Php 达到5时如何重置计数器

有人能帮我重置mysql计数器吗。这是密码

$host="localhost"; // Host name 
$username=""; // Mysql username 
$password=""; // Mysql password 
$db_name="test"; // Database name 
$tbl_name="members"; // Table name 

// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect to server "); 
mysql_select_db("$db_name")or die("cannot select DB");

$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql);

$rows=mysql_fetch_array($result);
$counter=$rows['counter'];

// if have no counter value set counter = 1
if(empty($counter)){
$counter=1;
$sql1="INSERT INTO $tbl_name(counter) VALUES('$counter')";
$result1=mysql_query($sql1);
}

echo "You 're visitors No. ";
echo $counter;

// count more value
$addcounter=$counter+1;



// reset counter if 5 has been reached
If (counter==5){

echo "counter=5 ";
// now im getting an error here//
counter=0;
}


$sql2="update $tbl_name set counter='$addcounter'";
$result2=mysql_query($sql2);

mysql_close();
?>

the error is:
Parse error: syntax error, unexpected '=' in C:\xampp\htdocs\test\counter.php on line  
 and based on // counter=0

你忘了美元：

计数器本身不是一个变量，另外，在更新查询中，如果达到5，则最有可能在0之后返回值

If ($counter==5){
  echo "counter=5 ";
  $counter=0;
  $addcounter=0;
}

---

这是正确的工作代码

<?php
$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password=""; // Mysql password 
$db_name="gametest"; // Database name 
$tbl_name="counter"; // Table name 

// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect to server "); 
mysql_select_db("$db_name")or die("cannot select DB");

$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql);

$rows=mysql_fetch_array($result);
$counter=$rows['counter'];

// if have no counter value set counter = 1
if(empty($counter)){
$counter=1;
$sql1="INSERT INTO $tbl_name(counter) VALUES('$counter')";
$result1=mysql_query($sql1);
}

echo "You 're visitors No. ";
echo $counter;

// count more value
$addcounter=$counter+1;


// reset counter if 5 has been reached
If ($counter==5){
$counter=0;
$addcounter=0;
}


$sql2="update $tbl_name set counter='$addcounter'";
$result2=mysql_query($sql2);

mysql_close();
?>

---

我尝试过，它会删除错误，但不会重置。有什么新的想法吗？我怎么看你修改过的帖子？我刚刚修改了回复，你需要重置$addcounter和$counterpress我迷路了，我怎么做。如果你有这段代码，它将非常有用，只要看看我刚刚修改的回复，你的sql查询是update$tbl_name set counter='$addcounter'；但您没有将$addcounter重置为0；在我添加的代码中，如果$counter==5，则重置$addcounter=0；

<?php
$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password=""; // Mysql password 
$db_name="gametest"; // Database name 
$tbl_name="counter"; // Table name 

// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect to server "); 
mysql_select_db("$db_name")or die("cannot select DB");

$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql);

$rows=mysql_fetch_array($result);
$counter=$rows['counter'];

// if have no counter value set counter = 1
if(empty($counter)){
$counter=1;
$sql1="INSERT INTO $tbl_name(counter) VALUES('$counter')";
$result1=mysql_query($sql1);
}

echo "You 're visitors No. ";
echo $counter;

// count more value
$addcounter=$counter+1;


// reset counter if 5 has been reached
If ($counter==5){
$counter=0;
$addcounter=0;
}


$sql2="update $tbl_name set counter='$addcounter'";
$result2=mysql_query($sql2);

mysql_close();
?>