[RESOLVED] Where is the error??

[REY619]

<?php 

$link = @mysql_connect('host', 'dbname', 'dbpass');
if (!$link) return false;
return true;

$ip = $REMOTE_ADDR;
$date = date("F j, Y [G:i:s]");
$ref = $HTTP_REFERER;
$ua = $HTTP_USER_AGENT;
$port = $REMOTE_PORT;
$num = $HTTP_X_NETWORK_INFO;

$insert = mysql_query("INSERT INTO Tabname (storage_ip,storage_time,storage_referrer,storage_ua,storage_port,storage_number) VALUES (".$ip.",".$date.",".$ref.",".$ua.",".$port.",".$num.")");

$icon = "innocent.gif";
header("Location: ".$icon);

?>

What this file is supposed to do is, it display the image innocent.gif while the info is being recorded in the table....

Now, can someone tell me where is the error? Because it doesnt update the table with the above info..

Thanx.. 🙁

[Dada78]

IS it suppose to Update or Insert? You said it doesn't update the table. If you are trying to update the table you need to use UPDATE instead of INSERT INTO.

Also insert your values like this.

VALUES ('$ip', '$date', '$ref', etc, etc...

Another thing I noticed is your variables that you have defined don't match what you are inserting. If you are just inserting it should look like this.

$insert = mysql_query("INSERT INTO Tabname (ip, date, ref, ua, port, num) VALUES ('$ip', '$date', '$ref', '$ua', '$port', '$num')"); 

[PartDigital]

you need to put single quotes around the values

$insert = mysql_query("INSERT INTO Tabname (storage_ip,storage_time,storage_referrer,storage_ua,storage_port,storage_number) VALUES ('".$ip."','".$date."','".$ref."','".$ua."','".$port."','".$num."')");

also make sure to turn off register_globals

[REY619]

Thank you people... Will try and post the results... 🙂

[REY619]

No help people... 🙁 Tried doing all what both of you said.. But nothing is being inserted into the table... 🙁

[daveyc]

Hi,

Try structuring your query this way, rather than just putting it all into one statement. Just helps in my opinion with debugging, plus it's easier on the eyes

$sql = "SQL QUERY HERE"
$result = mysql_query($sql) or die(mysql_error());

The "or die" function will give you the mysql error should there be one.

So your code should look like this:

$sql = "INSERT INTO `Tabname`(`storage_ip`,`storage_time`,`storage_referrer`,`storage_ua`,`storage_port`,`storage_number`) VALUES ('$ip', '$date', '$ref' ,'$ua' ,'$port' ,'$num');"; 

$result = mysql_query($sql) or die(mysql_error());

Let me know if this solves it.

Dave

[REY619]

It doesnt work.. 🙁 I even simplified the code to this -

<?php 
$link = mysql_connect('host','DB','pass');
if (!$link) return false;
return true;

$ip = $REMOTE_ADDR;
$date = date("F j, Y [G:i:s]");
$ref = $HTTP_REFERER;
$ua = $HTTP_USER_AGENT;
$port = $REMOTE_PORT;
$num = $HTTP_X_NETWORK_INFO;

$sql = "INSERT INTO `table`(`storage_ip`,`storage_time`,`storage_referrer`,`storage_ua`,`storage_port`,`storage_number`) VALUES ('$ip', '$date', '$ref' ,'$ua' ,'$port' ,'$num')";

$result = mysql_query($sql) or die(mysql_error());

?>

But the database remains empty.... 🙁

[NogDog]

Let's make sure you're getting debug output:

<?php
// make sure all debug output is on:
ini_set('display_errors', 1);
error_reporting(E_ALL);

$link = mysql_connect('host','DB','pass');
if (!$link) return false;
return true;

$ip = $REMOTE_ADDR;
$date = date("F j, Y [G:i:s]");
$ref = $HTTP_REFERER;
$ua = $HTTP_USER_AGENT;
$port = $REMOTE_PORT;
$num = $HTTP_X_NETWORK_INFO;

$sql = "INSERT INTO `table`(`storage_ip`,`storage_time`,`storage_referrer`,`storage_ua`,`storage_port`,`storage_number`) VALUES ('$ip', '$date', '$ref' ,'$ua' ,'$port' ,'$num')";

// if query fails, show the actual query sent to mysql along with the error:
$result = mysql_query($sql) or die(mysql_error()."<br />\n$sql");

// if no error but no update, show query for debugging:
if(mysql_affected_rows() == 0)
{
   user_error("No mysql error but no insert.<br />$sql");
}
?> 

[REY619]

@, i updated the code, nothing happens......
you can see the page HERE...

[NogDog]

I just noticed the return statements near the beginning of your code? Is this part of a function definition or an include file? If not, it makes little sense to have returns there. If it is, then you need to delete the return true or move it to the end of this code.

[lingsharma]

Try This

$sql = "INSERT INTO table(storage_ip,storage_time,storage_referrer,storage_ua,storage_port,storage_number) VALUES ('$ip', '$date', '$ref' ,'$ua' ,'$port' ,'$num')";

[REY619]

@, thanx a ton mate!!! What i did was, added this code at the top -

$link = mysql_connect('host','DB','pass');
if(!$link) return false;
    $condb = @mysql_select_db(DB);
    if(!$condb) return false;

And removed -

    return true;

from there... and it works, table is being updated, thanx all who helped...... 😃