"if ($row['client_ip'] = $remote) {"-issue

[aog]

Hello all.

 

I am working and googling my keyboard out the window here with a script. I'm not really skilled in PHP or mySQL, but I try.

My problem here is in the script you see below.

If I enter the page from computer with IP:127.0.0.1 (witch is the lockalhost), the script works as I want.

But if I enter the page from computer with ip 192.168.x.x (a computer on the LAN), the script does not work.

I do get the echo ("You are $remote");, but the rest of the script wont work regardless of the 192.168.x.x can be found in $result = mysql_query("SELECT * FROM $table2 WHERE client_ip='$remote'"); or not.

 

What am I doing wrong?

 

EDIT: Fixed the = to ==

 

<?php 

  include ("login.php");
  include ("db_info.php");


$remote = $_SERVER['REMOTE_ADDR'];

  echo ("You are $remote");


mysql_select_db($db, $con);

$result = mysql_query("SELECT * FROM $table2
WHERE client_ip='$remote'");
while($row = mysql_fetch_array($result))
  	{

  if ($row['client_ip'] == $remote) {

	$ip = $row['client_ip'];
	$name = $row['client_name'];

  echo $ip . " " . $name;
  echo "<br/>";

	}

  else {
  echo "NOT VALID!";
	}
  }

mysql_close($con);

?>

[trq]

== is the comparison operator not =

[aog]

Yes, you are right.

How ever, it didn't change my problem.

[quelle]

Your line

 echo ("You are $remote");

would go like this

 echo "You are " . $remote;

[aog]

Works both ways, but I guess you know what you're talking about.

Fixed it, but it won't fix my main problem.

[aog]

Ok. I found one problem.

 

Some how the last number of the ip 192.168.x.x was missing in the DB. (Sorry..)

Once I fixed that, I discovered that else {  echo "NOT VALID!"; } does not work either on local or remote computer.

So, if IP is in table2, it works as intended, but not if the IP is missing in the table.

 

What am I missing here? 

[BlueSkyIS]

$sql = "SELECT * FROM $table2 WHERE client_ip='$remote'";
echo "sql: $sql <br />";
$result = mysql_query($sql) or die(mysql_error());

 

any change? is sql what you expect?

[BlueSkyIS]

if there should only be 1 result, don't use a while loop

 

$row = mysql_fetch_array($result);

 

and since the SQL asks for only those records where client_ip='$remote', there is no need for an if() to check whether it is in a record.

[BlueSkyIS]

so...

 

$sql = "SELECT * FROM $table2 WHERE client_ip='$remote'";
$result = mysql_query($sql) or die(mysql_error());
if (mysql_numrows($result) > 0) {
     $row = mysql_fetch_array($result);
     $ip = $row['client_ip'];
     $name = $row['client_name'];
     echo $ip . " " . $name;
     echo "<br/>";
} else {
  echo "NOT VALID!";
}

[aog]

Sweet, that works. Thanks man.