_POST['reward1'];
$reward2 = $_POST['reward2'];
$reward3 = $_POST['reward3'];
$query = "UPDATE Points_Rewards Set Bank = ('$bank' + '$bankdb'), Reward_1 = ('$reward1' + '$reward1'), Reward_2 = ('$reward2' + '$reward2'), Reward_3 = ('$reward3' + '$reward3') WHERE Member_ID = '$memid'";
$result = mysql_query($query) or die(mysql_error());
if(mysql_query($query)){
echo "updated";}
else{
echo "fail";}
//close the connection
mysql_close($dbhandle);
?>
好的。你为什么不创建表单? – 2013-04-04 23:09:24
欢迎来到StackOverflow!请不要使用mysql_ *函数前进http://bit.ly/phpmsql。它们不再被维护,并且正式被弃用https://wiki.php.net/rfc/mysql_deprecation。了解有关Prepared语句http://j.mp/T9hLWi,并使用PDO http://php.net/pdo或MySQLi http://php.net/mysqli。本文:http://j.mp/QEx8IB可以帮助您决定使用哪一个。 – Revent 2013-04-04 23:12:12