Question

At first I'm beginner at php and MySQL. Now I'm working on my first bigger website with Administration panel. I want to do a simple visit counter with mysql and php. I made this code, where visitors IP save in $ip and Insert in DB. Looks easy, but it doesn't work :D I spend about 2 hours on it, but I can't find a mistake in my code. It doesn't show me any information in DB.

include 'connect_db.php';
$ip = $_SERVER['REMOTE_ADDR'];
$sql = mysql_query("SELECT * FROM table WHERE visits='$ip'");
if (!$ip == $sql) {
    $insert = mysql_query("INSERT INTO table (visits) VALUES '$ip'");
}

I have 2 items in my table. 1) ID with Auto Increment and 2) visits defined as VARCHAR 50. Server connection is ok, doesn't show any error.

Can you help me please?

Answer 1

试试这个。我希望它能提供帮助

include 'connect_db.php';
$ip = $_SERVER['REMOTE_ADDR']; 
if (isset($ip)){
  $sql = mysql_query("SELECT * FROM table WHERE visits = '$ip'");
  if(mysql_num_rows($sql)==0){
    $insert = mysql_query("INSERT INTO table (visits) VALUES ('$ip')");
 }
}

Answer 2

Thank you, I found the mistake .. one of them was statement (Thank you ImAtWar) and other mistake was after "VALUES" in INSERT must be in () ...

But it didn´t work properly. It insert IP in db again even when is inserted once. :l

Repaired code:

$sql = mysql_query("SELECT * FROM visitcont WHERE visits='$ip'");
if ($ip !== $sql) {
$insert = mysql_query("INSERT INTO visitcount (visits) VALUES ('$ip')");

Answer 3

你必须使用mysqli或PDO而不是mysql。但是现在试试这个：

include 'connect_db.php';
$ip = $_SERVER['REMOTE_ADDR'];
$sql = mysql_query("SELECT * FROM table WHERE visits= $ip");
if (!empty($ip) && ($ip != $sql)) {
    $insert = mysql_query("INSERT INTO table (visits) VALUES $ip");
}else{
   echo "already inserted";
}

Visit counter with mysql/php

3 个答案: