[SOLVED] mysql_fetch_array

[eon201]

Hi,

 

Im currently pulling in a load of data from my sql table. I found that using the mysql_fetch_array function was the way to pull it all in as an array.

 

The only problem? I cant manage to get it to store a value into another name eg $ip = $row['ip'] and count how many instance's are contained within it.

 

I need to do similar things to the 'url' and the 'time' but as I cant access the data there isnt much I can do!

 

Can anyone help me?? I havnt used the 'while' statement or the mysql_fetch_array function before so im feeling quite lost.

 

$row = mysql_fetch_array( $result );

while($row = mysql_fetch_array($result)){
echo $row['ip']. " - ". $row['url']. " - ". $row['date'];
echo "<br />";


}

 

Thanks as always. Eon201

[kratsg]

Try the list function, may be of more use...

 

$row = mysql_fetch_array( $result );
while(list($ip,$url,$date) = mysql_fetch_array($result)){
echo $ip." - ".$url." - ".$date;
echo "<br />";
}

 

The list function works like this, if each set of a row is an array, then list the array elements in order from 0th index to nth index with $ip, $url, $date (as the 0th, 1st, and 2nd).

[eon201]

Thanks Kratsg,

 

so far the code is now like this..

$row = mysql_fetch_array( $result );

while(list($ip,$url,$date) = mysql_fetch_array($result)){
echo $ip." - ".$url." - ".$date;
echo "<br />";
}

$visitstoday = count($ip); //This is what I need to do. It currently returns a big fat 0
echo "$visitstoday";
mysql_close();

 

 

Good tip on the list function - very handy

What I need to do though is count how many ip's occur within the array without having to display the whole thing!

 

Any ideas??

[rajivgonsalves]

try this

 

<?

 

$row = mysql_fetch_array( $result );

$arrData = array()

while(list($ip,$url,$date) = mysql_fetch_array($result)){

if (!isset($arrData[$ip])) { $arrData[$ip] = array(); }

$arrData[$ip][] = array("url"=>$url,"date"=>$date);

}

 

$visitstoday = count($arrData); //This is what I need to do. It currently returns a big fat 0

echo "$visitstoday";

mysql_close();

 

?>

[kenrbnsn]

Please show us the MySQL query you're using.

 

Ken

[eon201]

Ok here is the full code... (without sql login info!)

 

mysql_connect("localhost",$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
echo 'CONNECT TO MYSQL SUCCESSFUL<br/><br/> ';

// Retrieve all the data from the "log" table
$result = mysql_query("SELECT * FROM logtable")
or die(mysql_error());  
echo "I have got the data from the table - logtable <br/><br/>";

// store the record of the "log" table into $row
$row = mysql_fetch_array( $result );

while($row = mysql_fetch_array( $result )) {

echo $row['ip']. " -- ";
echo $row['url']. " -- ";
echo $row['date']. "<br/><br/>";

$ip[] = $row['ip'];
$url[] = $row['url'];
$date[] = $row['date'];
} 

 

Literally all I would like to do is be able to:

 

Count each value

compare each value to another ie $url / $ip

and echo each value

 

It should be so so simple. Yet this all behaves differently now im calling it back from the mysql.

 

Thanks for the help so far

[kenrbnsn]

Remove this line

<?php
$row = mysql_fetch_array( $result );
?>

It is causing your code to miss the first row of data.

 

Ken

[eon201]

Ok thanks for that.

 

But does anyone know how to solve my original problem???

[eon201]

Hmmm... I Still havnt figured this out. Is it becuase im using the wron function of mysql_fetch_array in the first place???

 

Thanks.

 

Eon201

[eon201]

Im an idiot! It works fine now! Must have just been a problem in my syntax somewhere!!

 

Thanks to all who have lent a hand!