0

有人可以帮助我重置 mysql 计数器。这是代码

$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
4

2 回答 2

7

你忘了$

$counter=0;

计数器本身不是变量,另外在您的更新查询中,如果您达到 5,您很可能在 0 值之后

If ($counter==5){
  echo "counter=5 ";
  $counter=0;
  $addcounter=0;
}
于 2012-09-28T13:40:59.103 回答
1

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();
?>
于 2012-09-28T14:24:54.340 回答