[英]PHP/Ajax Chat/MYSql - Create fake users in chat room
I am using Ajax Free Chat on my server. 我在服务器上使用Ajax Free Chat。 I would like to have a couple of fake users logged in to get things going. 我希望有几个假用户登录才能使事情进展顺利。 But I dont know how to implement these using PHP/mysql. 但是我不知道如何使用PHP / mysql来实现它们。
I have found there is a database table called 'online' which displays the logged in users on the user list within the chat. 我发现有一个名为“在线”的数据库表,该表在聊天中的用户列表上显示已登录的用户。 Maybe its this table I need to edit? 也许我需要编辑此表? Any help reatly appreciated. 任何帮助深表感谢。
function addToOnlineList() {
$sql = 'INSERT INTO '.$this->getDataBaseTable('online').'(
userID,
userName,
userRole,
channel,
dateTime,
ip
)
VALUES (
'.$this->db->makeSafe($this->getUserID()).',
'.$this->db->makeSafe($this->getUserName()).',
'.$this->db->makeSafe($this->getUserRole()).',
'.$this->db->makeSafe($this->getChannel()).',
NOW(),
'.$this->db->makeSafe($this->ipToStorageFormat($_SERVER['REMOTE_ADDR'])).'
);';
According to your comments that what you should do 根据您的意见,您应该做什么
First you need to get sure that you are connected to the database if not use this 首先,如果不使用此功能,则需要确保已连接到数据库
$connection = mysql_connect("localhost","username","password");
mysql_select_db("dbname",$connection);
next 下一个
function addToOnlineList() {
$sql = 'INSERT INTO '.$this->getDataBaseTable('online').'(
userID,
userName,
userRole,
channel,
dateTime,
ip
)
VALUES (
'.$this->db->makeSafe($this->getUserID()).',
'.$this->db->makeSafe($this->getUserName()).',
'.$this->db->makeSafe($this->getUserRole()).',
'.$this->db->makeSafe($this->getChannel()).',
NOW(),
'.$this->db->makeSafe($this->ipToStorageFormat($_SERVER['REMOTE_ADDR'])).'
);';
//here what you should add
mysql_query($sql) or die(mysql_error());
}
If you know what the values like $this->getUserID()
the above will work if not you should replace them with some values from you Good Luck if you need more help just comment 如果您知道像$this->getUserID()
这样的值,那么上面的方法可以正常工作,如果您需要更多帮助,则应该用好运中的一些值替换它们,只需注释
Edit this is how you can generate a simple data 编辑这就是您可以生成简单数据的方式
$names = array("namee1","name2","name3","will");
echo $names[array_rand($names)];
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