PHP第一次工作然后失败

Question

Basically what this does is makes it so after the user presses a button, they have to wait 2 minutes to press it again. When you press it the first time, it works, then if you press it again before 2 minutes, it says "please wait 2 minutes." but after 2 minutes, you can press it as many times as you like without the error, not what I want. Here is the code

<?php
if('sub') {

$client = $_SERVER['REMOTE_ADDR'];
$username="user";
$password="pass";
$database="db";
mysql_connect(localhost,$username,$password);
$time = strtotime("2 minutes");
$now = strtotime("now");
@mysql_select_db($database) or die( "Unable to select database");
$query = "SELECT Time FROM log WHERE IP='$client'";
$result= mysql_query($query);
$num=mysql_num_rows($result);


while($rows=mysql_fetch_array($result)){
$timestamp = $rows['Time'];
}
echo n; 
echo $now;
echo t;
echo $timestamp;


if($now < $timestamp)
{
echo "<center><h2 style=\"color:red\" class=\"texts\" id=\"homeLink\">Please wait 2 minutes.</h2></center>";

}
else{
//some other code

$query1 = "INSERT INTO log VALUES ('$client','$time','$username')";
$query2 = "UPDATE `log` SET `Time`='$time' SET `Username`='$username' WHERE `IP`='$client'";

if($num == 0)
{
mysql_query($query1);
}
else
{
mysql_query($query2);
}

mysql_close();
}
?>

As you can see, if there is no row for the IP of the user, it makes a new one, if there is it will update it. After the first time, it no longer works. Hope someone can help. Thanks!

Answer 1

Your update statement is a bit off, so the update fails;

UPDATE `log` SET `Time`='$time' SET `Username`='$username' WHERE `IP`='$client'"

You should only use SET once, like this;

UPDATE `log` SET `Time`='$time', `Username`='$username' WHERE `IP`='$client'"

Answer 2

I'm pretty sure your database does not contain entries you think it contains. Can you check it (IPs and stuff on after button press)?