xyz91

I solved my problem. You can delete the topic. Thanks.

I do not know nothing working. Here code: <?php $dbhost="localhost"; $dbuser="root"; $dbpassword="xxxxxx"; $dbname="database1"; $conn = mysqli_connect($dbhost, $dbuser, $dbpassword, $dbname); if (!$conn) { echo "Error connection with MySQL" . PHP_EOL; echo "Errno: " . mysqli_connect_errno() . PHP_EOL; echo "Error: " . mysqli_connect_error() . PHP_EOL; exit; } $port = '80'; $result = mysqli_query($conn, "SELECT * FROM domens") or die ("Erro connection: $dbname"); print "<TABLE CELLPADDING=10 BORDER=1>"; print "<TR><TD>idt</TD><TD>Name</TD><TD>Working</TD><TD>Number of attempts</TD><TD>Time</TD></TR>\n"; while ($row = mysqli_fetch_array ($result)) { $idt = $row[0]; $naame = $row[1]; $quantity = $row[2]; $time = $row[3]; $host = $name; $numb_attemp = $quantity; $fp = @fsockopen($host, $port, $errno, $errstr, 30); if($fp){ print "<TR><TD>$idt</TD><TD>$name<TD>OK</TD><TD>0</TD><TD>$time</TD</TR>\n"; $sql = "UPDATE addresses SET quantity = 0"; } else { print "<TR><TD>$idt</TD><TD>$nazwa<TD>OK</TD><TD>$numb_attemp</TD><TD>$time</TD</TR>\n"; $sql = "UPDATE addresses SET quantity = quantity+1"; } if (mysqli_query($conn, $sql)) { echo "sql update"; } else { echo "sql no update"; } print "</TABLE>"; mysqli_close($conn); ?>

Hello, I have a problem with writing a function that will display the number of attempts to communicate with the host that have not been answered, if the host starts responding again, the table should display, for example 0. The website addresses are taken from the MySQL database. The next column should display the time from which the communication with the host was lost and if it responds, it should display, e.g. OK. Please, give me some hints how to write it. Thanks in advance.